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This is a modification of an unanswered problem on the math StackExchange.

When is the product $(1+1)(1+4)…(1+n^2)$ a perfect square?

If $(1+1)(1+4)…(1+n^2)=k^2$ then one possibility is $n=3$, $k=10$. Could there be other integer solutions for $(n,k)$?

Surely the answer is NO, but I am unable to prove that $n=3$, $k=10$ is the only possibility.

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up vote 30 down vote accepted

Javier Cilleruelo has shown that $n=3$ is the only solution: see .

A couple more comments: apparently Chebyshev already showed that the largest prime factor of $\prod_{j=1}^{n} (1+j^2)$ is bigger than $Cn$ for any constant $C$ provided that $n$ is large. This would solve the problem for all large $n$ (taking $C=2$ in Chebyshev's work). The problem of finding the largest prime factor of $\prod_{j=1}^{n} f(j)$ for a polynomial $f$ has been studied by many authors (Erdos, Hooley, Heath-Brown ...); for a first improvement of Chebyshev's work see Erdos's paper .

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In fact, the first to improve Chebushev's result was Nagell (reference [4] from Cilleruelo paper), who showed that the largest prime factor of the product $\prod f(j)$ is $\ge n\log n.$ Ciruello's proof goes along Nigell's one by making the estimate precise and using computer search for small $n.$ – Alvin Dec 27 '13 at 15:33
Why does the fact that the largest prime factor exceeds 2n shows the product is not a perfect square? Presumably it should be pretty obvious but I just don't see it. – John Jiang Dec 28 '13 at 0:04
@JohnJiang: If $p$ divides $a^2+1$ and $b^2+1$ then $p$ divides $|a-b|$ or $(a+b)$. If $p>2n$ and $a$ and $b$ are distinct numbers below $n$, this is impossible. So a prime larger than $2n$ that divides the product, only divides to exponent $1$. – Lucia Dec 28 '13 at 0:10

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