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).
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