I have no clue where to start. Would I negate the proof and try to prove it? Any help appreciated
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n mod 12 must be one of 1, 5, 7, 11 as the only integers in [0,11] that are not divisible by either 2 or 3. The squares are 1, 25, 49 and 121, respectively, and each square is congruent to 1 (mod 12).
You could also observe that n must be equal to 6k ± 1, for some integer k, as a simpler characterization of all integers not divisible by either 2 or 3. That makes:
n² = (6k ± k)² = 36k² ± 12k + 1 = 12(3k² ± k) + 1
...and n² is clearly 1 more than a multiple of 12.
You could also observe that n must be equal to 6k ± 1, for some integer k, as a simpler characterization of all integers not divisible by either 2 or 3. That makes:
n² = (6k ± k)² = 36k² ± 12k + 1 = 12(3k² ± k) + 1
...and n² is clearly 1 more than a multiple of 12.