Two whole numbers $N$ and $M$ are each divided by 7. The first leaves remainder 3 and the second leaves remainder 5. What is the remainder when $N^2+M^2$ is divided by 7?

Remainders may be squared and added just like the numbers themselves. $N^2$ behaves like $3^2=9$, which leaves 2, and $M^2$ behaves like $5^2=25$, which leaves 4. So $N^2+M^2$ leaves 2 + 4 = 6. (Check with $N=10$ and $M=12:100+144=244=7\times 34+6$.) Adding the remainders first and then squaring — 3 + 5 = 8, leaving 1 — is the trap that produces the answer 1.