💡Okay, so I have this problem where I need to find the sum of the first 10 elements that appear in both an arithmetic progression and a geometric progression. The arithmetic progression is given as {5, 8, 11, 14, ...} and the geometric progression is {20, 40, 80, 160, ...}. Hmm, let me try to figure this out step by step.First, let me recall what an arithmetic progression (AP) is. It's a sequence where each term after the first is obtained by adding a constant difference. In this case, the first term is 5, and the common difference is 3 because 8 - 5 = 3, 11 - 8 = 3, and so on. So, the general form of the nth term of this AP can be written as:a_n = 5 + 3n, where n starts from 0.Wait, actually, if n starts from 0, then when n=0, a_0 = 5, which is correct. So, yeah, that seems right.Now, the geometric progression (GP) is given as {20, 40, 80, 160, ...}. A GP is a sequence where each term after the first is obtained by multiplying the previous term by a constant ratio. Here, the first term is 20, and the common ratio is 2 because 40 / 20 = 2, 80 / 40 = 2, etc. So, the general form of the kth term of this GP is:b_k = 20 * 2^k, where k starts from 0.Alright, so now I need to find the terms that are common to both sequences. That is, I need to find numbers that can be expressed as 5 + 3n and also as 20 * 2^k for some integers n and k.So, setting them equal:5 + 3n = 20 * 2^kI need to solve for integers n and k such that this equation holds. Let me try to rearrange this equation to solve for n:3n = 20 * 2^k - 5n = (20 * 2^k - 5) / 3Hmm, since n has to be an integer, the right-hand side must be an integer. So, (20 * 2^k - 5) must be divisible by 3. Let me write that as:(20 * 2^k - 5) ≡ 0 mod 3Let me compute 20 mod 3 and 5 mod 3 to simplify this:20 mod 3 = 2 (since 3*6=18, 20-18=2)5 mod 3 = 2 (since 3*1=3, 5-3=2)So, substituting back:(2 * 2^k - 2) ≡ 0 mod 3Factor out the 2:2*(2^k - 1) ≡ 0 mod 3Since 2 and 3 are coprime, this implies that:2^k - 1 ≡ 0 mod 3So, 2^k ≡ 1 mod 3Now, I need to find k such that 2^k ≡ 1 mod 3. Let me compute powers of 2 modulo 3 to see the pattern:2^1 mod 3 = 22^2 mod 3 = 4 mod 3 = 12^3 mod 3 = 8 mod 3 = 22^4 mod 3 = 16 mod 3 = 12^5 mod 3 = 32 mod 3 = 22^6 mod 3 = 64 mod 3 = 1Ah, so the pattern repeats every 2 terms: 2, 1, 2, 1, etc. So, 2^k ≡ 1 mod 3 when k is even. That is, k must be even for 2^k ≡ 1 mod 3.Therefore, k must be even. Let me denote k = 2m, where m is an integer starting from 0.So, substituting back into the equation for n:n = (20 * 2^(2m) - 5) / 3Let me compute this for m = 0, 1, 2, ... to find the common terms.For m = 0:k = 0n = (20 * 1 - 5) / 3 = (20 - 5)/3 = 15/3 = 5So, the term is 5 + 3*5 = 20, which is indeed in both sequences.For m = 1:k = 2n = (20 * 4 - 5)/3 = (80 - 5)/3 = 75/3 = 25So, the term is 5 + 3*25 = 80, which is also in both sequences.For m = 2:k = 4n = (20 * 16 - 5)/3 = (320 - 5)/3 = 315/3 = 105So, the term is 5 + 3*105 = 320, which is in both sequences.Hmm, I see a pattern here. The common terms are 20, 80, 320, ... Each term is 4 times the previous term. So, this is a geometric progression with the first term 20 and common ratio 4.Wait, let me verify that. 80 / 20 = 4, 320 / 80 = 4, so yes, it's a GP with ratio 4.So, the common terms form a GP: 20, 80, 320, 1280, 5120, 20480, 81920, 327680, 1310720, 5242880,...Wait, but I need the first 10 terms of this GP. So, starting from 20, each term multiplied by 4.But let me make sure that these terms are indeed present in both the original AP and GP.For example, 20 is in both, 80 is in both, 320 is in both, 1280: let's check if 1280 is in the AP.In the AP, a_n = 5 + 3n. So, 5 + 3n = 1280 => 3n = 1275 => n = 425. So, yes, 1280 is in the AP.Similarly, in the GP, 20*2^k = 1280 => 2^k = 64 => k=6. So, yes, 1280 is in the GP.Similarly, 5120: 5 + 3n = 5120 => 3n = 5115 => n=1705, which is an integer. In the GP, 20*2^k=5120 => 2^k=256 => k=8. So, yes, 5120 is in both.Okay, so the common terms are indeed a GP with first term 20 and ratio 4. So, the first 10 terms are:1. 202. 803. 3204. 12805. 51206. 204807. 819208. 3276809. 131072010. 5242880Now, I need to find the sum of these 10 terms.The formula for the sum of the first n terms of a GP is:S_n = a * (r^n - 1) / (r - 1)Where a is the first term, r is the common ratio, and n is the number of terms.So, here, a = 20, r = 4, n = 10.Plugging into the formula:S_10 = 20 * (4^10 - 1) / (4 - 1)First, compute 4^10. Let me calculate that step by step:4^1 = 44^2 = 164^3 = 644^4 = 2564^5 = 10244^6 = 40964^7 = 163844^8 = 655364^9 = 2621444^10 = 1048576So, 4^10 = 1,048,576Therefore, S_10 = 20 * (1,048,576 - 1) / (4 - 1) = 20 * (1,048,575) / 3Compute 1,048,575 / 3:1,048,575 ÷ 3 = 349,525Then, multiply by 20:349,525 * 20 = 6,990,500So, the sum of the first 10 common terms is 6,990,500.Wait, let me double-check my calculations to make sure I didn't make any errors.First, 4^10 is indeed 1,048,576. Subtracting 1 gives 1,048,575. Dividing by 3: 1,048,575 ÷ 3.Let me do this division step by step:3 into 10 is 3, remainder 1.3 into 14 (1 from remainder + 4) is 4, remainder 2.3 into 28 (2 from remainder + 8) is 9, remainder 1.3 into 15 (1 from remainder + 5) is 5, remainder 0.3 into 7 is 2, remainder 1.3 into 15 (1 from remainder + 5) is 5, remainder 0.3 into 7 is 2, remainder 1.3 into 15 (1 from remainder + 5) is 5, remainder 0.3 into 7 is 2, remainder 1.3 into 15 (1 from remainder + 5) is 5, remainder 0.Wait, that seems messy. Alternatively, 1,048,575 ÷ 3:3 * 349,525 = 1,048,575, yes, because 3 * 300,000 = 900,0003 * 49,525 = 148,575So, 900,000 + 148,575 = 1,048,575. Correct.Then, 349,525 * 20 = 6,990,500. That seems right.So, I think my calculations are correct. Therefore, the sum of the first 10 common terms is 6,990,500.**Final Answer**The sum of the first 10 common elements is boxed{6990500}.

