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

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.

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].
• 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. – Dietrich Burde Jan 29 '20 at 10:42
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.