Unique square root that is also a quadratic residue I'm trying to understand the following proof of Lemma 1 without much success, as the logical leaps are non trivial to me.

Explanations why the following claims hold would be very appreciated:


*

*"Since $N \equiv 1~ mod~4$, their Jacobi symbols satisfy $(+x/N) = (-x/N)$ and $(+y/N) = (-y/N).$"

*"Since $P \equiv 3~mod~4$, $(+x/N) \neq (+y/N)$": 
I understand why $gcd(x+y,N) = P$ and $gcd(x-y,N)=Q$ like they say, but why does it follow that $P \equiv 3~mod~4$, $(+x/N) \neq (+y/N)$?

*"Exactly one of these roots has Jacobi symbol +1 with respect to both P and Q, because $P \equiv 3~mod~4$": 
I understand what they're trying to say - that the other root is not a real quadratic residue, otherwise it would have Jacobi symbols of 1 in both primes, but why does it follow from $P \equiv 3~mod~4$?


Many thanks in advance.
 A: There are a couple of important facts to remember: Jacobi (and Legendre) symbols are multiplicative, in particular $(a/N) := (a/P)(a/Q)$ and $(-a/P)=(-1/P)(a/P)$. For primes we have:
$$
(-1/p) =
\begin{cases}
1 & p\equiv 1 \pmod 4 \\
-1 & p\equiv 3 \pmod 4
\end{cases}
$$
That rule also works for non-prime numbers (i.e. Jacobi symbols), but I wont use that fact. In particular $(-a/P)=(-1/P)(a/P)=-(a/P)$, since $P\equiv 3\pmod 4$ (and similarly for $Q$). Now:
1.
$$
(-x/N) = (-x/P)(-x/Q) 
= (-1)^2(x/P)(x/Q)
= (x/N)
$$
2. We have:
$$
x+y=kP \iff x\equiv -y\pmod P \implies (x/P)=(-y/P)=-(y/P)
$$
where the last equation uses $P\equiv 3\pmod 4$. By similar arguments, $(x/Q) = (y/Q)$. Thus:
$$
(x/N) = (x/P)(x/Q) = -(y/P)(y/Q) = -(y/N)
$$
3. $(x/N)=(x/P)(x/Q)=1$ implies that $(x/P)=(x/Q)=\pm 1$, and similarly for $-x$. But $(-x/P) = -(x/P)$ and $(-x/Q)=-(x/Q)$, so if $(x/P)=(x/Q)=+ 1$ then $(-x/P)=(-x/Q)=-1$ and vice versa. 
A: Claim 1 follows from $$\begin{pmatrix}-1\\N\\\end{pmatrix}=\begin{pmatrix}-1\\P\\\end{pmatrix}\begin{pmatrix}-1\\Q\\\end{pmatrix}=-1\times -1=1$$
Claim 2 follows from $$\begin{pmatrix}x\\N\\\end{pmatrix}=\begin{pmatrix}x\\P\\\end{pmatrix}\begin{pmatrix}x\\Q\\\end{pmatrix}=\begin{pmatrix}-y\\P\\\end{pmatrix}\begin{pmatrix}y\\Q\\\end{pmatrix}=-\begin{pmatrix}y\\P\\\end{pmatrix}\begin{pmatrix}y\\Q\\\end{pmatrix}=-\begin{pmatrix}y\\N\\\end{pmatrix}$$
Claim 3 similarly depends upon the fact that
$$\begin{pmatrix}-1\\P\\\end{pmatrix}=-1$$ because then 
$$\begin{pmatrix}x\\P\\\end{pmatrix}\ne \begin{pmatrix}-x\\P\\\end{pmatrix}$$
