A number theory problem by Angaddeep Singh

Any number can be written as $6k, 6k+1, 6k+2, 6k+3, 6k+4,$ or $6k+5.$

Note that $6k, 6k+2, 6k+3$ , and $6k+4$ cannot be prime (excluding 2 and 3) since they are multiples of 2 or 3. Thus we will only consider primes of the form $p= 6k+1$ and $p= 6k+5$ .

If $p= 6k+1$ then this is the same as saying $p \equiv 1$ (mod 6).Then $p^2 \equiv 1^2 = 1$ (mod 6).

If $p= 6k+5$ then this is the same as saying $p \equiv 5$ (mod 6). Then $p^2 \equiv 5^2 = 25 = 1$ (mod 6).

In both cases, a prime number greater than 3 squared will yield a remainder of $\boxed{1}$ when divided by 6 because it is of the form $6k+1.$

In a table:

we may assume any prime number after 3 to get the answer also the square of each prime no after 3 is (multiple of 6 + 1)