<Question>
Let $(a_1, a_2, \dots, a_n)$ be a sequence of positive real numbers, such that
\[\sum_{i = 1}^n a_i = 96, \quad \sum_{i = 1}^n a_i^2 = 144, \quad \sum_{i = 1}^n a_i^3 = 216.\]Find the sum of all possible values of $n.$
</Question>

<think>
<Plan_1>
Okay, let's see. I need to find the sum of all possible values of n given the three conditions on the sequence of positive real numbers (a₁, a₂, …, aₙ). The sums are:

1. The sum of the numbers is 96.
2. The sum of their squares is 144.
3. The sum of their cubes is 216.

Hmm. So, we have three equations involving the same sequence. The variables are the a_i's and n. Since all a_i are positive real numbers, maybe there's a way to relate these equations to find possible n's.

Let me think. If all the a_i's are equal, that might simplify things. Let's check if that's possible.

<prompt_1.1> Assume all \( a_i \) are equal and solve for \( n \). </prompt_1.1>
-----

Is there another way the a_i's could be arranged without all being equal? For example, maybe some are 3 and some are different. Let's see.

Suppose not all a_i are equal. Then there must be at least two different values among them. Let's consider the case where two of them are different.

<prompt_1.2> Consider the case where not all \( a_i \) are equal and analyze the possibility of other solutions. </prompt_1.2>
-----

Alternatively, maybe there's another approach using Cauchy-Schwarz or Hölder's inequality? Let me check.

<prompt_1.3> Explore alternative methods using inequalities like Cauchy-Schwarz or Hölder's inequality. </prompt_1.3>

Alternatively, maybe using the relationship between the sums. Let's denote S1 = 96, S2 = 144, S3 = 216.

<prompt_1.4> Explore alternative approaches using the relationship between sums and verify if \( n = 24 \) is the only solution. </prompt_1.4>
</Plan_1>

<Execution_1>
<execution_1.1>
If all a_i are equal, then each a_i would be 96/n. Then, the sum of squares would be n*(96/n)² = 96² / n = 144. Let's solve for n:

96² / n = 144

So, n = 96² / 144. Let's compute that.

96 squared is 9216. 9216 divided by 144: 144*64 = 9216, so n = 64.
</execution_1.1>

<execution_1.2>
Let’s suppose k of them are 3 and (n - k) of them are some other value, say x.

Then, the sum would be 3k + x(n - k) = 96.

The sum of squares would be 9k + x²(n - k) = 144.

The sum of cubes would be 27k + x³(n - k) = 216.

Now, that gives us three equations with two variables k and x (but k must be an integer between 1 and n-1). Let me see if this is possible.

From the first equation: x(n - k) = 96 - 3k.

From the second equation: x²(n - k) = 144 - 9k.

From the third equation: x³(n - k) = 216 - 27k.

Let me denote m = n - k, so m is a positive integer less than n.

Then, we have:

x*m = 96 - 3k,

x²*m = 144 - 9k,

x³*m = 216 - 27k.

Let's divide the second equation by the first: (x²*m)/(x*m) = (144 - 9k)/(96 - 3k).

Simplifies to x = (144 - 9k)/(96 - 3k) = [9(16 - k)]/[3(32 - k)] = [3(16 - k)]/[32 - k].

Similarly, divide the third equation by the second: (x³*m)/(x²*m) = (216 - 27k)/(144 - 9k).

Simplifies to x = (216 - 27k)/(144 - 9k) = [27(8 - k)]/[9(16 - k)] = [3(8 - k)]/[16 - k].

So now we have two expressions for x:

x = [3(16 - k)]/[32 - k] and x = [3(8 - k)]/[16 - k].

Set them equal:

[3(16 - k)]/[32 - k] = [3(8 - k)]/[16 - k].

Cancel the 3:

(16 - k)/(32 - k) = (8 - k)/(16 - k).

Cross-multiply:

(16 - k)² = (32 - k)(8 - k).

Expand both sides:

Left side: (16 - k)² = 256 - 32k + k².

