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.
Impossible Diophantine equation.
2
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number-theory
diophantine-equations
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0*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). – 2017-02-20
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0if $n=35$ then $6+n(7n^2-1)$ is divisible by 36. – 2017-02-20
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2Factorising as $(n+1)(7n^2-7n+6)=k^2$ might help. Perhaps proving the only solution is $n=-1$ could be the way forward. – 2017-02-20
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1*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 ? – 2017-02-22
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2This 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. – 2017-02-22
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0@Gerry Myerson : Thanks for your answer! :-) – 2017-02-23