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Suppose I were given the task of proving that for all negative integers $3n^{2} \equiv 3n \pmod{6}$. The original intent was to use negative induction. But, I was wondering if another, perhaps simpler approach might be to exploit the fact that congruence is an equivalence relation, and thus with the proper transformation of terms, might the original statement might be proven with forward induction?

user37611
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Since $3\mid 3n^2-3n$, it suffices you show $2\mid n^2-n$ for any $n$. Can you do this? Note that $n^2-n=n(n-1)$ is a product of consecutive integers.

Pedro
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  • Directly, because the product of any two consecutive integers is even. And thank you for the simplification! – user37611 Oct 26 '13 at 23:54