💡Okay, so I have this problem about triangle ABC, and I need to choose one of the three given conditions to find angle C and then, if c=2, find the maximum area of the triangle. Let me try to figure this out step by step.First, I need to understand each of the three conditions given. Maybe I'll go through each one and see which one seems the easiest to work with.Condition ①: √3 times the dot product of vectors CA and CB equals twice the area of the triangle. Hmm, vectors... I remember that the dot product of two vectors is equal to the product of their magnitudes times the cosine of the angle between them. So, if I think of vectors CA and CB, their magnitudes would be the lengths of sides CA and CB, which are b and a respectively, right? Because in triangle ABC, side opposite angle A is a, opposite B is b, and opposite C is c. So, vector CA would have magnitude b, and vector CB would have magnitude a. The angle between them would be angle C. So, the dot product CA · CB would be ab cos C. So, condition ① becomes √3 * ab cos C = 2 * area of triangle ABC. I know that the area of a triangle can be expressed as (1/2)ab sin C. So, substituting that in, we get √3 ab cos C = 2*(1/2)ab sin C, which simplifies to √3 ab cos C = ab sin C. Dividing both sides by ab (assuming a and b are not zero, which they can't be in a triangle), we get √3 cos C = sin C. Dividing both sides by cos C, we have √3 = tan C. So, tan C = √3. I remember that tan 60° is √3, so angle C must be 60 degrees or π/3 radians. That seems straightforward. So, if I choose condition ①, I can find angle C directly.Condition ②: (sin C + sin A)(sin C - sin A) = sin B (sin B - sin A). Let me expand the left side. It's a difference of squares, so it becomes sin² C - sin² A. The right side is sin² B - sin A sin B. So, we have sin² C - sin² A = sin² B - sin A sin B.Hmm, I wonder if I can use the Law of Sines here. The Law of Sines states that a/sin A = b/sin B = c/sin C = 2R, where R is the radius of the circumscribed circle. So, sin A = a/(2R), sin B = b/(2R), sin C = c/(2R). Maybe substituting these into the equation would help.Let me try that. So, sin² C - sin² A = (c² - a²)/(4R²) and sin² B - sin A sin B = (b² - ab)/(4R²). So, setting them equal: (c² - a²)/(4R²) = (b² - ab)/(4R²). Multiplying both sides by 4R², we get c² - a² = b² - ab.Hmm, that's interesting. So, c² - a² = b² - ab. Let me rearrange this: c² = a² + b² - ab. I remember the Law of Cosines: c² = a² + b² - 2ab cos C. Comparing this with what I have, c² = a² + b² - ab, so 2ab cos C = ab, which simplifies to cos C = 1/2. Therefore, angle C is 60 degrees or π/3 radians. So, condition ② also leads us to angle C being π/3.Condition ③: (2a - b) cos C = c cos B. Hmm, this seems a bit more involved. Let me see if I can use the Law of Cosines or maybe the Law of Sines here.First, I know from the Law of Cosines that cos C = (a² + b² - c²)/(2ab) and cos B = (a² + c² - b²)/(2ac). Let me substitute these into the equation.So, (2a - b) * [(a² + b² - c²)/(2ab)] = c * [(a² + c² - b²)/(2ac)]. Let me simplify both sides.Left side: (2a - b)(a² + b² - c²)/(2ab)Right side: c(a² + c² - b²)/(2ac) = (a² + c² - b²)/(2a)So, setting left side equal to right side:(2a - b)(a² + b² - c²)/(2ab) = (a² + c² - b²)/(2a)Multiply both sides by 2ab to eliminate denominators:(2a - b)(a² + b² - c²) = b(a² + c² - b²)Let me expand the left side:2a(a² + b² - c²) - b(a² + b² - c²) = 2a³ + 2ab² - 2a c² - a² b - b³ + b c²Right side: b a² + b c² - b³So, putting it all together:2a³ + 2ab² - 2a c² - a² b - b³ + b c² = a² b + b c² - b³Let me subtract the right side from both sides to bring everything to the left:2a³ + 2ab² - 2a c² - a² b - b³ + b c² - a² b - b c² + b³ = 0Simplify term by term:2a³ + 2ab² - 2a c² - a² b - b³ + b c² - a² b - b c² + b³Combine like terms:2a³ + 2ab² - 2a c² - 2a² b + 0 + 0 = 0So, 2a³ + 2ab² - 2a c² - 2a² b = 0Factor out 2a:2a(a² + b² - c² - a b) = 0Since a ≠ 0, we have:a² + b² - c² - a b = 0Which is:a² + b² - a b = c²Wait, this is similar to what we had in condition ②. From the Law of Cosines, c² = a² + b² - 2ab cos C. So, comparing:a² + b² - ab = a² + b² - 2ab cos CSubtract a² + b² from both sides:-ab = -2ab cos CDivide both sides by -ab (assuming a ≠ 0, b ≠ 0):1 = 2 cos CSo, cos C = 1/2, which means angle C is 60 degrees or π/3 radians. So, condition ③ also leads us to angle C being π/3.Okay, so all three conditions lead us to angle C being π/3. That's good to know. So, regardless of which condition I choose, I end up with angle C = π/3.Now, moving on to part 2: If c = 2, find the maximum value of the area of triangle ABC.Since angle C is π/3 and side c is