Prove that the equation $7^x=1+y^2+z^2$ has no solutions over positive integers.
Problem
Source: 2019 Baltic Way P19
Tags: number theory
MarkBcc168
18.11.2019 14:29
Let $x=2^m\cdot n$ where $n$ odd. Since $7^n-1\equiv 6\pmod 8$, there exists a prime $p\equiv 3\pmod 4$ such that $\nu_p(7^n-1)$ is odd. By LTE,
$$\nu_p(7^x-1) = \nu_p(7^n-1) + \nu_p(2^m) = \nu_p(7^n-1)$$which is odd. This implies that $7^x-1$ cannot be written as sum of two squares, done.
ITryMyBest
18.11.2019 17:23
@above $7^n-1\equiv 6\pmod 8$ I think this is wrong. Take n=2 for example. @below Sorry. My bad.
AKS_9_54_61
18.11.2019 17:35
ITryMyBest wrote: @above $7^n-1\equiv 6\pmod 8$ I think this is wrong. Take n=2 for example. I guess he mentioned that n is odd but I am not sure
MarkBcc168
19.11.2019 03:14
ARNAV200205 wrote: ITryMyBest wrote: @above $7^n-1\equiv 6\pmod 8$ I think this is wrong. Take n=2 for example. I guess he mentioned that n is odd but I am not sure It was mentioned in the first sentence that $n$ is odd.
CANBANKAN
19.11.2019 03:52
The solution is correct