What can be said about the smallest $N$ where $N^2 \not\equiv 1$ mod $p_x$; for all $x : 1 \leq x \leq y$

Question

Choose any value for $y : y \in \mathbb{N}$

Let

$N(y)$ be the smallest natural number that satisfies the following system of quadratic congruences:

$N^2 \not\equiv 1$ mod $p_x$; for all $x : 1 \leq x \leq y$.

($p_x$ are consecutive prime numbers)

Is it true that

$N(y) \leq p_y^2$

Answer 1

It's likely (although of course not proven) that there is always a twin prime pair $q, q+2$ with $p_y < q < p_y^2-1$, in which case $N(y) \le q+1 < p_y^2$. Conversely, all prime factors of $N(y)\pm 1$ are greater than $p_y$, so if $N(y) < p_y^2$, $N(y)\pm 1$ will be twin primes. Thus your conjecture implies that there are infinitely many twin primes. Don't expect to see a proof any time soon!

Empirically, $N(y)$ tends to be quite close to $p_y$.