If $p$ and $q=2p−1$ are primes, and $N=pq,$ then $N$ is pseudo-prime for all possible bases $b$ that are quadratic residues modulo $2p−1.$ ... Being $n \in \mathbb Z,$ $n$ is said to be pseudo-prime with respect to the base $b$ if it is compound and also verifies the congruence $b^{n−1}\equiv 1 \pmod n$  where $n|b^{n-1}-1$.
Show that if $p$ and $q=2p−1$ are primes, and $N=pq,$ then $N$ is pseudo-prime for all possible bases $b$ that are quadratic residues modulo $2p−1$.
I can solve it when given concrete examples, but I am struggling with being able to generalize this, as asked here.
For example, $91\equiv 13\cdot 7$, then $13\equiv 2\cdot 7-1$,  $$\begin{equation*}         \begin{cases}         4^2 = 9^2 \equiv 3 \pmod{13}\\         5^2 = 8^2 \equiv 12 \pmod{13})\\         6^2 = 7^2 \equiv 10 \pmod{13}\\         \end{cases}     \end{equation*}$$
so \begin{equation*}         \begin{cases}         4^2 = 9^2 \equiv 3 \pmod{13}\\         5^2 = 8^2 \equiv 12 \pmod{13}\\         6^2 = 7^2 \equiv 10 \pmod{13}.\\         \end{cases}     \end{equation*}
But I do not know the general case for $p$ and $2p-1$.

We want to show that if $p$ and $q:=2p-1$ are prime, then $N:=pq$ satisfies the property that if $$a \equiv u^2 \pmod{q},$$
and $p,q\nmid a$, 
then 
$$a^{N-1}\equiv 1 \pmod{N}.$$
One solution is to use the Chinese remainder theorem. In this context,
it says that it suffices to show both 
$$a^{N-1}\equiv 1 \pmod{p}$$
and
$$a^{N-1}\equiv 1 \pmod{q}.$$
For the first statement,
$$
a^{N-1}=a^{p(2p-1)-1}=a^{2p^2-p-1}
=a^{(2p+1)(p-1)} \equiv 1\pmod{p},$$
since $a^{p-1}\equiv 1\pmod{p}$ by Fermat's little theorem.
For the second statement, 
$$
a^{N-1}=a^{q(q+1)/2-1}
\equiv
u^{q(q+1)-2}
= u^{q^2+q-2}
= u^{(q-1)(q+2)}
\equiv 1\pmod{q}
$$
again by Fermat's little theorem, since $u^{q-1}\equiv 1\pmod{q}$.
