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QUESTION

Prove that gcd$(n^3+n-1,n+1)$ is either 1 or 3. Can both
possibilities occur?



ANSWER

By long division, $n^3+n-1=(n^2-n+2)(n+1)-3$, so by lemma 1.9
gcd$(n^3+n-1,n+1)$=gcd$(n+1,3)$ and as the only divisors of 3 are
1 and 3, this nust be 1 or 3.

Both possibilities occur: e.g. $n=1$ gives gcd=1 and $n=2$ gives
gcd=3.




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