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Here,

Any other prime numbers that satisfy this condition?

I asked if $a=2$ and $b=3$, then $a^2-1$ is an integer multiple of $b$. Is there any other pair of primes $a$ and $b$ that satisfy this relationship? The answer is yes, there are infinite pairs: set $a= \text{any prime} =p$ and $b$ to any prime divisor of $p^2-1$. If we add the condition that $a<b$, are there any other pairs?

More Clearly,

Are there primes $a,b$ where $a<b$ such that $a^2 -1$ is an integer multiple of $b$, aside from $a=2,b=3$?

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No. $a^2-1=(a+1)(a-1)$ certainly has no prime divisors $>a+1$.

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Hagen von Eitzen gave a very good succinct explanation, but I had some doubts and tried to attack it differently.

I noted that $a^2-1$ is necessarily even, so if $kb=a^2-1$, we must have $2pb=a^2-1$, where $p,k \in \mathbb{N}$. I tried to show this implies $b\leq a$, but didn't get very far.

Still having doubts, I wrote a Python script to computationally check the proposition. I know this isn't a proof at all, I'm just sharing this hoping that it might be somehow helpful.

The primes.txt file is just a copy-paste of this list of the first 1000 primes, though you could of course do it with a larger list of primes if you wanted to. The code for my script is available here.

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