Okay, let's see. I need to find the smallest positive integer value of 'a' such that the sum of the geometric series 4 + 12/a + 36/a² + ... is a perfect square. Hmm, geometric series, right. Let me recall how geometric series work.

First, a geometric series has the form S = a₁ + a₁*r + a₁*r² + a₁*r³ + ..., where a₁ is the first term and r is the common ratio. The sum of an infinite geometric series is S = a₁ / (1 - r), provided that |r| < 1. So, in this problem, the first term is 4, and the common ratio seems to be the factor between consecutive terms. Let me check that.

The first term is 4. The second term is 12/a. So, the ratio r would be (12/a) divided by 4, which is (12/a)/4 = 3/a. Then the third term is 36/a², which divided by the second term (12/a) gives (36/a²)/(12/a) = 3/a. So yes, the common ratio r is 3/a. That makes sense.

So, the sum S of this infinite geometric series is S = 4 / (1 - r) = 4 / (1 - 3/a). We need this sum to be a perfect square. So, 4 / (1 - 3/a) must be a perfect square. Let me write that down:

S = 4 / (1 - 3/a) = 4 / [(a - 3)/a] = 4a / (a - 3)

So, S = 4a / (a - 3). We need this to be a perfect square. Let me denote S = k², where k is a positive integer. Therefore:

4a / (a - 3) = k²

We need to solve for a, which is a positive integer. Let's rearrange this equation:

4a = k²(a - 3)

Expanding the right side:

4a = k²*a - 3k²

Bring all terms to one side:

4a - k²*a + 3k² = 0

Factor out 'a' from the first two terms:

a*(4 - k²) + 3k² = 0

Then:

a*(k² - 4) = 3k²

Therefore:

a = 3k² / (k² - 4)

So, a must be a positive integer. Therefore, (k² - 4) must divide 3k². Let's write that as:

a = 3k² / (k² - 4)

Since a must be a positive integer, the denominator (k² - 4) must be a positive divisor of 3k². Also, since the denominator is positive, k² - 4 > 0, so k² > 4. Therefore, k must be an integer greater than 2 (since k is positive). So, k ≥ 3.

Our goal is to find the smallest positive integer a. Therefore, we need to find the smallest k ≥ 3 such that (k² - 4) divides 3k², and then compute a accordingly.

Alternatively, we can write:

a = 3k² / (k² - 4) = 3 + 12 / (k² - 4)

How did I get that? Let me check:

3k² / (k² - 4) = [3(k² - 4) + 12] / (k² - 4) = 3 + 12 / (k² - 4)

Yes, that's right. So, a = 3 + 12 / (k² - 4). Since a must be an integer, 12 / (k² - 4) must be an integer. Therefore, (k² - 4) must be a positive divisor of 12. Let's list the positive divisors of 12:

1, 2, 3, 4, 6, 12

Therefore, (k² - 4) must be one of these numbers. So, k² - 4 ∈ {1, 2, 3, 4, 6, 12}

Therefore, k² ∈ {5, 6, 7, 8, 10, 16}

But k² must be a perfect square, right? Because k is an integer. So, looking at the numbers 5, 6, 7, 8, 10, 16, which of these are perfect squares?

16 is a perfect square (4²). The others: 5,6,7,8,10 are not perfect squares. Wait, but k² must be such that (k² - 4) is a divisor of 12, and k² must be a perfect square. Therefore, k² - 4 must be a divisor of 12, and k² must be a perfect square. So, k² - 4 must be in {1, 2, 3, 4, 6, 12}, and k² must be a perfect square.

Therefore, let's check which of the numbers 5, 6, 7, 8, 10, 16 are perfect squares. 16 is 4², which is a perfect square. The others are not. So, only k² = 16 is possible. Then, k² - 4 = 12, which is a divisor of 12. Therefore, k = 4.

So, k = 4. Then, a = 3 + 12 / (16 - 4) = 3 + 12 / 12 = 3 + 1 = 4. So, a = 4. But wait, let's check if that's correct.

