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Any integer solutions to $3n^{2}+3n+1=m^{3}$?

1 Answers 1

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We have that $$3n^2 + 3n + 1 = m^3$$ Adding $n^3$ to both sides, we get that $$n^3 + 3n^2 + 3n+1 = m^3 + n^3$$ $$(n+1)^3 = m^3 + n^3$$ Hence, no solution exists except for trivial solutions. (This is of-course the Fermat's theorem where the exponent is $3$. The proof for the exponent $3$ is actually not that hard.)

And the trivial solutions are

$$n=0 \text{ gives us }m=1$$

$$n=-1 \text{ gives us }m=1$$