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We have $24n^3+6n+7$ and we need to find all natural $n$ for which this equation is a sixth power of natural number.

My try: let $k \in N$

$24n^3+6n+7=k^6$

$24n^3+6n+6=k^6-1$

$6(2n+1)(2n^2-n+1)=(k^3+1)(k^3-1)$ let $m=k^3-1$

$6(2n+1)(2n^2-n+1)=m(m+2)$

since $2n^2-n+1>2n+1$ for $n \ge 2$ and for $n=1$ we have $37\neq k^6$ then

$\begin{cases} m=6(2n+1)\\m+2=2n^2-n+1\end{cases}$

solving we have $n=7$ but then we have $8281$ which can't be represented as $k^6$ what's wrong here ?

Gregor
  • 708
  • The problem is that $m = k^3 - 1$ does not play any role in the last system of equations. – Karolis Juodelė Mar 16 '14 at 16:26
  • You have shown that (under certain assumptions that need not hold) we have $m=90$. But $90$ is not of the form $k^3-1$. – André Nicolas Mar 16 '14 at 16:27
  • OK thanks a lot Andre – Gregor Mar 16 '14 at 16:32
  • A sixth power is a square, and it is a square when $n=7$. Asking for it to be just a square makes it an elliptic curve, and someone who knows about those might be able to get a finite list of $n$ where it's a square. Then one could check whether any of those were also cubes. – coffeemath Mar 17 '14 at 09:34

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