I have edited the title.
Thanks to Dietrich Burde, we now have that there is no $n\geq4$ such that $a^n-b^n$ is a perfect square for coprime $a,b$. It shows for $5^n-3^n,7^n-3^n$ and $10^{2m}-6^{2m}=2^{2m}\cdot(5^m-3^m)$.
Now, my question is how can I prove that there is no odd integer $n\geq5$ such that $10^n-6^n$ is square.

I have three questions:

  • Find all $n\in\mathbb N$ such that $5^n-3^n$ is a perfect square.
  • Find all $n\in\mathbb N$ such that $7^n-3^n$ is a perfect square.
  • Find all $n\in\mathbb N$ such that $10^n-6^n$ is a perfect square.

I checked up to $n\leq 10000$ then only found these.

  • For $5^n-3^n$ : $n=2$
  • For $7^n-3^n$ : $n=1$
  • For $10^n-6^n$ : $n=1,2,3$

I tried to prove it in the same way as the answer to this question, Does there exist $n\in\mathbb{N}$ such that $5^n-2^n$ is a perfect square?, which I asked two days ago. But it didn't work for this.

I will appreciate any help. Thank you.


Having now several cases of the same type asked, it is perhaps useful to ask for a general solution. Here we can consider the generalized Fermat equation $$ x^p+y^q=z^r $$ in the hyperbolic case with $$ \frac{1}{p}+\frac{1}{q}+\frac{1}{r}<1. $$ For us, we are in this case with $x^2+3^n=7^n$, or $x^2+3^n=5^n$ or $x^2+6^n=10^n$, and $n\ge 5$. Now we can use the results on the generalized Fermat equation, e.g., listed by Bennett. It follows that there are no solutions for all $n\ge 5$.

Conjecturally all solutions with coprime $x,y,z$ are given by $1^p+2^3=3^2$ and the $9$ solutions listed on page $2$, with proofs in certain cases, see Theorem $1$ and $2$ on page $3$ and lateron. Our case of exponents $(2,n,n)$ has been solved by Darmon-Merel [33] and Poonen [57].

  • 1
    $\begingroup$ Why not? We have $gcd(x,7,3)=gcd(x,5,3)=1$ for example. For $gcd(x,10,6)$ we can perhaps reduce it to a coprime case, but you are right, one should check if it reduces. $\endgroup$ – Dietrich Burde Jan 29 '20 at 10:42

This is base 10/6. The normal base-rules apply, along with additional factors corresponding to that 2 divides both.

There are very few sevenites in this base, so finding an example that if $p$ divides, so does $p^2$, is rare. It works for $n=1,\ 2,\ 3$, but for larger values, additional unpaired primes appear, which cahses the next working example to be very large.

Have a look at the factorisations in the "Carmichael Project" and ye would see that in general $b^n-a^n$ tends to accumulate anomonously large primes and are not compact squares, cubes etc, except in the very small cases.


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.