Wait, but if k = 4, then S = k² = 16. Let's compute S using the formula we derived: S = 4a / (a - 3). If a = 4, then S = 4*4 / (4 - 3) = 16 / 1 = 16, which is 4², indeed a perfect square. So, that works. But the problem asks for the smallest possible value of a. So, a = 4. But wait, the question is asking for the smallest positive integer a. But is there a smaller a that works?

Wait, but according to our earlier analysis, k must be ≥ 3, and the only possible k is 4 because k² - 4 must be a divisor of 12 and k² must be a perfect square. But maybe I missed something here. Let's double-check.

We have (k² - 4) divides 12. So, (k² - 4) must be a positive divisor of 12, and k² must be a perfect square. So, the possible (k² - 4) values are 1, 2, 3, 4, 6, 12. Then, k² = (k² - 4) + 4. So, k² = 5, 6, 7, 8, 10, 16. Among these, only 16 is a perfect square. Therefore, only k = 4 is possible. Therefore, a = 4. But wait, maybe there are other divisors? Wait, 12 has divisors 1, 2, 3, 4, 6, 12, but also negative ones? But since (k² - 4) is positive (as k² > 4), only positive divisors are considered. So, only k = 4. Therefore, a = 4.

But wait, let's check if there's another possible value. Suppose k² - 4 = 12. Then k² = 16, so k = 4. Then a = 3 + 12/12 = 4. Alternatively, if k² - 4 = 6, then k² = 10, which is not a square. Similarly, k² - 4 = 4, k² = 8, not square. k² - 4 = 3, k² = 7, not square. k² - 4 = 2, k² = 6, not square. k² - 4 = 1, k² = 5, not square. So, only k = 4 works.

Therefore, a = 4 is the only possible value. But wait, that seems too straightforward. Let me verify again.

Original series: 4 + 12/a + 36/a² + ... The sum is 4 / (1 - 3/a). For the sum to be a perfect square, 4a / (a - 3) must be a perfect square. Then, we set 4a / (a - 3) = k², leading to a = 3k² / (k² - 4). Then, since a must be integer, (k² - 4) must divide 3k². Then, we wrote a as 3 + 12/(k² - 4). Therefore, (k² - 4) must divide 12. So, the possible (k² - 4) is a divisor of 12. Then, as above, only 12 is possible, leading to k = 4. Therefore, a = 4. Therefore, the answer is 4. But wait, the problem says "the smallest possible value of a". But maybe there's a mistake here. Let me test with a = 4.

If a = 4, then the series is 4 + 12/4 + 36/16 + 108/64 + ... Let's compute the sum. The first term is 4, the common ratio is 3/4. So, sum S = 4 / (1 - 3/4) = 4 / (1/4) = 16. 16 is 4², so that's a perfect square. So, a = 4 works. But is there a smaller a?

Wait, suppose a = 1. Then the common ratio is 3/1 = 3, which is greater than 1, so the series diverges. So, sum is infinity. Not acceptable. a = 2. Then the common ratio is 3/2, which is greater than 1, so again the series diverges. a = 3. Then the common ratio is 3/3 = 1, so the series becomes 4 + 4 + 4 + ... which diverges. Therefore, a must be greater than 3 to have |r| < 1. So, a must be at least 4. Therefore, a = 4 is the smallest possible positive integer. Therefore, the answer is 4.

Wait, but let me check if there are other possible k. If k is 5, then k² = 25. Then, (k² - 4) = 21. 21 is not a divisor of 12. So, 12/21 is not an integer, so a would not be integer. Similarly, k = 6, k² = 36, (36 -4) = 32, which is not a divisor of 12. So, 12/32 is not integer. Therefore, no other k values. Therefore, only k = 4. Therefore, a = 4.