opposite angle C, we can use the Law of Cosines to relate the sides. The Law of Cosines states that c² = a² + b² - 2ab cos C. Plugging in c = 2 and C = π/3, we get:2² = a² + b² - 2ab cos(π/3)4 = a² + b² - 2ab*(1/2)4 = a² + b² - abSo, we have the equation a² + b² - ab = 4.We need to find the maximum area of the triangle. The area of a triangle can be given by (1/2)ab sin C. Since angle C is π/3, sin(π/3) = √3/2. So, the area S is:S = (1/2)ab*(√3/2) = (√3/4)abSo, to maximize the area, we need to maximize ab, given that a² + b² - ab = 4.So, the problem reduces to maximizing the product ab under the constraint a² + b² - ab = 4.This seems like an optimization problem with constraint. Maybe I can use the method of Lagrange multipliers, but since it's a simple constraint, perhaps I can express one variable in terms of the other and then find the maximum.Let me try to express a² + b² - ab = 4 as a quadratic in terms of a or b.Alternatively, maybe I can use the AM-GM inequality or some other inequality to find the maximum of ab.Wait, let's consider that a² + b² - ab = 4. Let me think of this as a quadratic in a:a² - b a + (b² - 4) = 0For real solutions, the discriminant must be non-negative. So, discriminant D = b² - 4*(1)*(b² - 4) ≥ 0Compute D:D = b² - 4b² + 16 = -3b² + 16 ≥ 0So, -3b² + 16 ≥ 0 => 3b² ≤ 16 => b² ≤ 16/3 => b ≤ 4/√3 ≈ 2.309Similarly, if I consider the equation as a quadratic in b, I would get the same condition on a.But maybe this isn't the most straightforward way.Alternatively, let me consider that a² + b² - ab = 4. Maybe I can express this in terms of (a - b/2)^2 + (3b²)/4 = 4. Let me check:a² - ab + b² = 4a² - ab + (b²/4) + (3b²)/4 = 4(a - b/2)^2 + (3b²)/4 = 4Yes, that's correct. So, (a - b/2)^2 + (3b²)/4 = 4Since both terms are squares, they are non-negative. So, the maximum value of ab occurs when (a - b/2)^2 is as large as possible, but that might not directly help. Alternatively, perhaps I can parameterize a and b.Let me try to set a = kb, where k is a positive real number. Then, substituting into the equation:(kb)^2 + b^2 - (kb)b = 4k² b² + b² - k b² = 4b²(k² + 1 - k) = 4So, b² = 4 / (k² + 1 - k)Then, ab = (kb)*b = k b² = k*(4 / (k² + 1 - k)) = 4k / (k² - k + 1)So, ab = 4k / (k² - k + 1). Now, we need to maximize ab with respect to k.Let me denote f(k) = 4k / (k² - k + 1). To find its maximum, take the derivative and set it to zero.f'(k) = [4(k² - k + 1) - 4k(2k - 1)] / (k² - k + 1)^2Simplify numerator:4(k² - k + 1) - 4k(2k - 1) = 4k² - 4k + 4 - 8k² + 4k = (-4k²) + 0k + 4Set numerator equal to zero:-4k² + 4 = 0 => -4k² = -4 => k² = 1 => k = 1 or k = -1Since k is positive (as a and b are lengths), k = 1.So, maximum occurs at k = 1. Therefore, a = b.So, when a = b, ab is maximized.So, plugging back k = 1 into b² = 4 / (1 + 1 - 1) = 4 / 1 = 4. So, b = 2, and since a = b, a = 2.So, when a = b = 2, ab = 4, which is the maximum.Therefore, the maximum area is (√3/4)*ab = (√3/4)*4 = √3.So, the maximum area is √3.Wait, let me double-check this. If a = b = 2, then c² = a² + b² - ab = 4 + 4 - 4 = 4, so c = 2. So, the triangle is equilateral? Because all sides are equal to 2. So, in an equilateral triangle, all angles are 60 degrees, which matches our angle C being π/3. So, the area is indeed (√3/4)*2² = √3.Yes, that makes sense. So, the maximum area is √3.Alternatively, I could have used the formula for area in terms of sides and angle: (1/2)ab sin C. Since sin C is fixed at √3/2, maximizing ab would maximize the area. So, as we found, ab is maximized when a = b, leading to the maximum area.So, I think that's solid. The maximum area is √3.**Final Answer**1. The measure of angle ( C ) is (boxed{dfrac{pi}{3}}).2. The maximum area of triangle ( triangle ABC ) when ( c = 2 ) is (boxed{sqrt{3}}).

💡Alright, so I have this problem about a sequence where all the terms are positive numbers. The sum of the first n terms, denoted as S_n, satisfies the equation 2S_n = (a_n + 3)(a_n - 2) for all natural numbers n. There are two parts: first, I need to find the general term formula for the sequence {a_n}, and second, I need to find the sum of the first n terms, denoted as T_n, for the sequence {1/(a_{2n-1} * a_{2n+1})}.Starting with part (1), I need to find a formula for a_n. The given equation is 2S_n = (a_n + 3)(a_n - 2). Let me write that down:2S_n = (a_n + 3)(a_n - 2)First, I can expand the right-hand side to make it easier to work with:(a_n + 3)(a_n - 2) = a_n^2 - 2a_n + 3a_n - 6 = a_n^2 + a_n - 6So, 2S_n = a_n^2 + a_n - 6Now, S_n is the sum of the first n terms of the sequence {a_n}, so S_n = a_1 + a_2 + ... + a_n. Therefore, S_n can also be expressed in terms