What integer for A (a^2+110a+18) will yield a perfect square?

Question

I'm interested in a way to figure out when this equation would yield a perfect square other than to try one integer after another in a test sort of fashion. Does anyone know a method for finding out A where it will yield a perfect square? Formulas, identities, methods, etc.

Answer 1

a² + 110 a + 18 is a perfect square when a = 9, yielding 33². It's not always possible to form perfect integer squares with quadratic functions of integers.

How to do this? Let the quadratic equation = k², where k is an integer, and use the expression for the roots of the expression. The radical must result in an integer. It will be in the form √(m - k²), where m will be some integer based on the coefficients, and this must equal to another integer n. Rearranging gets you:

m = k² - n² = (k + n)(k - n)

Break down m into 2 factors, even if m is a prime (m = m * 1), and solve for k and n. If factors are both odd or both even, then k and n will be integers, and you'll have a solution. If one factor is even and the other odd, then k and n will be half integers, which won't make for a solution. For example, if m = 6, there is no way to factorize it into odd-only or even-only pairs, so no solution would be possible.

Hey, Benoit's got a good answer, is he showing me up? Well, anyway, something to chew on. I see that Benoit's answer is the special case where 3007 is broken down to 3007 * 1, but it can also be broken down to 97 * 31.

Answer 2

If it's a square, then
a² + 110a + 18 = n² for some n.
→ a² + 110a + 18 - n² = 0

Quadratic formula:
a = [-110 ± sqrt(110² - 4(18-n²))] / 2
= -55 ± sqrt(3007 + n²)
→ 3007 + n² is a square.
→ 3007 + n² = k² for some integer k
→ 3007 = k² - n² = (k-n)(k+n)

Setting (k-n) and (k+n) as pairs of factors of 3007 and solving for k and n gives us...
Factors 1 and 3007 → k=±1504, n=±1503
Factors 31 and 97 → k=±64, n=±33

Plugging n=1503 and n=33 in the quadratic result
a = -55 ± sqrt(3007 + n²) we get the four solutions:

1) a = 1449 → 449² + 110*1449 + 18 = 2,259,009 = 1503²
2) a = -1559 → (-1559)² - 110*1559 + 18 = 1503²
3) a = 9 → 9² + 110*9 + 18 = 1089 = 33²
4) a = -119 → (-119)² - 110*119 + 18 = 1089 = 33²