The square of any odd integer, when divided by 8, always leaves remainder:

Write the odd integer as $2k+1$. Then $(2k+1{)}^2=4k^2+4k+1=4k(k+1)+1$. Of the consecutive numbers $k$ and $k+1$ one is even, so $4k(k+1)$ is a multiple of 8. The remainder is therefore always 1 — check 9 ÷ 8, 25 ÷ 8, 49 ÷ 8.