Prove that if n is an integer, then n^2 mod 5 is either 0,1, or 4

Question

anonymous · Answer

there are only five possible ways for n.

first case: $n \equiv 0(mod 5)$, then $n^2 \equiv 0(mod 5)$

second case: $n \equiv 1(mod 5)$), then $n^2 \equiv 1(mod 5)$

third case:$n \equiv 2(mod 5)$), then $n^2 \equiv 4(mod 5)$

fourth case:$n \equiv 3(mod 5)$), then $n^2 \equiv 6(mod 5)$, which is equal to $n^2 \equiv 1(mod 5)$, as  $6 \equiv 1(mod 5)$

fifth case:$n \equiv 4(mod 5)$), then $n^2 \equiv 16(mod 5)$
which is equal to $n^2 \equiv 1(mod 5)$, as  $16 \equiv 1(mod 5)$

so we conclude that if n is an integer, n^2 mod 5 is either 0,1, or 4

anonymous · Answer

sorry, for fourth case, it's $n^2 \equiv 9 (mod 5)$ and it's equal to $n^2 \equiv 4 (mod 5)$

anonymous · Answer

Excellent, but how do we know that there are only five possible values for n?

anonymous · Answer

because the remainder for any integer k divided by an integer n has only n-1 remainders.
e.g. for 2, you can only get 0 or 1 as remainder. any other remainder means that the division is incomplete.

anonymous · Answer

sorry it'snot n-1, it's n. but the range is 0,1,2,3,...,n-1