$x^3 - 6 = 25y^2 + 35y$ has no non-zero integer solutions

Question

I need to show that $x^3 - 6 = 25y^2 + 35y$ has no non-zero integer solutions. I tried moving things around and factor. Or use modulo 5 and 7 reductions for something shady to become a quadratic residue of a prime. I also couldn't really come up with something to use infinite descent. Hints would be appreciated.

This was a problem for non-linear diophantine equations (Pythagorean triples, rational points on curves, Fermat's last theorem, Pell's equations, etc.) in last year's elementary number theory exam.

Where did you get this problem--homework, textbook, research question? That is good for giving people context. — Mike, Feb 02 '21 at 03:31
@Hypernova Those 2 are relatively prime so can I say that they should both be cubes? Can that be a contradiction? Assuming that two cubes can't be that close? — Zara, Feb 02 '21 at 03:37
@Zara That's not a bad idea but the solution is more general — Hypernova, Feb 02 '21 at 04:29

score 7 · Accepted Answer · answered Feb 02 '21 at 03:37

7

We have $$x^3=25y^2+35y+6=(5y+6)(5y+1)$$Now, not both $5y+6$ and $5y+1$ are cubes, since there are no two positive cubes that differ by less than $7$, so there is a prime $p$ that divides both factors. But then $p$ divides their difference, and $p=5$. But this is absurd, since $p\nmid 5y+1$.

answered Feb 02 '21 at 03:37

saulspatz

53,824

1

i.e. by Euclid $(5y+6,5y+1) = (5,5y+1) = 1,,$ so being coprime factors of a cube they are both cubes, contra their difference $= 5.\ $ – Bill Dubuque Feb 02 '21 at 08:48

$x^3 - 6 = 25y^2 + 35y$ has no non-zero integer solutions

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