Right side: (32 - k)(8 - k) = 256 - 32k -8k + k² = 256 -40k +k².

Set equal:

256 -32k +k² = 256 -40k +k².

Subtract 256 and k² from both sides:

-32k = -40k.

Which gives 8k = 0 => k = 0.

But k is the number of terms equal to 3, which we assumed to be at least 1. So k = 0 is not allowed here. Therefore, this case leads to a contradiction, meaning there are no solutions where only one of the a_i is different from 3. Hence, the only possible solution is when all a_i are equal to 3.
</execution_1.2>

<execution_1.3>
By Cauchy-Schwarz, we have (sum a_i)^2 ≤ n * sum a_i^2. Let's check:

96² = 9216 ≤ n * 144. Therefore, n ≥ 9216 / 144 = 64.
</execution_1.3>

<execution_1.4>
Maybe we can find some relation between these sums. For example, Cauchy-Schwarz says that (Σa_i)^2 ≤ n Σa_i². Let's check:

(96)^2 ≤ n * 144 ⇒ 9216 ≤ 144n ⇒ n ≥ 9216 / 144 = 64. So n ≥ 64.

Similarly, Hölder's inequality relates the sums of different orders. Hölder's with exponents p, q such that 1/p + 1/q = 1. Maybe not sure. Alternatively, using the inequality that relates L^p norms. The L^3 norm is less than or equal to the L^2 norm, but here L^3 is 216, L^2 is 144. Wait, but 216^2 = 46656, and 144*96 = 13824. Wait, not sure.

Alternatively, consider that for positive numbers, the ratio of sums can give information about the number of terms. Let's think in terms of the Power Mean inequality.

The Power Mean inequality states that for positive real numbers, the mean of the k-th powers is at least the (k-th root of the mean of the m-th powers) for m < k. So, for example, the quadratic mean is greater than or equal to the arithmetic mean. Let's compute:

Quadratic mean (QM) = sqrt(S2 / n) = sqrt(144 / n)

Arithmetic mean (AM) = S1 / n = 96 / n

QM ≥ AM ⇒ sqrt(144 / n) ≥ 96 / n ⇒ sqrt(144 / n) ≥ 96 / n ⇒ sqrt(144n) ≥ 96 ⇒ sqrt(144n) = 12√n ≥ 96 ⇒ √n ≥ 8 ⇒ n ≥ 64.

Which is the same result as before. Similarly, cubic mean (CM) = (S3 / n)^(1/3) = (216 / n)^(1/3)

Compare with AM: 96 / n

Power Mean for k=3: CM ≥ AM ⇒ (216 / n)^(1/3) ≥ 96 / n ⇒ 6 / n^(1/3) ≥ 96 / n ⇒ Multiply both sides by n^(1/3):

6 ≥ 96 / n^(2/3) ⇒ 6n^(2/3) ≥ 96 ⇒ n^(2/3) ≥ 16 ⇒ n ≥ 16^(3/2) ⇒ n ≥ 64.

Again, same result.
</execution_1.4>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Let me check if this is possible. If all a_i = 3, then:

Sum: 3n = 96 ⇒ n = 32. Wait, but 3n = 96 ⇒ n = 32.
But then sum of squares would be 9n = 9*32 = 288, which is not 144. Hmm, that's a problem.

So, my initial assumption that all a_i are equal might not hold. Wait, but if all a_i are equal, then we can solve for n. Let's try that.

<prompt_2.1> Re-evaluate the assumption that all \( a_i \) are equal and solve for \( n \) again. </prompt_2.1>
-----

Based on execution_1.2:
But wait, maybe there's another case where two different values are taken, but not in the way I considered. For example, maybe more than two different values?

<prompt_2.2> Consider the possibility of more than two different values and analyze if it's possible. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
Let’s suppose all a_i are equal. Then each a_i = 96/n. Then, the sum of squares would be n*(96/n)^2 = 96² / n. This is given to be 144. So:

96² / n = 144 ⇒ n = 96² / 144.

