# If n > 3 and (n + 1) is a square, is there any n that is a prime?

I am looking at properties of squares and came about this property. I am investigating the difference of squares in relation to primes.

• I meant to say is there any n which is prime? I cannot find a prime that has this properties. How can I edit this? – caliper Dec 4 '13 at 18:21
• $n+1 = a^2 \iff n = a^2 - 1 = (a-1)(a+1)$. – Daniel Fischer Dec 4 '13 at 18:23
• Thanks but I'm looking for a prime n and this is a difference of square. I meant to ask Is n a composite? – caliper Dec 4 '13 at 18:28

Does there exist a prime of the form $a^2-1$ when $a^2-1>3$?
No, because $$a^2-1=(a-1)(a+1)$$ and both $a-1$ and $a+1$ are non-trivial factors.
• That's what I mean. Here $n=a^2-1$ so $m=a^2$ which is a square. Fermat Primes are of the form $2^n+1$ (Mersenne Primes are of the form $2^n-1$), so these are different. – Rebecca J. Stones Dec 4 '13 at 18:55