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November 14th, 2006, 01:04 AM   #1
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4ab-a-b is a square

Seen on another forum:

Find all positive integers a,b>0 such that 4ab-a-b is a perfect square.

Enjoy!
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November 14th, 2006, 06:26 AM   #2
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After solving this problem, I realized that it should have been placed in the College Number Theory section. Anyhow, too late. I copy-paste my solution in Latex below. Since Latex is not installed on this forum, you will have to do with the code. The solution for this problem will be posted along with my archives of 250+ interesting (college-level) problems on this site in a short while (probably in a few days).

Write $4ab-a-b=u^{2}$. Then $a=\frac{u^{2}+b}{4b-1}$, meaning that $4b-1$ divides $4u^{2}+1$. If such is the
case, then $4u^{2}+1$ must have at least one prime divisor which equals $3$ modulo $4$. If we can show that this
is absurd, i.e that all prime divisors of $(2u)^{2}+1$ have the form $4k+1$, then we will be done.\\
By Euler's criterion, $-1$ is a quadratic residue modulo $p$, where $p$ is an odd prime, if and only if
$(-1)^{\frac{p-1}{2}}=1 \ mod \ p$. This implies that $p$ must equal $3$ modulo $4$. In particular, $(2u)^{2}+1$
has only prime factors of the form $4k+1$, which solves the problem.
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November 14th, 2006, 08:00 AM   #3
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Euler first noted that this expression, along with others such as 8ab - 3a - 3b, is never a perfect square for positive a and b. It is also never a triangular number.
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