Calculating that: 96 squared is 9216, so 9216 / 144. Let me divide 9216 by 144. 144*60 = 8640, subtract that from 9216: 9216 - 8640 = 576. Then 144*4 = 576. So total is 60 + 4 = 64. Therefore, n = 64. Then, the sum of cubes would be n*(96/n)^3 = 96³ / n². Let's compute that: 96³ = 96*96*96. Let me compute 96*96 first: 9216. Then 9216*96. Let's do 9216*100 = 921600, subtract 9216*4 = 36864, so 921600 - 36864 = 884736. So 96³ = 884736. Then divide by n² = 64² = 4096. So 884736 / 4096. Let's divide numerator and denominator by 16: 884736 ÷16=55296; 4096 ÷16=256. So 55296 / 256. Divide numerator and denominator by 16 again: 55296 ÷16=3456; 256 ÷16=16. So 3456 /16=216.
</execution_2.1>

<execution_2.2>
But that might complicate things. Let's see.

Alternatively, maybe some of the a_i are 3, and others are different. Wait, but if we have some 3's and others not, then the sum of squares and cubes would be a combination. But as we saw earlier, trying two different values leads to a contradiction unless k=0 or m=0, which isn't allowed. So maybe the only solution is when all a_i are equal, but that led to a problem earlier. Wait, when I assumed all a_i equal, I got n=64 for the sum of squares, but then the sum would be 3*64=192, which is not 96. Wait, no, that was a miscalculation. Wait, if all a_i are equal, then each a_i = 96/n. Then sum of squares is n*(96/n)^2 = 96² / n. We set that equal to 144, so n = 96² / 144 = 9216 / 144 = 64. Then sum of cubes would be n*(96/n)^3 = 96³ / n². Let's compute that: 96³ / 64². 96/64 = 1.5, so 96³ = (1.5*64)^3 = 1.5³ *64³. Then 96³ /64² = 1.5³ *64 = 3.375*64 = 216. So indeed, if all a_i are equal to 96/64 = 1.5, then the sum of cubes would be 64*(1.5)^3 = 64*3.375 = 216. So that works. So in this case, all a_i are 1.5, and n=64. So that's a valid solution.

Wait, so in my earlier attempt, I thought of all a_i being 3, which didn't work, but if all a_i are 1.5, then it works. So that gives n=64.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
So n=64 is a possible value. But the problem says "the sum of all possible values of n". So is 64 the only possible value?

<prompt_3.1> Verify if \( n = 64 \) is the only possible value by considering other potential values for \( n \). </prompt_3.1>
-----

Based on execution_2.2:
But wait, the problem says "sum of all possible values of n". So is 64 the only possible value? Let's check again.

<prompt_3.2> Verify if \( n = 64 \) is the only possible solution by considering other potential values for \( n \). </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
Let me check the sum of cubes again with n=64. Each a_i is 96/64 = 1.5. Then, sum of cubes is 64*(1.5)^3. 1.5 cubed is 3.375. 64*3.375: 64*3 = 192, 64*0.375 = 24, so total 192 +24 = 216.
</execution_3.1>

<execution_3.2>
Suppose there exists another n where the numbers are not all equal, but still satisfy the three equations. But earlier, when I tried assuming two different values, it led to a contradiction. So maybe 64 is the only possible value. But let me verify.

Alternatively, maybe there's another approach. Let's consider the Cauchy-Schwarz inequality. For the sequences (a_i) and (1), we have:

(Σa_i)^2 ≤ n Σa_i².

Which gives 96² ≤ n*144 ⇒ n ≥ 96² /144 = 64. So n must be at least 64. But in our earlier case, n=64 is possible when all a_i are equal. So if n=64 is possible, and the inequality shows that n cannot be less than 64, but can it be more than 64?

Wait, the problem states that the numbers are positive real numbers. So if n >64, can we have such numbers?
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.1:
But wait, maybe there are other values of n? Let me think.

Suppose not all a_i are equal. Then, perhaps some of them are different. But how would that affect the sums?

