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I'm not sure if this is true but, I've tried with many different values of $i, j$ and didn't get any contradictions. The question again, here

Prove that there are no natural numbers, $i, j$ such that $ 3i^2+3i+7=j^3$

Any help would be appreciated.

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    Of course it does not bother me/us ! I'm just curious.2012-11-07

1 Answers 1

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Let $(i,j)$ be a solution. As $3i(i+1)+7 = 1 [3]$, $j^3 = 1 [3]$ thus $j = 1 [3]$. Let write $j=3k+1$. A straight forward computation shows that: $ j^3-1 = 9k(3k^2+3k+1) $ Thus $3i(i+1)+6 = 9k(3k^2+3k+1)$, and then $3$ divides $i(i+1)+2$.

This is not possible because $i(i+1)+2$ is always equal to $1$ or $2$ modulus 3.