nice problem, nice solution. thanks...

This time the official solution is simpler; it runs along these lines. Denote $a_\ell = \ell^2 + 1$. Then $\dfrac{m} {n} = \dfrac {\sigma_{p-1}(a_0,a_1,\ldots,a_{p-1})} {\sigma_{p}(a_0,a_1,\ldots,a_{p-1})}$, where $\sigma_{j}(a_0,a_1,\ldots,a_{p-1})$ is the elementary symmetric polynomial of degree $j$. The polynomial of roots $a_\ell$ is $\displaystyle P(x) = \prod_{\ell=0}^{p-1} (x - 1 - \ell^2)$; with the substitution $t^2 = x-1$ we have $\displaystyle P(t^2+1) = \prod_{\ell=0}^{p-1} (t^2 - \ell^2) = \left ( \prod_{\ell=0}^{p-1} (t - \ell) \right )\left ( \prod_{\ell=0}^{p-1} (t + \ell) \right )$, which seen in $\mathbb{F}_p[t]$ is $(t^p-t)^2 = t^{2p} - 2t^{p+1} + t^2$. Performing the reverse substitution, $P(x) \equiv (x-1)^p - 2(x-1)^{(p+1)/2} + (x-1) = x^p + \cdots + (p + 2(p+1)/2 + 1)x -4$, thus by Viète's relation getting $\sigma_{p-1} \equiv 2 \pmod{p}$ and $\sigma_{p} \equiv 4 \pmod{p}$, therefore $p\mid 2\sigma_{p-1} - \sigma_p$. Since $p\nmid \sigma_p$, that is equivalent to $p\mid 2m - n$.

The numbers from $2$ to $p-2$ can be arranged in pairs of the form $\left (x,\frac{1}{x}\right ) \pmod{p}$ (the numbers in the pair are different, as the only solutions to $x\equiv \frac{1}{x} \pmod{p}$ are $1$ and $-1$). For any pair defined as above we have $\frac{1}{x^2+1}+\frac{1}{(1/x)^2+1} \equiv 1 \pmod{p}$ so we have that $\frac{1}{2^2+1}+\cdots +\frac{1}{(p-2)^2+1}\equiv \frac{p-3}{2} \pmod{p}$, so our sum is $\frac{p-3}{2}+1+2\cdot\frac{1}{2}=\frac{p+1}{2}$ modulo $p$, and we get the conclusion.

This link http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=490409 points to a unification with Problem 4, Day 1, Junior League.

Drytime wrote: The numbers from $2$ to $p-2$ can be arranged in pairs of the form $\left (x,\frac{1}{x}\right ) \pmod{p}$ (the numbers in the pair are different, as the only solutions to $x\equiv \frac{1}{x} \pmod{p}$ are $1$ and $-1$). For any pair defined as above we have $\frac{1}{x^2+1}+\frac{1}{(1/x)^2+1} \equiv 1 \pmod{p}$ so we have that $\frac{1}{2^2+1}+\cdots +\frac{1}{(p-2)^2+1}\equiv \frac{p-3}{2} \pmod{p}$, so our sum is $\frac{p-3}{2}+1+2\cdot\frac{1}{2}=\frac{p+1}{2}$ modulo $p$, and we get the conclusion. Very nice and simple solution DRYTIME! Congratulations !!!

Re-write it as (looking the equation in $\mathbb{F}_p$) $$2 \cdot \left(\sum_{\left(\frac{x}{p}\right)} \frac{1}{1+x}\right)=\frac{m-n}{n} (\bigstar)$$Quadratic residues are roots of polynomial $P(X)=x^{\frac{p-1}{2}}-1$ in $\mathbb{F}_p[X]$. $\frac{P'(-1)}{P(-1)}=(-1) \sum_{\left(\frac{x}{p}\right)} \frac{1}{1+x}=\frac{1}{2}$ $$(\bigstar) \implies 1=\frac{m-n}{n}$$from where we finish.

Not too hard. Let $g$ be the primitive root modulo $p$. Thus, \[ \sum_{i = 0}^{p - 1} \frac{1}{i^2 + 1} \equiv 1 + 2 \left( \sum_{k = 0}^{\frac{p - 1}{2}} \frac{1}{g^{2k} + 1} \right) \ (\text{mod} \ p) \]Now, notice that $\frac{1}{g^a + 1} + \frac{1}{g^{p - 1 - a} + 1} \equiv \frac{1}{g^a + 1} + \frac{1}{g^{-a} + 1} \equiv 1 \ (\text{mod} \ p)$. Thus, we have \[ \frac{m}{n} \equiv 1 + 2 \left( \sum_{k = 0}^{\frac{p - 1}{2}} \frac{1}{g^{2k} + 1} \right) \equiv 1 + \sum_{k = 0}^{\frac{p - 1}{2}} \left( \frac{1}{g^{2k} + 1} + \frac{1}{g^{p - 1 - 2k} + 1} \right) \equiv 1 + \sum_{k = 0}^{\frac{p - 1}{2}} 1 \equiv \frac{p + 1}{2} \]Thus, we have $2m \equiv (p + 1)n \equiv n \ (\text{mod} \ p)$, finishing the proof. $\textbf{Remark.}$ This argument won't work for $p \equiv 1 \ (\text{mod} \ 4)$ as we know that $p \mid g^{\frac{p - 1}{2}} + 1$, in this case which inverse doesn't exist. However, suppose $g^{2k} \equiv -1 \ (\text{mod} \ p)$ for some $p \equiv 3 \ (\text{mod} \ 4)$, then $p \mid (g^k)^2 + (1)^2$, and it is well known that there are no primes of the form $4k + 3$ which divides $x^2 + 1$.

Solved with Elliott Liu Observe that if the LHS was $\frac{1}{2}\mod p$, we would be done, since then \[\frac{2m-n}{n} \equiv 2\cdot\frac{m}{n} - 1 \equiv 1 - 1\equiv 0\mod p\]which implies $p \mid 2m-n$. Now, we have the identity \[(k^{2} + 1)(k^{p-3} - k^{p-5} + \ldots - k^{2} + 1) = k^{p-1} + 1 \equiv 2\mod p\]for $k\neq 0$. Finally, we get the LHS is \[1 + \sum_{k=1}^{p-1} \frac{1}{k^{2} + 1}\equiv 1 + \frac{1}{2}\sum_{k=1}^{p-1}k^{p-3} - k^{p-5} + \ldots - k^{2} + 1 \equiv 1 + \frac{p-1}{2}\equiv \frac{1}{2}\mod p\]

All the statements are considered in $[F_p]$: Consider the pairs $(x,\frac{1}{x})$ and clearly $x^2 \equiv 1 \pmod p$ has no solutions in the interval $[2,.......,p-2]$ so all such pairs are distinct. Observe that $\frac{1}{x^2+1}+\frac{1}{(1/x)^2+1} \equiv 1 \pmod{p}$ so $\frac{1}{2^2+1}+\cdots +\frac{1}{(p-2)^2+1}\equiv \frac{p-3}{2} \pmod{p}$, hence $\frac{p-3}{2}+1+2\cdot\frac{1}{2}=\frac{p+1}{2}$,and $2(p+1)-2=2-2=0$

A similar sol post for storage: $2*LHS-1=1+2\sum\limits_{i=1}^{p-1}\frac{1}{i^2+1}\equiv1+\sum\limits_{i=1}^{p-1}(\frac{1}{i^2+1}+\frac{1}{(\frac{1}{i})^2+1})\equiv 1+p-1\equiv 0(mod\ p)$ Then $\frac{2m-n}{n}\equiv 0(mod\ p)$ Done $\square$