# For any integer $N$, can you find an odd prime $p$ such that $(p^2-1)/4$ is coprime to $N$?

For any integer $N$, can you find an odd prime $p$ such that $\frac{p^2-1}{4}$ is coprime to $N$?

This is equivalent to asking: Does there exist a prime $q$ which divides $\frac{p^2-1}{4}$ for every odd prime $p$?

• Yes, q = 2. Bu I assume you mean q is odd and p > 3. – fleablood Oct 1 '15 at 7:29
• Which is to say p^2 - 1/4 is never a power of 2. Which... it never is unless p =3. – fleablood Oct 1 '15 at 7:34
• As 24 divides p^2 - 1 for all p >= 5, maybe you meant (p^2 - 1)/24? Then for example 7 => 2, 11 =>5, 13 => 7 and 17 => 12=3*4, the first composite number and ... um, now I lost track what the question was supposed to be. N to not have a coprime generated N must be a product of all the primes and prime factors generated requiring only a finite number of primes and prime factors generated. Is this the case? I doubt it but I doubt I can prove it. – fleablood Oct 1 '15 at 8:06

Not if $N$ happens to be even.
• And if $N$ is odd and divisible by $3$, $p=3$ is the only solution. – Robert Israel Oct 1 '15 at 7:15
$\frac{p^2 - 1}{4} = \frac{p-1}{2}\frac{p + 1}{2}$ is always even. So if N is even, no.
$\frac{3^2 - 1}{4} = 2$. So f N is odd, yes.