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I was wondering if for some natural numbers $p,q \in \mathbb{N}$ the expression $$pq(p+q)(p-q)$$ is a perfect square?

What I tried, is that without loss of generality, we can assume $(p,q)=1$ else we can factorise $(p,q)^2$.

Now if the power of a prime $r$ in $p$ or $q$ is odd, then $r$ divides either $p + q$ or $p-q$ which implies $r$ divides both $p$ and $q$. Thus we must also have $p,q$ are perfect squares.

I encountered this while trying to prove that $1$ is not a congruent number. I have been trying this for a while but without much progress.

Gary
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1 Answers1

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As you've noted, we can assume $p, q$ are perfect squares. So, in order for $pq(p+q)(p-q)$ to be a perfect square, we need $(p+q)(p-q) = p^{2} - q^{2}$ to be a perfect square. This would mean $p, q$ are part of a Pythagorean triple, where $p, q$ are perfect squares. However, as shown in the answers to the question Pythagorean triples and perfect squares, a Pythagorean triple can have at most one perfect square in it. So this is impossible.

ckefa
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