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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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Why are you interested in this question ? –  Lierre Nov 7 '12 at 9:31
    
I was asked this around 3 or 4 years ago but didn't know how to prove it. I just tried using a simple programmed loop. Now as I knew this website, I found the chance to trigger every question in my mind. I hope it wouldn't bother you. –  Tariq Nov 7 '12 at 10:59
    
Of course it does not bother me/us ! I'm just curious. –  Lierre Nov 7 '12 at 13:00

1 Answer 1

up vote 8 down vote accepted

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.

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