Show that
$$p \equiv 3\pmod 4~\land~\pmatrix{\frac{a}{p}}=-1~\rightarrow~ \pmatrix{\frac{a-1}{p}}=1$$
where p is a prime number and $\pmatrix{\frac{a}{p}}$ is a Legendre symbol.
This was incurred during I was studying to prove that
$$p: \text{odd prime}~\rightarrow~\text{There exist}~x,y:~x^2+y^2+1\equiv0 \pmod p~\land~0 \le x,y \le \frac{p-1}{2} $$
In the book, first it proves this theorem by using the concept of sets. Later, it also proves this theorem by using the Legendre symbol. On the way, it tries to show that when $p \equiv 3 \pmod4$, there exist $a$ which is a quadratic nonresidue of p. So $(-a)$ is a quadaratic residue of p, which means for some x,
$$x^2 \equiv -a \pmod p$$
Now we have to show that for some y,
$$y^2 \equiv a-1 \pmod p$$
which is exactly the same statement of the question I suggested. Actually, the book assumed that $a$ is the least possitive quadratic nonresidue, but I doubt if I should add that assumption. So it would be glad if someone could say to me if this assumption is required or not, too.
What I've tried is to solve this is by using Euler's Criterion, which means
$$(a-1)^{(p-1)/2} \equiv 1 \pmod p$$
But I got stuck... It seems to be really easy but... I failed, sadly.
Thanks!
Lest it remains unanswered:
The assumption that $a$ is the least (positive) quadratic nonresidue is important. With that assumption, we have that $1,\, \dotsc,\, a-1$ are quadratic residues by definition of the least positive nonresidue, so the existence of a $y$ with $y^2 \equiv a-1 \pmod{p}$ follows directly.
Without that assumption, $a-1$ can be a quadratic nonresidue too, for example
$$\left(\frac{8}{11}\right) = \left(\frac{7}{11}\right) = -1.$$