The smallest prime

A number $a$ is called a "quadratic residue" modulo $p$ if there exists an $n$ such that $n^2 \equiv a \pmod{p}$ . For example, $2$ is a quadratic residue modulo $7$ , because $3^2 = 9 \equiv 2 \pmod{7}$ , but $3$ is not a quadratic residue modulo $7$ .

For an odd prime $p$ , Euler's criterion gives a way of testing whether any $a$ coprime to $p$ is a quadratic residue modulo $p$ .

The criterion is as follows: evaluate $a^{\frac{p-1}{2}}$ . If and only if $a$ is a quadratic residue, $a^{\frac{p-1}{2}} \equiv 1 \pmod{p}$ . If and only if $a$ is not a quadratic residue, $a^{\frac{p-1}{2}} \equiv -1 \pmod{p}$ .

This problem is equivalent to finding the smallest prime $p$ such that none of the values $\{2,3,6\}$ are quadratic residues modulo $p$ (if such a prime exists).

For $p=2$ or $p=3$ , $p|f(p)$ . So now let's look at $p>3$ .

Let's assume that neither $2$ nor $3$ is a quadratic residue modulo $p$ . Then, by Euler's criterion, $2^{\frac{p-1}{2}} \equiv 3^{\frac{p-1}{2}} \equiv -1 \pmod{p}$ .

But multiplying these two expressions gives $6^{\frac{p-1}{2}} \equiv 1 \pmod{p}$ ; so $6$ is a quadratic residue modulo $p$ .

In the same way, we can show that at least one of $\{2,3,6\}$ must be a quadratic residue modulo $p$ for every prime $p$ , and so the answer to the problem is $\boxed{-1}$ .

Since,legendre's symbol is completely multiplicative $(\frac{6}{p})=(\frac{2}{p})(\frac{3}{p})$ Hence,if both 2 and 3 are not quadratic residue modulo any prime p,then 6 has to be one.hence for all prime some n(infinitely many) can be found such that p divides f(n).