Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 9 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.