Originally Posted by

**flashylightsmeow** Thanks Plato! This makes sense.

So am I right that, for $4x\equiv4(\bmod 11)$, we have to find integers $T$ such that $4x = 11T + 4$ or $x = 11T/4 +1$. Given that $11T/4$ is an integer, T could be any multiple of $4$.

This implies that x can take infinitely many values ($12,23,34$ etc).

Doesn't that mean, in the words of the author, that mod $11$ DOES suffer from the same defect as mod $12$ in this case?

I get very nervous saying the author is wrong though. So I'm sure I've misunderstood something key to answering this.