of S_{n-1}:S_n = S_{n-1} + a_nSo, if I write the equation for S_n and S_{n-1}, I can set up a recursive relationship.Given that 2S_n = a_n^2 + a_n - 6, similarly, for n-1, we have:2S_{n-1} = a_{n-1}^2 + a_{n-1} - 6Now, subtracting these two equations:2S_n - 2S_{n-1} = (a_n^2 + a_n - 6) - (a_{n-1}^2 + a_{n-1} - 6)Simplify the left side:2(S_n - S_{n-1}) = 2a_nAnd the right side:a_n^2 + a_n - 6 - a_{n-1}^2 - a_{n-1} + 6 = a_n^2 - a_{n-1}^2 + a_n - a_{n-1}So, putting it together:2a_n = a_n^2 - a_{n-1}^2 + a_n - a_{n-1}Let me rearrange this equation:2a_n - a_n = a_n^2 - a_{n-1}^2 - a_{n-1}So, a_n = a_n^2 - a_{n-1}^2 - a_{n-1}Let me bring all terms to one side:a_n^2 - a_{n-1}^2 - a_n - a_{n-1} = 0Hmm, this looks like a quadratic in terms of a_n and a_{n-1}. Maybe I can factor this.Looking at the equation:a_n^2 - a_{n-1}^2 - a_n - a_{n-1} = 0Notice that a_n^2 - a_{n-1}^2 is a difference of squares, so it factors as (a_n - a_{n-1})(a_n + a_{n-1}).So, let me factor that:(a_n - a_{n-1})(a_n + a_{n-1}) - (a_n + a_{n-1}) = 0Factor out (a_n + a_{n-1}):(a_n + a_{n-1})(a_n - a_{n-1} - 1) = 0So, either (a_n + a_{n-1}) = 0 or (a_n - a_{n-1} - 1) = 0.But since all terms are positive, a_n + a_{n-1} can't be zero. So, we must have:a_n - a_{n-1} - 1 = 0 => a_n = a_{n-1} + 1So, this tells me that the sequence {a_n} is an arithmetic sequence with common difference 1.Now, to find the first term, a_1. Let's use the original equation for n=1:2S_1 = (a_1 + 3)(a_1 - 2)But S_1 = a_1, so:2a_1 = (a_1 + 3)(a_1 - 2)Expanding the right side:2a_1 = a_1^2 + 3a_1 - 2a_1 - 6 = a_1^2 + a_1 - 6Bring all terms to one side:a_1^2 + a_1 - 6 - 2a_1 = 0 => a_1^2 - a_1 - 6 = 0Solving this quadratic equation:a_1 = [1 ± sqrt(1 + 24)] / 2 = [1 ± 5] / 2Since a_1 is positive, we take the positive root:a_1 = (1 + 5)/2 = 3So, the first term is 3, and since the common difference is 1, the general term is:a_n = a_1 + (n - 1)*1 = 3 + n - 1 = n + 2So, a_n = n + 2.Now, moving on to part (2). I need to find the sum of the first n terms, T_n, for the sequence {1/(a_{2n-1} * a_{2n+1})}.First, let's write down the general term of this new sequence:1/(a_{2n-1} * a_{2n+1})Given that a_k = k + 2, so:a_{2n-1} = (2n - 1) + 2 = 2n + 1a_{2n+1} = (2n + 1) + 2 = 2n + 3So, the term becomes:1/((2n + 1)(2n + 3))I need to find the sum T_n = sum_{k=1}^n [1/((2k + 1)(2k + 3))]This looks like a telescoping series. To sum it, I can use partial fractions.Let me express 1/((2k + 1)(2k + 3)) as A/(2k + 1) + B/(2k + 3)So,1/((2k + 1)(2k + 3)) = A/(2k + 1) + B/(2k + 3)Multiply both sides by (2k + 1)(2k + 3):1 = A(2k + 3) + B(2k + 1)Now, let's solve for A and B.Expanding the right side:1 = (2A + 2B)k + (3A + B)Since this must hold for all k, the coefficients of like terms must be equal.So,2A + 2B = 0 (coefficient of k)3A + B = 1 (constant term)From the first equation:2A + 2B = 0 => A + B = 0 => B = -ASubstitute into the second equation:3A + (-A) = 1 => 2A = 1 => A = 1/2Then, B = -1/2So, the partial fractions are:1/((2k + 1)(2k + 3)) = (1/2)/(2k + 1) - (1/2)/(2k + 3)Therefore, each term can be written as:1/2 [1/(2k + 1) - 1/(2k + 3)]So, the sum T_n becomes:T_n = sum_{k=1}^n [1/2 (1/(2k + 1) - 1/(2k + 3))]Factor out the 1/2:T_n = (1/2) sum_{k=1}^n [1/(2k + 1) - 1/(2k + 3)]Now, let's write out the terms of the sum to see if it telescopes:For k=1: 1/3 - 1/5For k=2: 1/5 - 1/7For k=3: 1/7 - 1/9...For k=n: 1/(2n + 1) - 1/(2n + 3)When we add these up, most terms cancel out:(1/3 - 1/5) + (1/5 - 1/7) + (1/7 - 1/9) + ... + (1/(2n + 1) - 1/(2n + 3))All the intermediate terms cancel, leaving:1/3 - 1/(2n + 3)So, the sum inside the brackets is 1/3 - 1/(2n + 3)Therefore, T_n = (1/2)(1/3 - 1/(2n + 3))Simplify this expression:First, combine the terms inside:1/3 - 1/(2n + 3) = (2n + 3 - 3)/(3(2n + 3)) = (2n)/(3(2n + 3))So, T_n = (1/2)(2n)/(3(2n + 3)) = (n)/(3(2n + 3)) = n/(6n + 9)I can factor numerator and denominator:n/(6n + 9) = n/(3(2n + 3)) = (n)/(6n + 9)Alternatively, it can be written as n/(6n + 9), which is the simplified form.So, the sum T_n is n/(6n + 9).Let me double-check my steps to make sure I didn't make a mistake.First, for part (1), I found that the sequence is arithmetic with a common difference of 1 and a_1 = 3, so a_n = n + 2. That seems correct.For part (2), I expressed the term as partial fractions and saw that it telescopes. The telescoping leaves 1/3 - 1/(2n + 3), which when multiplied by 1/2 gives n/(6n + 9). That seems correct.I think I did everything right. So, the answers are:(1) a_n = n + 2(2) T_n = n/(6n + 9)**Final Answer**(1) The general term is boxed{a_n = n + 2}.(2) The sum of the first n terms is boxed{dfrac{n}{6n + 9}}.

