Quadratic residue

Show that there are infinitely many pairwise coprime integers $d$ for which there is at least one integer $c$ so that $a^2 + b^2 \equiv c \pmod d$ has no integer solutions.

I see that c must be from $\{0,1,\dots,d-1\}$ and $a^2, b^2$ from $Q_{d}$ (quadratic residue modulo d) under modulo d.

So I need to find such d that any sum of two numbers from $Q_{d}$ won't cover at least one number from $\{0,1,\dots,d-1\}$. However I am stuck at this point.

Solution 1:

It suffices to show that if $p\equiv 3\pmod 4$ is prime, then $$ a^2+b^2\equiv p\pmod{p^2} \tag{1} $$ does not have integer solutions. To this end, notice that (1) implies $$ a^2\equiv -b^2\pmod p \tag{2}, $$ whence $$ a\equiv b\equiv 0\pmod p \tag{3}; $$ for if we had, for instance, $b\not\equiv 0\pmod p$, then (2) would lead to $(a/b)^2\equiv-1\pmod p$, while $-1$ is a quadratic non-residue mod $p$. Finally, (3) leads to $a^2+b^2\equiv 0\pmod{p^2}$, contradicting (1).