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when I deal a number theory problem,I conjecture :

Let $p$ is give postive integer number,show that:there exist infinitely many postive integer numbers $k$,such $p\cdot (3k+1) $ is cube?

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  • $\begingroup$ Have you tried anything? $\endgroup$ – Wojowu Jul 27 '16 at 10:39
  • $\begingroup$ @Wojowu,I have conjecture.I have try some idea.such there exist $k?$such $3k+1=p^2l^3$? $\endgroup$ – function sug Jul 27 '16 at 10:43
  • $\begingroup$ Hint: try your last idea the other way around - can you find $l$ such that for some $k$ we have $p^2l^3=3k+1$? $\endgroup$ – Wojowu Jul 27 '16 at 10:46
  • $\begingroup$ @Wojowu,That's a idea,But How to prove it $\endgroup$ – function sug Jul 27 '16 at 10:50
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    $\begingroup$ I've just realized your conjecture is wrong - take $p=3$. Can you see why your conjecture fails? $\endgroup$ – Wojowu Jul 27 '16 at 10:54
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If you write $p=3^m\cdot n$ , $n$ not divisible by $3$, then you have infinite many integers $k$, such that $p(3k+1)$ is a cube, if and only if $m$ is divisible by $3$ (it can also be $0$).

It is clear that we have no solution if $m$ is not divisble by $3$. If $m$ is divisble by $3$, you can choose an arbitary positive integer $s$ that is congruent to $1$ modulo $3$.

Then, $s^3n^2$ is of the form $3k+1$ and $p\cdot s^3n^2=s^3\cdot 3^m\cdot n^3$ is a cube.

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