, 2015

Solution:
Answer: $225$

By the Chinese remainder theorem, we want the product of the answers modulo $5$ and modulo $7$ (i.e. when $35$ is replaced by $5$ and $7$, respectively).

First we do the modulo $7$ case. Since $x^2+1$ is irreducible modulo $7$ (or more conceptually, in $\mathbb{F}_7[x]$), exactly half of the nonzero residues modulo $x^2+1$ and $7$ (or just modulo $x^2+\overline{1}$ if we're working in $\mathbb{F}_7[x]$) are quadratic residues, i.e. our answer is $1+\frac{7^2-1}{2}=25$ (where we add back one for the zero polynomial).

Now we do the modulo $5$ case. Since $x^2+1$ factors as $(x+2)(x-2)$ modulo $5$ (or more conceptually, in $\mathbb{F}_5[x]$), by the polynomial Chinese remainder theorem modulo $x^2+\overline{1}$ (working in $\mathbb{F}_5[x]$), we want the product of the number of polynomial quadratic residues modulo $x \pm \overline{2}$. By centering/evaluating polynomials at $\mp \overline{2}$ accordingly, the polynomial squares modulo these linear polynomials are just those reducing to integer squares modulo $5$. So we have an answer of $\left(1+\frac{5-1}{2}\right)^2=9$ in this case.

Our final answer is thus $25 \cdot 9=225$.