I am trying to prove that the following Diophantine equation is impossible in positive integers, but i keep failing. $6+n(7n^2-1)=k^2$.

NOTE: it has a negative solution so i don't think working with modulos will help.

  • $\begingroup$ Note: $\frac{6+n(7n^2-1)}{6}=\frac{(n-1)n(n+1)}{6}+(n-1)n(n+1)+n+1\in\mathbb{N}$ --- Now check if $6+n(7n^2-1)$ can be devided by $6^2$ (which is necessary for a solution). $\endgroup$
    – user90369
    Feb 20, 2017 at 15:47
  • $\begingroup$ if $n=35$ then $6+n(7n^2-1)$ is divisible by 36. $\endgroup$ Feb 20, 2017 at 15:51
  • 2
    $\begingroup$ Factorising as $(n+1)(7n^2-7n+6)=k^2$ might help. Perhaps proving the only solution is $n=-1$ could be the way forward. $\endgroup$
    – Old Peter
    Feb 20, 2017 at 19:55
  • 1
    $\begingroup$ Second note: It seems to be that for $\frac{6+n(7n^2-1)}{36}\in\mathbb{N}$ it's always divisible by $8$ . This means: If the result is odd there is no positive solution possible. If the result is even, it's divisible by $4^2$ but sorry I don't know how to continue because at the latest now one has to show that there is always a prime factor which has no even power. It would be interesting if such a prime number exist for all possible values of $\enspace 6+n(7n^2-1)$. Maybe I think too complicate, what other methods are useful here ? $\endgroup$
    – user90369
    Feb 22, 2017 at 10:57
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    $\begingroup$ This is what's known as an elliptic curve. Elementary methods are rarely successful in determining all the integer points on an elliptic curve, but there is a large literature on more advanced methods. $\endgroup$ Feb 22, 2017 at 11:57

1 Answer 1


With substitution $m=n+1$, the equation becomes $m(7m^2-21m+20)=k^2$. Put $m=du^2$, $7m^2-21m+20=dv^2$, with $d$ a positive divisor of $20$ and $u$, $v$ coprime positive integers. Then we get $7du^4-21u^2+20/d=v^2$.

When $d=1$, the last equation is impossible because $5$ is quadratic non-residue modulo $7$. The same argument works for $d=2$ or $4$. If $d=5$, then note that $u$ must be odd, otherwise $v$ would be also even, in contradiction with the requirement $\gcd (u,v)=1$. But for odd $u$ one has $35u^4-21u^2+4 \equiv 2 \pmod 4$, while any even square is multiple of $4$. A similar argument eliminates the possibility $d=10$.

It remains to solve in coprime integers the equation $140u^4 -21u^2+1=v^2$. Multiplication by $(140u)^2$ results in the elliptic equation $Y^2=X^3-21X^2+140X$, whose integer points are obtained with the following SAGE lines

sage: E = EllipticCurve([0,-21,0,140,0])

sage: E.integral_points(both_signs=True)

The output [(0 : 0 : 1)] means that there are no non-zero solutions to the elliptic equation and therefore for the initial equation.


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