Question

Prove: The square of a number that is

Answer 1

Let 3n + 1 denote the "number" in question. The claim is that

(3n + 1)² = 3m + 1

for some integer m.

Now,

(3n + 1)² = (3n)² + 2 (3n) + 1²

… = 9n² + 6n + 1

… = 3n (3n + 2) + 1

… = 3m + 1

where we take m = n (3n + 2).