Therefore, the smallest possible a is 4. So, the answer is 4. But let me check again. Suppose someone claims a = 12. Then, common ratio is 3/12 = 1/4. Then sum S = 4 / (1 - 1/4) = 4 / (3/4) = 16/3, which is not a perfect square. So, 16/3 is not a perfect square. So, a = 12 doesn't work. Similarly, a = 6. Then ratio is 3/6 = 1/2. Sum S = 4 / (1 - 1/2) = 8, which is 2³, not a perfect square. So, not. a = 5. Then ratio 3/5. Sum S = 4 / (1 - 3/5) = 4 / (2/5) = 10, which is not a perfect square. a = 8. Ratio 3/8. Sum S = 4 / (1 - 3/8) = 4 / (5/8) = 32/5, not a square. a = 9. Ratio 1/3. Sum S = 4 / (1 - 1/3) = 4 / (2/3) = 6, not a square. a = 10. Ratio 3/10. Sum S = 4 / (7/10) = 40/7 ≈ 5.714, not a square.

So, indeed, a = 4 is the only value where the sum is a perfect square. Therefore, the answer is 4. But the problem says "the smallest possible value of a where a is a positive integer". Since a must be greater than 3, the smallest possible is 4. So, 4 is the answer.

Wait, but let me check once more. Suppose someone thinks that maybe k can be 2. Then k² = 4. Then, (k² -4) = 0, which would make a undefined. So, k must be at least 3. So, k = 3. Let's check k = 3. Then, k² = 9, so (k² -4) = 5. 5 divides 12? No, 5 doesn't divide 12. So, 12/5 is 2.4, which is not integer. Therefore, a would be 3*9 / 5 = 27/5, which is not integer. So, no. Then k =4 is the first possible.

Therefore, the answer is 4. But the problem says "smallest possible value of a". So, 4 is the answer. Therefore, the final answer is \boxed{4}.

Wait, but wait! Wait a minute. Let me check again. Because in the initial steps, when we set S = k², and then express a as 3 + 12/(k² -4). So, for a to be integer, (k² -4) must divide 12. So, the possible divisors of 12 are 1,2,3,4,6,12. Then, k² -4 must be one of these. So, k² must be 5,6,7,8,10,16. But k² must be a perfect square. So, among these, only 16 is a perfect square. So, k² =16, so k=4. Then, a=3 +12/12=4. Therefore, only possible a is 4. So, yes, 4 is the answer. Therefore, the answer is \boxed{4}.

But hold on, in the problem statement, the series is written as 4 + 12/a + 36/a² + ... The general term is 4*(3/a)^n, starting from n=0. So, the first term is 4, common ratio is 3/a. So, the sum is 4/(1 - 3/a) = 4a/(a -3). So, that's correct. Then, set 4a/(a -3) = k². Then, solving for a gives a = 3k²/(k² -4). Then, to have a integer, (k² -4) must divide 3k². Then, write a = 3 + 12/(k² -4). Therefore, 12/(k² -4) must be integer. Therefore, (k² -4) divides 12. Then, since k² must be a perfect square, k² -4 must be a divisor of 12 and k² must be a perfect square. So, the possible k² -4 are 1,2,3,4,6,12. Then, k² =5,6,7,8,10,16. Among these, only 16 is a perfect square. Therefore, k=4. Then, a=4. So, 4 is the answer. Therefore, the answer is 4.

But let me check again if there's any other possible k. Suppose k² -4 is 12, then k²=16, k=4. If k² -4 is 6, then k²=10, not square. If k² -4 is 4, k²=8, not square. If k² -4 is 3, k²=7, not square. If k² -4 is 2, k²=6, not square. If k² -4 is 1, k²=5, not square. So, only k² -4=12 gives k=4. Therefore, a=4. Therefore, the answer is 4.

Therefore, I think the answer is 4. So, the smallest possible value of a is 4.

**Final Answer**
The smallest possible value of \( a \) is \boxed{4}.
Therefore, the smallest possible value of $a$ where $a$ is a positive integer is $\boxed{4}$.
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