Prove that the sum of the squares of any two consecutive numbers always leaves a remainder of 1 when divided by 4.

(n)^2 + (n + 1)^2n2 + n2 + 2n + 12n2 + 2n + 12(n2+ n) + 12n(n+1) + 1Since the numbers are consecutive, either n or (n+1) are even and divisible by 2. There is already a factor of 2 so 2n(n+1) is a multiple of 4, leaving the +1 as a remainder.