<prompt_4.1> Consider if there are other possible values for \( n \) by assuming the sequence is not constant. </prompt_4.1>
-----

Based on execution_3.2:
Let's suppose n=96. Let's see if it's possible. Let's assume n=96. Then, the average of the a_i's is 96/96=1. The sum of squares would be 144, so average square is 144/96=1.5. The sum of cubes is 216, so average cube is 216/96=2.25.

<prompt_4.2> Check if \( n = 96 \) is possible by considering the average, sum of squares, and sum of cubes. </prompt_4.2>

Similarly, check n=144. Then average a_i=96/144=2/3≈0.666. Sum of squares=144, so average square=144/144=1. Sum of cubes=216, average cube=216/144=1.5.

<prompt_4.3> Check if \( n = 144 \) is possible by considering the average, sum of squares, and sum of cubes. </prompt_4.3>

Wait, but how do we know that these values of n are possible? For example, for n=96, can we actually construct a sequence of 96 positive real numbers with sum 96, sum of squares 144, and sum of cubes 216?

<prompt_4.4> Verify the possibility of \( n = 96 \) and \( n = 144 \) by attempting to construct sequences that satisfy the given conditions. </prompt_4.4>
</Plan_4>

<Execution_4>
<execution_4.1>
For example, maybe some are higher and some are lower. But how would that affect the sums?

Wait, but let's think about the relationship between the sums. If all a_i are equal, we have a unique solution for n. If they are not all equal, maybe there are other solutions. But according to the earlier analysis when trying to set two different values, it leads to a contradiction unless all are equal. So perhaps n=64 is the only possible value.
</execution_4.1>

<execution_4.2>
Let me see if such numbers can exist. Let's denote the numbers as a_i. Let's use the Cauchy-Schwarz inequality again. If the average is 1, and the sum of squares is 1.5, then the variance would be E[X²] - (E[X])² = 1.5 -1=0.5. So variance is positive, meaning the numbers are spread out around 1. But then the sum of cubes is 2.25. Let's see if this is possible. Let's consider if all numbers are equal to 1 except some. Suppose most are 1, and a few are different. Let me try a simple case: suppose one number is different. Let’s say 95 numbers are 1, and one number is x. Then:

Sum: 95*1 + x = 96 ⇒ x = 1.

Sum of squares: 95*1 + x² = 95 + x² = 144 ⇒ x² = 49 ⇒ x=7 (since positive). But then x=7, so sum would be 95*1 +7=102≠96. Contradiction. So that doesn't work. Alternatively, maybe two numbers are different. Let’s say 94 numbers are 1, and two numbers are x and y. Then:

Sum: 94 + x + y =96 ⇒ x + y =2.

Sum of squares: 94*1 +x² + y² =144 ⇒x² + y²=50.

Sum of cubes:94*1 +x³ + y³=216 ⇒x³ + y³=122.

So we have:

x + y =2,

x² + y²=50,

x³ + y³=122.

From x + y =2, we have (x + y)^2 =4= x² + y² +2xy=50 +2xy ⇒ 4=50 +2xy ⇒2xy= -46 ⇒xy= -23.

But x and y are positive real numbers, so their product can't be negative. Hence, impossible. Therefore, n=96 is not possible.

Similarly, trying n=144. Let's see. Let me check if n=144 works. Let's assume 143 numbers are 2/3 and one number is different. Wait, sum would be 143*(2/3) + x =96. Let's compute 143*(2/3)= (143*2)/3≈286/3≈95.333. So x≈96 -95.333≈0.666. Then sum of squares:143*(4/9) + x²≈143*(0.444) +0.444≈63.333 +0.444≈63.777, which is less than 144. So that's not enough. So maybe this approach isn't working. Alternatively, perhaps using Lagrange multipliers or some optimization. Let me think.

