# Impossible Diophantine equation.

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.

• 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). Feb 20, 2017 at 15:47
• if $n=35$ then $6+n(7n^2-1)$ is divisible by 36. Feb 20, 2017 at 15:51
• 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. Feb 20, 2017 at 19:55
• 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 ? Feb 22, 2017 at 10:57
• 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. Feb 22, 2017 at 11:57

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