2
$\begingroup$

Find all pairs $(m,n)\in{\mathbb{N^2}}$ such that

$$(m^2-1)^3-n^2=2$$

Is $(2,5)$ the only solution?

$\endgroup$
  • $\begingroup$ How did you obtain (2,5)? Why does this not give rise to any other solution? What else have you tried, to obtain an alternative solution(s)? $\endgroup$ – Nij Jul 1 '16 at 23:16
  • $\begingroup$ First, you could find all cubes that are 2 greater than perfect squares. Then see if the root of the cubes can be written as $m^2-1$. $\endgroup$ – scott Jul 1 '16 at 23:18
  • $\begingroup$ Easy congruence data: $m$ must be even (in fact, it must be $\equiv 2\bmod 4$) and $n$ odd. $\endgroup$ – Steven Stadnicki Jul 1 '16 at 23:25
  • 1
    $\begingroup$ @Nij Obviously, $(m^2-1)$ and $n$ are of the same parity (since their powers will have the same parity as they do and the difference of those powers is even). If they're both even, then $m^2-1$ is a multiple of 8 (because squares of odd numbers are congruent to 1 mod 8) and thus so is its cube, and $n^2$ is obviously a multiple of 4, so their difference can't be 2. $\endgroup$ – Steven Stadnicki Jul 2 '16 at 0:03
  • 1
    $\begingroup$ @Nij From there, we know (again) that $n^2\equiv 1\bmod 8$ (because $n$ is odd) and so $(m^2-1)^3\equiv 3\bmod 8$, so $m^2-1\equiv 3\bmod 8$ and $m^2\equiv 4\bmod 8$. But if $m$ were a multiple of 4, its square would be a multiple of 16 and thus $\equiv 0\bmod 8$, so $m\equiv 2\bmod 4$. $\endgroup$ – Steven Stadnicki Jul 2 '16 at 0:05
5
$\begingroup$

Fermat claimed to have proved there is only one positive integer solution $x^3-y^2=2$, $(x,y)=(3,5)$. It is unknown if he actually had a proof.

The most common proof uses that $\mathbb Z[\sqrt{-2}]$ is a unique factorization domain.

That pretty much solves your question, setting $x=m^2-1,y=n$.

There might be an easier way to prove your subset has only one solution.

You could start by writing it as $$m^6-n^2=3(m^4-m^2+1)$$ or $$(m^2+1)(m^3-n)(m^3+n)=3(m^6+1)$$

Not sure where to go from there.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

Maybe this helps

$(m^2-1)^3-n^2 =2\iff (m^2-1)^3-27=n^2-25$

$\iff (m^2-4)( (m^4-2m^2+1)+3(m^2-1)+9)=(n-5)(n+5)$

$\iff(m+2)(m-2)(m^4+m^2+7)=(n-5)(n+5)$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.