Given ODD primes $p \neq q:$ and $$ \color{magenta}{p \equiv 3 \pmod 4} $$
Lemma: $$ (-p|q) = (q|p). $$
Lemma: If $$ a^2 + p \equiv 0 \pmod q, $$ THEN $$ (q|p) = 1. $$
Let $$ F_1 = 4 + p, $$
$$ F_2 = 4 F_1^2 + p, $$
$$ F_3 = 4 F_1^2 F_2^2 + p, $$
$$ F_4 = 4 F_1^2 F_2^2 F_3^2 + p, $$
$$ F_5 = 4 F_1^2 F_2^2 F_3^2 F_4^2 + p, $$
and so on.
These are all of the form $a^2 + p$ and are odd, so the only primes than can be factors are quadratic residues for $p.$ Next, all the $F_j$ are prime to $p$ itself. Finally, these are all coprime. So, however they factor, we get an infinite list of primes that are quadratic residues of $p.$
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Given ODD primes $p \neq q:$ and $$ \color{magenta}{p \equiv 1 \pmod 4} $$
Lemma: $$ (p|q) = (q|p). $$
Lemma: If $$ a^2 - p \equiv 0 \pmod q, $$ THEN $$ (q|p) = 1. $$
FIND an even square $$ W = 4^k = \left( 2^k \right)^2 $$ such that
$$ \color{magenta}{ W > p.} $$
Let $$ F_1 = W - p, $$
$$ F_2 = W F_1^2 - p, $$
$$ F_3 = W F_1^2 F_2^2 - p, $$
$$ F_4 = W F_1^2 F_2^2 F_3^2 - p, $$
$$ F_5 = W F_1^2 F_2^2 F_3^2 F_4^2 - p, $$
and so on. As $p \equiv 1 \pmod 4$ and $W \equiv 0 \pmod 4,$ we know $W - p \equiv 3 \pmod 4 $ and so $W-p \geq 3. $ So the $F_j$ are larger than $1$ and strictly increasing.
These are all of the form $a^2 - p$ and are odd, so the only primes than can be factors are quadratic residues for $p.$ Next, all the $F_j$ are prime to $p$ itself. Finally, these are all coprime. So, however they factor, we get an infinite list of primes that are quadratic residues of $p.$
If $p=2$, both products are $1$, so we assume that $p$ is odd. By rules of exponents, the product of the quadratic residues is $$R=g^{(p-1)(p+1)/4}$$ and the product of the quadratic non-residues is $$N=g^{(p-1)(p-1)/4}.$$ Note that the product of all the residues is $-1$. (Each element $x$ cancels with its inverse unless $x$ is its own inverse, and $x^2=1$ has only the solutions $x=\pm 1$.)
If $n\equiv1\pmod4$, then $\frac{p-1}{4}$ is an integer $k$, so that $N=\left(g^{p-1}\right)^k=1$ and therefore, $R=-1$. Similarly, if $n\equiv3\pmod4$, then $\frac{p+1}{4}$ is an integer, so $R=1$ and $N=-1$.
Best Answer
First of all, it is enough to show the existence when $x=1$, since given $1=a+b$ where $a$ is a quadratic residue and $b$ is not, $x=ax+bx$ must be such that one of the $ax,bx$ is a quadratic residue and the other isn't.
When $p=5$ the existence can be checked manually: $1=2+4$. Suppose for $p\equiv1\pmod4$ and $p>5$ that there exists no quadratic residue $a$ and non-quadratic residue $b$ such that $a+b=1$.
Let $1<a<p-1$ be such that $\bar a\in\mathbb F_p^\times$ is a quadratic residue. Such an integer exists, since there are at least $(p-1)/2>2$ quadratic residues.
Then $\bar a+1$ must also be a quadratic residue, since otherwise $(\bar a+1)+(-\bar a)=1$ would be a counter-example.
The same argument shows that $\bar a-1$ must also be a quadratic residue since $(1-\bar a)$.
Thus, inductively, all $1\le a\le p-1$ must be a quadratic residue, a contradiction.