Find all $n\in\mathbb N$ such that $10^n-6^n$ is a perfect square

Question

EDIT:
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.

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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.

Answer 1

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].

Answer 2

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.