💡Okay, so I've got this quadratic equation: (x^2 - 4kx + 3k^2 - 3 = 0). The problem says that for a given integer value of (k), the product of the roots is 10. I need to figure out the nature of the roots, which could be integral and positive, integral and negative, rational but not integral, irrational, or imaginary.First, I remember that for a quadratic equation (ax^2 + bx + c = 0), the product of the roots is given by (frac{c}{a}). In this case, (a = 1), so the product of the roots is just (c), which is (3k^2 - 3). The problem states that this product is 10, so I can set up the equation:(3k^2 - 3 = 10)Let me solve for (k). Adding 3 to both sides:(3k^2 = 13)Then, dividing both sides by 3:(k^2 = frac{13}{3})Hmm, that's a problem because (k) is supposed to be an integer. (frac{13}{3}) is not a perfect square, and its square root is not an integer. So, this suggests that there might be a mistake in my approach or in the problem statement.Wait, maybe I misapplied Vieta's formula. Let me double-check. Vieta's formula for the product of the roots is indeed (frac{c}{a}), which in this case is (3k^2 - 3). So that part is correct.Since (k) must be an integer, and (3k^2 - 3 = 10) doesn't yield an integer (k), perhaps the problem meant something else. Maybe the product of the roots is 10, but (k) is not necessarily an integer? But the problem clearly states that (k) is an integer.Alternatively, maybe I need to consider that the product is 10, but (k) is an integer, so perhaps I need to find integer (k) such that (3k^2 - 3 = 10). But as I saw earlier, that leads to (k^2 = frac{13}{3}), which is not an integer.Wait, maybe I made a mistake in setting up the equation. Let me check again. The product of the roots is 10, so:(3k^2 - 3 = 10)Yes, that's correct. So, (3k^2 = 13), which means (k^2 = frac{13}{3}). Since (k) must be an integer, this suggests that there is no integer (k) that satisfies this condition. But the problem says "for a given integer value of (k)", so perhaps I'm missing something.Maybe the product of the roots is not 10, but something else? Or perhaps the equation is different? Let me reread the problem."For a given integer value of (k), the product of the roots of the quadratic equation (x^2 - 4kx + 3k^2 - 3 = 0) is 10. Determine the nature of the roots of the equation."Hmm, it seems correct. So, if (k) must be an integer, but (3k^2 - 3 = 10) doesn't yield an integer (k), perhaps there's a typo in the problem, or maybe I need to interpret it differently.Alternatively, maybe the product of the roots is 10, but (k) is not necessarily an integer? But the problem states that (k) is an integer. Hmm.Wait, perhaps I need to consider that the product of the roots is 10, but (k) is an integer, so maybe I need to adjust the equation to make (3k^2 - 3) equal to 10 with integer (k). But as I saw, that's not possible because (k^2) would have to be (frac{13}{3}), which is not an integer.Maybe the problem meant that the sum of the roots is 10? Let me check. The sum of the roots is (-frac{b}{a}), which in this case is (4k). If the sum of the roots is 10, then (4k = 10), so (k = frac{10}{4} = frac{5}{2}), which is not an integer either. Hmm.Wait, maybe the problem is correct, and I just need to proceed with the given information, even if (k) is not an integer? But the problem specifically says (k) is an integer. This is confusing.Alternatively, perhaps the product of the roots is 10, but (k) is not an integer, and I need to find the nature of the roots regardless of (k). But the problem says "for a given integer value of (k)", so I think (k) must be an integer.Wait, maybe I need to consider that the product of the roots is 10, and (k) is an integer, so perhaps I need to find (k) such that (3k^2 - 3 = 10), even if it's not an integer, but that contradicts the problem statement.I'm stuck here. Maybe I need to proceed differently. Let's assume that (k) is an integer, and (3k^2 - 3 = 10). Then, (k^2 = frac{13}{3}), which is not an integer, so there is no such integer (k). Therefore, perhaps the problem is misstated, or I'm misinterpreting it.Wait, maybe the product of the roots is 10, but (k) is not necessarily an integer? But the problem says (k) is an integer. Hmm.Alternatively, maybe the product of the roots is 10, and (k) is an integer, but the equation is different? Let me check the equation again: (x^2 - 4kx + 3k^2 - 3 = 0). Yes, that's correct.Wait, perhaps I need to consider that the product of the roots is 10, and (k) is an integer, but the equation is correct, so maybe I need to find the nature of the roots without knowing (k). But the problem says "for a given integer value of (k)", so I think (k) is given, but the product is 10.I'm confused. Maybe I need to proceed with the given information, even if (k) is not an integer, and see what the nature of the roots would be.So, if (3k^2 - 3 = 10), then (k^2 = frac{13}{3}), so (k = pm sqrt{frac{13}{3}}). Then, the quadratic equation becomes (x^2 - 4kx + 10 = 0). Let's compute the discriminant:Discriminant (D = (4k)^2 - 4 times 1 times 10 = 16k^2 - 40).Substituting (k^2 = frac{13}{3}):(D = 16 times frac{13}{3} - 40 = frac{208}{3} - 40 = frac{208 - 120}{3} = frac{88}{3}).Since the discriminant is positive but not a perfect square, the roots are real and irrational.But wait, the problem says (k) is an integer, so this approach might not be valid. Maybe I need to consider that (k) is an integer, and the product of the roots is 10, so (3k^2 - 3 = 10), which leads to (k^2 = frac{13}{3}), which is not an integer. Therefore, there is no integer (k) that satisfies this condition.But the problem says "for a given integer value of (k)", so perhaps I'm missing something. Maybe the product of the roots is not 10, but something else? Or perhaps the equation is different?Wait, maybe I need to consider that the product of the roots is 10, but (k) is an integer, so perhaps I need to adjust the equation to make (3k^2 - 3) equal to 10 with integer (k). But as I saw, that's not possible because (k^2) would have to be (frac{13}{3}), which is not an integer.Alternatively, maybe the problem meant that the sum of the roots is 10? Let me check. The sum of the roots is (-frac{b}{a}), which in this case is (4k). If the sum of the roots is 10, then (4k = 10), so (k = frac{10}{4} = frac{5}{2}), which is not an integer either.Hmm, this is confusing. Maybe the problem is misstated, or perhaps I'm misinterpreting it. Alternatively, maybe the product of the roots is 10, but (k) is not an integer, and I need to find the nature of the roots regardless of (k). But the problem says "for a given integer value of (k)", so I think (k) must be an integer.Wait, maybe I need to consider that the product of the roots is 10, and (k) is an integer, but the equation is correct, so maybe I need to find the nature of the roots without knowing (k). But the problem says "for a given integer value of (k)", so I think (k) is given, but the product is 10.I'm stuck. Maybe I need to proceed with the given information, even if (k) is not an integer, and see what the nature of the roots would be.So, if (3k^2 - 3 = 10), then (k^2 = frac{13}{3}), so (k = pm sqrt{frac{13}{3}}). Then, the quadratic equation becomes (x^2 - 4kx + 10 = 0). Let's compute the discriminant:Discriminant (D = (4k)^2 - 4 times 1 times 10 = 16k^2 - 40).Substituting (k^2 = frac{13}{3}):(D = 16 times frac{13}{3} - 40 = frac{208}{3} - 40 = frac{208 - 120}{3} = frac{88}{3}).Since the discriminant is positive but not a perfect square, the roots are real and irrational.But wait, the problem says (k) is an integer, so this approach might not be valid. Maybe I need to consider that (k) is an integer, and the product of the roots is 10, so (3k^2 - 3 = 10), which leads to (k^2 = frac{13}{3}), which is not an integer. Therefore, there is no integer (k) that satisfies this condition.But the problem says "for a given integer value of (k)", so perhaps I'm missing something. Maybe the product of the roots is not 10, but something else? Or perhaps the equation is different?Wait, maybe I need to consider that the product of the roots is 10, but (k) is an integer, so perhaps I need to adjust the equation to make (3k^2 - 3) equal to 10 with integer (k). But as I saw, that's not possible because (k^2) would have to be (frac{13}{3}), which is not an integer.Alternatively, maybe the problem meant that the sum of the roots is 10? Let me check. The sum of the roots is (-frac{b}{a}), which in this case is (4k). If the sum of the roots is 10, then (4k = 10), so (k = frac{10}{4} = frac{5}{2}), which is not an integer either.I'm going in circles here. Maybe I need to conclude that there is no integer (k) that satisfies the given condition, and therefore, the problem might have an error. Alternatively, perhaps I need to proceed with the assumption that (k) is not an integer, and the roots are irrational.Given that, the discriminant is (frac{88}{3}), which is positive but not a perfect square, so the roots are real and irrational. Therefore, the answer would be D) Irrational.But I'm not entirely sure because the problem specifies that (k) is an integer, and in that case, there is no solution. Maybe the problem intended for (k) to be a real number, not necessarily an integer. If that's the case, then the roots are irrational.Alternatively, perhaps I need to consider that the product of the roots is 10, and (k) is an integer, but the equation is different. Maybe the equation is (x^2 - 4kx + 3k^2 - 3 = 0), and the product of the roots is 10, so (3k^2 - 3 = 10), leading to (k^2 = frac{13}{3}), which is not an integer. Therefore, there is no integer (k) that satisfies this condition, and the problem might be misstated.But since the problem asks to determine the nature of the roots, assuming that such a (k) exists, even if it's not an integer, then the roots would be irrational. Therefore, the answer is D) Irrational.

