Show that there are infinitely many perfect squares of the form $50^m-50^n$, but no perfect square of the form $2020^m+2020^n$, where $m$ and $n$ are positive integers.
Problem
Source: Bundeswettbewerb Mathematik 2020, Round 1 - Problem 1
Tags: number theory, number theory proposed
EmilXM
17.11.2020 22:55
I will prove that, $(m,n) = (2x,2x-1)$ works.
$$50^{2x}-50^{2x-1} = 50^{2x}\cdot49 = (7\cdot50^x)^2$$$$2020^{2x}+2020^{2x-1} = 2021\cdot2020^{2x-1}$$since $gcd(2021, 2020^{2x-1})=1$, if their product is a perfect square, then each of $2021$ and $2020^{2x-1}$ must be a perfect square. But obviously they are not. Hence, there is infinitle many such $(m,n)$.
cadaeibf wrote: however the $m$ and $n$ can be different in the second case What do you mean?
cadaeibf
17.11.2020 23:46
EmilXM wrote:
I will prove that, $(m,n) = (2x,2x-1)$ works.
$$50^{2x}-50^{2x-1} = 50^{2x}\cdot49 = (7\cdot50^x)^2$$$$2020^{2x}+2020^{2x-1} = 2021\cdot2020^{2x-1}$$since $gcd(2021, 2020^{2x-1})=1$, if their product is a perfect square, then each of $2021$ and $2020^{2x-1}$ must be a perfect square. But obviously they are not. Hence, there is infinitle many such $(m,n)$.
The first part was correct, however the $m$ and $n$ can be different in the second case, so here is the solution. Assuming $m-n=d>0$ ($d=0$ doesn't work), we can write $2020^n(2020^d+1)=a^2$ so by gcd argument both of the factors must be squares. The first one can easily be a square by having even $n$, however we'll see that $2020^d+1$ can't be a square $b^2$. This is because $2020^d+1\equiv 1^d+1=2\pmod{3}$ which is never a square mod 3, so we are done.