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

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

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

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

• 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)? – Nij Jul 1 '16 at 23:16
• 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 $m^2-1$. – scott Jul 1 '16 at 23:18
• Easy congruence data: $m$ must be even (in fact, it must be $\equiv 2\bmod 4$) and $n$ odd. – Steven Stadnicki Jul 1 '16 at 23:25
• @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. – Steven Stadnicki Jul 2 '16 at 0:03
• @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 $m\equiv 2\bmod 4$. – Steven Stadnicki Jul 2 '16 at 0:05

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.

Maybe this helps

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

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

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