Alternatively, consider that for fixed sum and sum of squares, the sum of cubes is maximized or minimized when the variables are equal or as spread out as possible. Wait, but in our earlier case, when all variables are equal, we get the sum of cubes as 216, which matches. So if we have variables unequal, the sum of cubes might be different. But in the problem, it's given as 216, so maybe the only solution is when all variables are equal. Hence, n=64 is the only possible value.
</execution_4.2>

<execution_4.3>
Wait, but how does that relate to each other? Maybe there's a relationship between the averages. Let's see. For n=64, average is 1.5, average square is 2.25, average cube is 3.375. Which is exactly 1.5, 2.25, 3.375, each term being (3/2) times the previous. But for n=96, average is 1, average square is 1.5, average cube is 2.25. So each term is 1.5 times the previous. Similarly, for n=144, average is 2/3, average square is 1, average cube is 1.5. Each term is 1.5 times the previous. So maybe there's a pattern here.

But how can we construct such sequences? Let's think. If we have a sequence where all elements are equal, then the ratios between the averages would be consistent. But if we have a mixture, maybe we can get different ratios. However, in the case where all elements are equal, the ratios are fixed. If we have some elements different, perhaps the ratios can vary. But in the cases above, even with different n's, the ratios between the averages are consistent. For example, in n=96, the average, average square, average cube are 1, 1.5, 2.25, which is 1, 1.5, 2.25 – each term is 1.5 times the previous. Similarly for n=144: 2/3, 1, 1.5 – again each term is 1.5 times the previous. So maybe these are the only possible solutions where all elements are scaled versions of each other, but with different numbers of elements. Wait, but how does that work?

Wait, perhaps the key is that the ratios between the sums are consistent. Let's denote that the sum of a_i is S1, sum of a_i² is S2, sum of a_i³ is S3. Then, if we have S1 = k * m, S2 = k * m², S3 = k * m³ for some k and m. Then, S1 = k*m, S2 = k*m², S3 = k*m³. Then, S2 = (S1)^2 / k, but S3 = S1^3 / k². Let's see:

From S1 = k*m, so k = S1 / m. Then S2 = k*m² = (S1 / m)*m² = S1*m. Therefore, m = S2 / S1. Similarly, S3 = k*m³ = (S1 / m)*m³ = S1*m². But since m = S2 / S1, then S3 = S1*(S2 / S1)^2 = S1*(S2² / S1²) = S2² / S1. Let's check with the given numbers.

Given S1=96, S2=144, S3=216. Let's compute S2² / S1: 144² / 96 = (20736)/96 = 216. Which matches S3=216. Therefore, this condition holds. Therefore, the only way this can happen is if all the a_i are equal. Because if the ratios hold such that S2 = (S1)^2 / k and S3 = S1^3 / k², which is the case when all a_i are equal (since each a_i = k, so sum is n*k, sum of squares is n*k², sum of cubes is n*k³). Therefore, in this case, the only solution is when all a_i are equal. Therefore, n must be 64, as calculated earlier.

Wait, but earlier I thought that maybe different n's could work, but according to this, since the ratios hold only when all a_i are equal, then n must be 64. Therefore, the sum of all possible values of n is 64.
</execution_4.3>

<execution_4.4>
Let me check the variance. The variance would be (sum of squares)/n - (mean)^2 = 144/96 - (1)^2 = 1.5 -1 = 0.5. So variance is 0.5. Since variance is positive, the numbers are not all the same. But is such a sequence possible? Similarly, for n=144, variance would be 144/144 - (2/3)^2 = 1 - 4/9 = 5/9 ≈0.555...
</execution_4.4>
</Execution_4>

<Plan_5>
Based on execution_4.1:
So, is there a way to have a different n?

Wait, earlier when I tried assuming two different values, it led to a contradiction, but maybe there's another way. Let me think. Suppose there are k numbers equal to some value a, and the remaining (n - k) numbers equal to another value b. Then, we have:

<prompt_5.1> Re-evaluate the possibility of different values for \( n \) using the given equations. </prompt_5.1>
-----

Based on execution_4.4:
But how do we know if such sequences exist? For example, in the case of n=96, with mean 1, variance 0.5. Let's see.

<prompt_5.2> Verify the possibility of sequences with \( n = 96 \) and \( n = 144 \) using variance and other methods. </prompt_5.2>
</Plan_5>

<Execution_5>
<execution_5.1>
1. ka + (n - k)b = 96,

2. ka² + (n - k)b² = 144,

3. ka³ + (n - k)b³ = 216.

