Suppose we have a semi-prime $N=pq$, where $p \ne q$, and $p>2$, $q>2$
Let $k$ be the multiplicative order of $2$ mod $N$, then either $p^{2} \bmod k \equiv 1$ or $q^{2} \bmod k \equiv 1$
Is there a simple proof (disproof) of this fact? There are counterexamples (see answer).
An interesting question is - Are there additional conditions that must hold for the conjecture to be true?
This is somewhat related to a question I asked previously, which was simply a special case of this conjecture when $N = 2^{m}-1$, where the multiplicative order in this case is $m$. See Conjecture involving semi-prime numbers of the form $2^{x}-1$
This conjecture is wrong. With $p=11, q=23\,$ we have $N=253\,$ and $k=\mathrm{ord}_{253}(2) = 110.$
But $11^2 \equiv 11 \pmod {110}\;$ and $23^2 \equiv 89 \pmod {110}.$
There are alot more counter-examples, e.g. $p=11,q=37\;$ or $p=19,q=23\dots$