TITLE: When is the product $(1+1)(1+4)…(1+n^2)$ a perfect square?
QUESTION [22 upvotes]: 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.

REPLY [37 votes]: Javier Cilleruelo has shown that $n=3$ is the only solution: see http://www.uam.es/personal_pdi/ciencias/cillerue/Papers/squares-sinlogo.pdf .  
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 http://www.renyi.hu/~p_erdos/1952-07.pdf .