Let me try this with k=1 and n=96. Then, we have:

a + 95b = 96,

a² + 95b² = 144,

a³ + 95b³ = 216.

Let me solve the first equation for a: a = 96 - 95b. Substitute into the second equation:

(96 - 95b)^2 + 95b² = 144.

Expanding (96 - 95b)^2: 96² - 2*96*95b + (95b)^2 = 9216 - 18,  96*95*2=18,800b? Wait, 96*95=9120, so 2*9120=18,240. So:

9216 - 18,240b + 9025b² + 95b² = 144.

Combine like terms:

9216 - 18,240b + 9025b² + 95b² = 9216 - 18,240b + 9120b² = 144.

So, 9120b² - 18,240b + 9216 = 144.

Subtract 144:

9120b² - 18,240b + 9072 = 0.

Divide all terms by 48 to simplify:

9120/48 = 190, 18,240/48=380, 9072/48=189. So:

190b² - 380b + 189 = 0.

Divide by 190:

b² - 2b + 189/190 = 0.

Compute discriminant: D = 4 - 4*1*189/190 = 4 - 756/190 = 4 - 3.9789 ≈0.0211. So sqrt(D) ≈0.145.

Solutions: b = [2 ±0.145]/2 ⇒ b≈(2.145)/2≈1.0725 or b≈1.855/2≈0.9275.

But since a =96 -95b, let's check if these b values make sense. If b≈1.0725, then a≈96 -95*1.0725≈96 -101.8875≈-5.8875, which is negative. Not allowed. If b≈0.9275, then a≈96 -95*0.9275≈96 -88.1125≈7.8875. Positive. So possible. Then check the third equation:

a³ +95b³ ≈ (7.8875)^3 +95*(0.9275)^3.

Compute 7.8875³: approx 7^3=343, 8^3=512, 7.8875 is close to 8. Let's compute:

7.8875 *7.8875 = approx 62.2187. Then multiply by 7.8875: 62.2187*7.8875≈491.5.

Similarly, 0.9275³≈0.8. So 95*0.8≈76.

Total≈491.5 +76≈567.5≠216. So that's way off. Hence, this case doesn't work. Therefore, n=96 is not possible with k=1. Maybe try another k? But this is getting too time-consuming. Maybe there's a better approach.
</execution_5.1>

<execution_5.2>
Let’s suppose we have 96 numbers. Let’s say m numbers are 1 + sqrt(0.5) and (96 - m) numbers are 1 - sqrt(0.5). But since the numbers must be positive, we need 1 - sqrt(0.5) > 0. sqrt(0.5) ≈0.707, so 1 - 0.707 ≈0.293 >0. So possible. Then, the sum would be m*(1 + sqrt(0.5)) + (96 - m)*(1 - sqrt(0.5)) =96 + m*(2*sqrt(0.5)) =96 + m*sqrt(2). But we need the sum to be 96, so m*sqrt(2)=0 ⇒m=0. Therefore, only possible if all numbers are 1. Hence, variance can't be achieved otherwise. Therefore, the only way to have variance 0.5 is if all numbers are equal. Hence, n=64 is the only solution.

Wait, but this contradicts the earlier idea that n could be 96 or 144. So maybe my previous reasoning was flawed. Let me clarify.

The key is that when we have the relation S3 = S2² / S1, as in this problem where 216 = 144² / 96, this implies that the sequence is such that all elements are equal. Because this identity holds only when all a_i are equal. If there's any variation among the a_i, then the sum of cubes would not satisfy this exact relation. Therefore, the only possible n is 64.
</execution_5.2>
</Execution_5>

<Plan_6>
Based on execution_5.2:
**Final Answer**
The sum of all possible values of \( n \) is \boxed{64}.
</Plan_6>

<Execution_6>
No prompt to execute.
</Execution_6>

<Final_answer>
Therefore, the sum of all possible values of \( n \) is \boxed{64}.
</Final_answer>
</think>