TUYMAADA 2012 S4

Let $p=4k+3$ be a prime. Prove that if $$\dfrac {1} {0^2+1}+\dfrac{1}{1^2+1}+\cdots+\dfrac{1}{(p-1)^2+1}=\dfrac{m} {n}$$ (where the fraction $\dfrac {m} {n}$ is in reduced terms), then $p \mid 2m-n$ .

Proposed by A. Golovanov