Let a,b,n be positive integers such that $(a,b) = 1$. Prove that if $(x,y)$ is a solution of the equation $ax+by = a^n + b^n$ then $$\left[\frac{x}{b}\right]+\left[\frac{y}{a}\right]=\left[\frac{a^{n-1}}{b}\right]+\left[\frac{b^{n-1}}{a}\right]$$
Problem
Source: Romania IMO TST 1990 p1
Tags: floor function, number theory
Sprites
09.08.2021 19:43
parmenides51 wrote: Let a,b,n be positive integers such that $(a,b) = 1$. Prove that if $(x,y)$ is a solution of the equation $ax+by = a$ then $$\left[\frac{x}{b}\right]+\left[\frac{y}{a}\right]=\left[\frac{a^{n-1}}{b}\right]+\left[\frac{b^{n-1}}{a}\right]$$ The version used here is wrong the actual version reads:- Romania IMO TST 1990 p1 wrote: Let $a,b,n$ be positive integers such that $(a,b) = 1$. Prove that if $(x,y)$ is a solution of the equation $ax+by = a^n+b^n$ then $$\left[\frac{x}{b}\right]+\left[\frac{y}{a}\right]=\left[\frac{a^{n-1}}{b}\right]+\left[\frac{b^{n-1}}{a}\right]$$ Source of this version:-Number theory concepts and problems by Titu andreesucu.
Rearrange it to get $a^n-ax=b^n-by=>a(a^{n-1}-x)=b(b^{n-1}-x)$,reducing $\pmod a$ and $\pmod b$ yields $y \equiv b^{n-1} \pmod a,x \equiv a^{n-1} \pmod b$
Thus we can find integers $c,d$ such that $y=b^{n-1}+ca,x=a^{n-1}+db$ with $c+d=0$
Putting this in the relation yeilds the result from the periodicity of the floor function.
parmenides51
13.08.2021 20:33
Thanks, typo has been corrected