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By factoring $p^2 − 1$, we have $(p + 1)(p - 1)$.

I know that $p = 2$ which gives $3$ is the only solution.

However, how do I prove that $p = 2$ is the only integer which gives a prime?

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  • $\begingroup$ Because of the $p-1$ factor... $\endgroup$ – abiessu Dec 3 '15 at 1:45
  • $\begingroup$ for any number>$2$, $p-1$ will give you a factor > $1$ $\endgroup$ – Kushal Bhuyan Dec 3 '15 at 1:46
  • $\begingroup$ any other solution can't be prime because it can be decomposed as a product of two numbers, neither of which equals 1 by the work you've already shown $\endgroup$ – cnick Dec 3 '15 at 1:46
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If $p>2$, both $p+1$ and $p-1$ are $>1$, hence $p^2-1$ is composite.

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In $p^2-1=(p-1)(p+1)$ give a prime number only and only if p-1 is 1 so that p+1 will be the only prime factor. So n must be 2 and its the only one which can give us a prime(3). I better suggest you to think about $$n^2-m=p$$ where 'n' is an natural number and m<$n^2$ such that 'p' will be a prime number.

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