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I'm trying to prove that $3$ divides $n^2 + n$ iff $n$ mod $3$ $\neq 1$ . I already tried it with proving a double implication, but I did not succeed. A tip or kickstart would be great.

Thank you

CPUFry
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2 Answers2

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Consider the following cases:

  • $n\equiv0\pmod3 \implies n^2+n\equiv0^2+0\equiv0\pmod3$
  • $n\equiv1\pmod3 \implies n^2+n\equiv1^2+1\equiv2\pmod3$
  • $n\equiv2\pmod3 \implies n^2+n\equiv2^2+2\equiv0\pmod3$
barak manos
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  • Thank you for your reply. So the proof would be 'by cases' ? – CPUFry Jan 19 '16 at 15:14
  • Well, you can construct the proof in any way you like, but the fact of the matter is that $n\not\equiv1\pmod3 \iff n\equiv0,2\pmod3$. You're welcome BTW :) – barak manos Jan 19 '16 at 15:20
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Just factor $n^{2}+n$ and it becomes quite obvious

citronrose
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