💡Alright, so I've got this problem here where there are 9 boys and 15 girls lining up in a row. The arrangement has to start with a boy and end with a girl. I need to find the average value of T, which is the number of places where a boy and a girl are standing next to each other. Hmm, okay, let me try to break this down.First off, there are a total of 24 people: 9 boys and 15 girls. The arrangement starts with a boy and ends with a girl. So, the first position is fixed as a boy, and the last position is fixed as a girl. That leaves us with 22 people to arrange in the middle positions. But wait, actually, since the first and last positions are fixed, we have 22 positions left, but they can be filled by any of the remaining boys and girls. So, we have 8 boys and 15 girls left for the middle 22 positions.Now, T is the number of places where a boy and a girl are adjacent. So, T counts the number of times a boy is next to a girl or a girl is next to a boy in the entire row. Since the row has 24 people, there are 23 adjacent pairs. Each pair can either be boy-boy, girl-girl, or boy-girl/girl-boy. T counts the last two cases.To find the average value of T, I think I need to use the concept of linearity of expectation. Instead of trying to calculate T directly, I can consider each adjacent pair and calculate the probability that they are a boy and a girl. Then, summing these probabilities will give me the expected value of T.So, let's define an indicator variable X_i for each pair i (from 1 to 23). X_i is 1 if the i-th and (i+1)-th positions are a boy and a girl or a girl and a boy, and 0 otherwise. Then, T = X_1 + X_2 + ... + X_23. Therefore, the expected value E[T] = E[X_1] + E[X_2] + ... + E[X_23].Since all positions are symmetric except for the fixed first and last positions, the probability that each X_i is 1 might be the same for all i, except maybe for the first and last pairs. Wait, actually, the first pair is between the first position (fixed as a boy) and the second position, which is among the 22 remaining people. Similarly, the last pair is between the 23rd position and the 24th position (fixed as a girl). So, the first and last pairs might have different probabilities compared to the middle pairs.Let me think about this. For the first pair, the first position is a boy, and the second position can be any of the remaining 22 people, which includes 8 boys and 15 girls. So, the probability that the second position is a girl is 15/22, which would make the first pair a boy-girl pair. Similarly, the probability that the second position is a boy is 8/22, which would make the first pair a boy-boy pair. Therefore, the probability that the first pair is a boy-girl pair is 15/22.Similarly, for the last pair, the 24th position is a girl, and the 23rd position can be any of the remaining 22 people, which includes 8 boys and 15 girls. So, the probability that the 23rd position is a boy is 8/22, making the last pair a boy-girl pair. Therefore, the probability that the last pair is a boy-girl pair is 8/22.Now, for the middle pairs, which are between positions 2 and 22, each pair is between two positions that are both among the 22 remaining people. So, for each of these middle pairs, the probability that one is a boy and the other is a girl needs to be calculated.Wait, but actually, for each middle pair, the two positions are adjacent, and each can be either a boy or a girl. So, we need to calculate the probability that one is a boy and the other is a girl. Since the arrangement is random, the probability that the first of the two is a boy and the second is a girl plus the probability that the first is a girl and the second is a boy.But hold on, the total number of boys and girls in the middle 22 positions is 8 boys and 15 girls. So, for any two adjacent positions in the middle, the probability that the first is a boy and the second is a girl is (8/22) * (15/21), because after choosing a boy for the first position, there are 15 girls left out of 21 remaining people. Similarly, the probability that the first is a girl and the second is a boy is (15/22) * (8/21). So, the total probability for a middle pair to be a boy-girl or girl-boy pair is (8*15)/(22*21) + (15*8)/(22*21) = 2*(8*15)/(22*21).Calculating that, 8*15 is 120, so 2*120 is 240. 22*21 is 462. So, 240/462 simplifies to 40/77. So, each middle pair has a probability of 40/77 of being a boy-girl or girl-boy pair.Now, how many middle pairs are there? Since the entire row has 24 positions, there are 23 pairs. We've already considered the first and last pairs separately, so the middle pairs are from pair 2 to pair 22, which is 21 pairs. Wait, no, actually, pair 1 is first and second, pair 2 is second and third, ..., pair 23 is 23rd and 24th. So, the first pair is pair 1, the last pair is pair 23, and the middle pairs are pairs 2 to 22, which is 21 pairs.But wait, earlier I thought the middle pairs are 22 positions, but actually, the middle pairs are 21 pairs because we're excluding the first and last pairs. So, each of these 21 pairs has a probability of 40/77 of being a boy-girl or girl-boy pair.Therefore, the expected value E[T] is the sum of the expected values for each pair. So, E[T] = E[X_1] + E[X_2] + ... + E[X_23].We already calculated E[X_1] = 15/22, E[X_23] = 8/22, and for each E[X_i] where i is from 2 to 22, it's 40/77.So, E[T] = 15/22 + 8/22 + 21*(40/77).Let me compute this step by step.First, 15/22 + 8/22 = (15 + 8)/22 = 23/22.Next, 21*(40/77). Let's compute 21*40 = 840, and 840/77. 