a) $(m,n)$= $(3q^2 , 1)$ b)http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=351573&hilit=jacobi

My soln for $(a)$ is the same as mahanmath ie the tuple $\{m,n,p\}=\{3q^2,1,3q\}$ always satisfies the 1st equation. for $(b)$ note that we get $m=\frac{p^2+n}{4n-1}$. Hence $4n-1|p^2+n => 4n-1|4(p^2+n)-4n+1=>4n-1|4p^2+1$ which is impossible as every factor of $x^2+1$ is of the form $4k+1$ ( when $x$ is even)

Set $4mn-m-n=x^2$ if we multiply by $4$ and than add $1$ we get $(4m-1)(4n-1)=4x^2+1$ now because $4m-1 \equiv 3(mod4) $ there must be prime factor $p$ of $4m-1$ such that $p\equiv 3(mod4) $ now we get $p|4x^2+1$ which now by well known theorem implies $p|(2x,1)$ which is $p|1$ which is obvious contradiction.

$m=\frac{t(t–1)}{2}$,$n=\frac{t(t+1)}{2}$ If we set this we see 1) have infinitely many solutions.

https://artofproblemsolving.com/community/c3046h1833823_representation_of_a_square_of_a_certain_shape

Wolowizard wrote: Set $4mn-m-n=x^2$ if we multiply by $4$ and than add $1$ we get $(4m-1)(4n-1)=4x^2+1$ now because $4m-1 \equiv 3(mod4) $ there must be prime factor $p$ of $4m-1$ such that $p\equiv 3(mod4) $ now we get $p|4x^2+1$ which now by well known theorem implies $p|(2x,1)$ which is $p|1$ which is obvious contradiction. could anyone pls explain that theorem