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

  • $\begingroup$ 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? $\endgroup$ – caliper Dec 4 '13 at 18:21
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    $\begingroup$ $n+1 = a^2 \iff n = a^2 - 1 = (a-1)(a+1)$. $\endgroup$ – Daniel Fischer Dec 4 '13 at 18:23
  • $\begingroup$ Thanks but I'm looking for a prime n and this is a difference of square. I meant to ask Is n a composite? $\endgroup$ – 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.

  • $\begingroup$ Thanks what I meant is say an integer m = n + 1 and m is a square. Is there any n that will meet this? Sorry but I'm not asking if there is prime of the form of mentioned, of course fermat primes (5 and 17) are examples of your forms. $\endgroup$ – caliper Dec 4 '13 at 18:47
  • $\begingroup$ 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. $\endgroup$ – Rebecca J. Stones Dec 4 '13 at 18:55
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    $\begingroup$ you are right. It is simply the difference of square hence n factors are (a + 1) and (a - 1). Thanks. $\endgroup$ – caliper Dec 4 '13 at 19:21

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