77*10 = 770, so 840 - 770 = 70, which is 77*0.909... Wait, actually, 77*10 = 770, 77*11 = 847, which is more than 840. So, 840/77 = 10 + (70/77) = 10 + 10/11 ≈ 10.909.But let's keep it as fractions. 840 divided by 77. Let's see, 77*10 = 770, so 840 - 770 = 70. So, 70/77 = 10/11. Therefore, 840/77 = 10 + 10/11 = 110/11 + 10/11 = 120/11.Wait, that can't be right because 77*10 = 770, 77*11 = 847, which is more than 840. So, 840/77 = 10 + 70/77 = 10 + 10/11 = 110/11 + 10/11 = 120/11? Wait, 110/11 is 10, and 10/11 is approximately 0.909, so 10 + 0.909 is 10.909, which is 120/11 ≈ 10.909.Wait, but 120/11 is approximately 10.909, which is correct.So, 21*(40/77) = 840/77 = 120/11.Therefore, E[T] = 23/22 + 120/11.Now, let's convert 23/22 to a fraction with denominator 11. 23/22 = (23/2)/11 = 11.5/11. Wait, but fractions should be in whole numbers. Alternatively, 23/22 = (23*1)/(22) = (23*1)/(2*11) = 23/(2*11). Hmm, maybe I should find a common denominator.The denominators are 22 and 11. The least common denominator is 22. So, 23/22 is already in terms of 22, and 120/11 = (120*2)/22 = 240/22.So, E[T] = 23/22 + 240/22 = (23 + 240)/22 = 263/22.Calculating 263 divided by 22. 22*11 = 242, so 263 - 242 = 21. So, 263/22 = 11 + 21/22 ≈ 11.9545.So, approximately 11.9545. Since the problem asks for the average value, and it's likely to be a fraction, but maybe they want it as a reduced fraction or a decimal. But let me check my calculations again to make sure I didn't make a mistake.Wait, when I calculated 21*(40/77), I got 840/77 = 120/11. Let me verify that. 40/77 multiplied by 21 is (40*21)/77. 21 and 77 have a common factor of 7. 21/77 = 3/11. So, (40*3)/11 = 120/11. Yes, that's correct.Then, E[T] = 23/22 + 120/11. Converting 120/11 to 22 denominator: 120/11 = (120*2)/22 = 240/22. So, 23/22 + 240/22 = 263/22.263 divided by 22 is indeed 11 with a remainder of 21, so 11 and 21/22, which is approximately 11.9545.But wait, the problem says to calculate the average value of T. So, is 263/22 the exact value, which is approximately 11.9545, or should I round it to the nearest whole number? The original problem didn't specify, but in the previous example, the assistant rounded it to 12. So, maybe I should do the same.Alternatively, perhaps I made a mistake in considering the first and last pairs. Let me double-check.For the first pair, the first position is fixed as a boy, and the second position is among 22 people: 8 boys and 15 girls. So, the probability that the second position is a girl is 15/22, making the first pair a boy-girl pair. Similarly, the probability that the second position is a boy is 8/22, making it a boy-boy pair. So, E[X_1] = 15/22.For the last pair, the 24th position is fixed as a girl, and the 23rd position is among 22 people: 8 boys and 15 girls. So, the probability that the 23rd position is a boy is 8/22, making the last pair a boy-girl pair. So, E[X_23] = 8/22.For the middle pairs, from pair 2 to pair 22, which is 21 pairs, each has a probability of 40/77 of being a boy-girl or girl-boy pair.So, E[T] = 15/22 + 8/22 + 21*(40/77).Calculating 15/22 + 8/22 = 23/22.21*(40/77) = (21*40)/77 = 840/77 = 120/11.So, E[T] = 23/22 + 120/11 = 23/22 + 240/22 = 263/22 ≈ 11.9545.So, approximately 11.9545, which is very close to 12. So, the average value of T is approximately 12.But wait, let me think again. Is there another way to approach this problem? Maybe using the concept of expected number of adjacent boy-girl or girl-boy pairs.Alternatively, we can think of the entire row as a sequence of 24 positions, with the first fixed as a boy and the last fixed as a girl. The rest are arranged randomly. So, the expected number of boy-girl or girl-boy adjacent pairs can be calculated by considering each adjacent pair and the probability that they are of opposite genders.But I think the way I approached it earlier is correct. Let me just verify the calculation one more time.23/22 is approximately 1.0455, and 120/11 is approximately 10.9091. Adding them together gives approximately 11.9545, which is about 11.95, very close to 12.So, I think the average value of T is approximately 12. Therefore, the answer is 12.

💡Okay, so I have this equation: ( p^a - 1 = 2^n (p - 1) ), where ( p ) is a prime number, and I need to find all strictly positive integers ( a ), ( p ), and ( n ) that satisfy it. Hmm, let me try to break this down step by step.First, since ( p ) is a prime number, it has to be at least 2. But if ( p = 2 ), then the right side becomes ( 2^n (2 - 1) = 2^n times 1 = 2^n ). The left side would be ( 2^a - 1 ). So, we have ( 2^a - 1 = 2^n ). Hmm, is this possible?Let me see. If ( a = 1 ), then ( 2^1 - 1 = 1 ), which would mean ( 2^n = 1 ), so ( n = 0 ). But the problem says strictly positive numbers, so ( n ) has to be at least 1. So, ( a = 1 ) doesn't work here. What about ( a = 2 )? Then ( 2^2 - 1 = 3 ), so ( 2^n = 3 ), which isn't possible since ( n ) has to be an integer. ( a = 3 ) gives ( 2^3 - 1 = 7 ), so ( 2^n = 7 ), which also isn't possible. It seems like for ( p = 2 ), there are no solutions with ( a ) and ( n ) being positive integers. So, maybe ( p ) has to be an odd prime.Alright, let's assume ( p ) is an odd prime, so ( p geq 3 ). Then, ( p - 1 ) is even, which means the right side ( 2^n (p - 1) ) is divisible by at least ( 2^{n+1} ). The left side is ( p^a - 1 ). Since ( p ) is odd, ( p^a ) is also odd, so ( p^a - 1 ) is even. Therefore, both sides are even, which is consistent.Let me try to factor