TUYMAADA 2012 J4

Let $p=1601$ . Prove that if $$\dfrac {1} {0^2+1}+\dfrac{1}{1^2+1}+\cdots+\dfrac{1}{(p-1)^2+1}=\dfrac{m} {n},$$ where we only sum over terms with denominators not divisible by $p$ (and the fraction $\dfrac {m} {n}$ is in reduced terms) then $p \mid 2m+n$ .

Proposed by A. Golovanov