If $p$ is $a$ prime number greater than 3, what is the remainder when $p^2$ is divided by 12?

Step 1 (Form of primes > 3): Any prime number $p>3$ must be coprime to both 2 and 3.
Therefore, modulo 12, $p$ can only be congruent to numbers coprime to 12. The possible residue classes are:

Step 2 (Evaluate p^2 mod 12): We square each of these possible residues:

• $1^2=1\equiv 1(\mathrm{mod}12)$

• $5^2=25=2\times 12+1\equiv 1(\mathrm{mod}12)$

• $7^2=49=4\times 12+1\equiv 1(\mathrm{mod}12)$

• $11^2=121=10\times 12+1\equiv 1(\mathrm{mod}12)$

Step 3 (Conclusion): For all prime numbers $p>3$, $p^2$ always leaves a remainder of $1$ when divided by 12.