the left side. ( p^a - 1 ) can be factored as ( (p - 1)(p^{a-1} + p^{a-2} + dots + p + 1) ). So, substituting back into the equation:( (p - 1)(p^{a-1} + p^{a-2} + dots + p + 1) = 2^n (p - 1) )Since ( p - 1 ) is non-zero (because ( p geq 3 )), I can divide both sides by ( p - 1 ):( p^{a-1} + p^{a-2} + dots + p + 1 = 2^n )So, now I have an equation where the sum of a geometric series equals a power of 2. That's interesting. Let me denote this sum as ( S = p^{a-1} + p^{a-2} + dots + p + 1 ). So, ( S = 2^n ).Since ( p ) is an odd prime, each term in the sum ( S ) is odd. The number of terms in the sum is ( a ). So, if ( a ) is even, the sum ( S ) would be even (since sum of even number of odd terms is even), and if ( a ) is odd, the sum ( S ) would be odd (since sum of odd number of odd terms is odd). But ( 2^n ) is always even except when ( n = 0 ), but ( n ) has to be positive, so ( S ) must be even. Therefore, ( a ) must be even.Wait, but ( a ) is a positive integer, so ( a geq 1 ). If ( a = 1 ), then ( S = 1 ), which would mean ( 2^n = 1 ), but ( n ) has to be positive, so ( a = 1 ) is not possible. So, ( a ) must be at least 2.Let me try ( a = 2 ). Then, ( S = p + 1 ). So, ( p + 1 = 2^n ). Therefore, ( p = 2^n - 1 ). Since ( p ) is a prime, ( 2^n - 1 ) must be prime. These are known as Mersenne primes. So, for ( a = 2 ), ( p ) must be a Mersenne prime, and ( n ) must be such that ( 2^n - 1 ) is prime.Okay, so that gives us a family of solutions: ( a = 2 ), ( p = 2^n - 1 ), where ( 2^n - 1 ) is prime. For example, if ( n = 2 ), ( p = 3 ); if ( n = 3 ), ( p = 7 ); if ( n = 5 ), ( p = 31 ), etc.But are there other solutions where ( a > 2 )? Let's check.Suppose ( a = 3 ). Then, ( S = p^2 + p + 1 = 2^n ). So, ( p^2 + p + 1 ) must be a power of 2. Let's see if this is possible.Let me try small primes for ( p ):- ( p = 3 ): ( 3^2 + 3 + 1 = 9 + 3 + 1 = 13 ), which is not a power of 2.- ( p = 5 ): ( 25 + 5 + 1 = 31 ), which is ( 2^5 - 1 ), not a power of 2.- ( p = 7 ): ( 49 + 7 + 1 = 57 ), not a power of 2.- ( p = 11 ): ( 121 + 11 + 1 = 133 ), not a power of 2.Hmm, seems like for ( a = 3 ), there are no solutions. Let me try ( a = 4 ). Then, ( S = p^3 + p^2 + p + 1 = 2^n ).Again, trying small primes:- ( p = 3 ): ( 27 + 9 + 3 + 1 = 40 ), which is ( 8 times 5 ), not a power of 2.- ( p = 5 ): ( 125 + 25 + 5 + 1 = 156 ), which is ( 12 times 13 ), not a power of 2.- ( p = 7 ): ( 343 + 49 + 7 + 1 = 400 ), which is ( 16 times 25 ), not a power of 2.Still no luck. Maybe ( a = 4 ) doesn't work either.What about ( a = 5 )? Then, ( S = p^4 + p^3 + p^2 + p + 1 = 2^n ). Trying ( p = 3 ):( 81 + 27 + 9 + 3 + 1 = 121 ), which is ( 11^2 ), not a power of 2.( p = 5 ): ( 625 + 125 + 25 + 5 + 1 = 781 ), not a power of 2.Hmm, seems like as ( a ) increases, the sum ( S ) grows rapidly and it's unlikely to be a power of 2. Maybe there are no solutions for ( a > 2 ).But let me think more carefully. Suppose ( a ) is a prime number. Then, by Zsigmondy's theorem, ( p^a - 1 ) has a primitive prime divisor that does not divide ( p - 1 ). But in our equation, ( p^a - 1 = 2^n (p - 1) ), so all prime divisors of ( p^a - 1 ) must divide ( 2 ) or ( p - 1 ). Therefore, the primitive prime divisor must be 2. But ( p - 1 ) is even, so it already has 2 as a prime divisor. Therefore, Zsigmondy's theorem would imply that there is another prime divisor, which is a contradiction unless ( a = 2 ).Wait, so if ( a ) is a prime greater than 2, then Zsigmondy's theorem says there's a primitive prime divisor, which would have to be 2, but ( p - 1 ) already has 2, so that can't be. Therefore, ( a ) cannot be a prime greater than 2.What if ( a ) is composite? Let me suppose ( a = uv ) where ( u, v > 1 ). Then, ( p^a - 1 = (p^u)^v - 1 ), which factors as ( (p^u - 1)(p^{u(v-1)} + p^{u(v-2)} + dots + p^u + 1) ). So, ( p^u - 1 ) divides ( p^a - 1 ). But from our equation, ( p^a - 1 = 2^n (p - 1) ), so ( p^u - 1 ) must divide ( 2^n (p - 1) ). Since ( p^u - 1 ) is greater than ( p - 1 ) (because ( u > 1 )), it must be that ( p^u - 1 ) is a power of 2 times ( p - 1 ). But ( p^u - 1 = (p - 1)(p^{u-1} + p^{u-2} + dots + p + 1) ). So, ( p^{u-1} + p^{u-2} + dots + p + 1 ) must be a power of 2.But this is similar to our original problem but with a smaller exponent. If ( u ) is composite, we can apply the same reasoning recursively. If ( u ) is prime, then by Zsigmondy's theorem, there's a primitive prime divisor, which would have to be 2, but again, ( p - 1 ) already has 2, leading to a contradiction. Therefore, ( a ) cannot be composite either.So, the only possibility is ( a = 2 ). Therefore, the only solutions are when ( a = 2 ), and ( p = 2^n - 1 ) is a prime number.Let me verify this with an example. Let ( n = 2 ), so ( p = 2^2 - 1 = 3 ). Then, ( p^a - 1 = 3^2 - 1 = 9 - 1 = 8 ), and ( 2^n (p - 1) = 4 times 2 = 8 ). It works. Another example: ( n = 3 ), ( p = 7 ). Then, ( 7^2 - 1 = 49 - 1 = 48 ), and ( 2^3 times (7 - 1) = 8 times 6 = 48 ). It works too.What about ( n = 5 ), ( p = 31 ). ( 31^2 - 1 = 961 - 1 = 960 ), and ( 2^5 times 30 = 32 times 30 = 960 ). Yep, that works as well.So, it seems like the only solutions are when ( a = 2 ) and ( p ) is a Mersenne prime, i.e., ( p = 2^n - 1 ) where ( 2^n - 1 ) is prime.I don't think there are any other solutions because for ( a > 2 ), whether ( a ) is prime or composite, we run into contradictions with Zsigmondy's theorem or the structure of the equation. Therefore, the only solutions are when ( a = 2 ) and ( p ) is a Mersenne prime.