Let N be an integer not divisible by 3. Prove N^2 = 3a + 1, where a is an integer

For N to be not divisible by 3, N can either be of the form 3k + 1 (1,4...) or 3k + 2 (2,5...), where k is an integer.
The proof can then be done by checking both 3k + 1 and 3k + 2 when N is squared, to see if they can be rearranged into the form 3a + 1.
N = 3k + 1, so N² = (3k + 1)² = 9k² + 6k + 1
This can then be rearranged to prove 3a + 1. Note that a can be made of any polynomial of k with integer powers, as k is an integer so its polynomial with integer powers will also be an integer for any value of k.
9k² + 6k + 1 = 3(3k² +2k) + 1, so true for N = 3k + 1
The same method can then be used to prove for N = 3k + 2
N= 3k + 2, N² = 9k² + 12k + 4 = 3(3k² + 4k + 1) + 1