Find all functions $f$ and $g$ defined from $\mathbb{R}_{>0}$ to $\mathbb{R}_{>0}$ such that for all $x, y > 0$ the two equations hold $$ (f(x) + y - 1)(g(y) + x - 1) = {(x + y)}^2 $$$$ (-f(x) + y)(g(y) + x) = (x + y + 1)(y - x - 1) $$ Note: $\mathbb{R}_{>0}$ denotes the set of positive real numbers.
Problem
Source: 2022 Pan-African Mathematics Olympiad Problem 4
Tags: functional equation, algebra, function, system of equations
rafaello
26.06.2022 16:04
Call first one $P$ and second $Q$. $P(1,1),Q(1,1)\implies f(1)=g(1)=2$. Then, $P(1,x)$ and $P(x,1)$ yield $f(x)=g(x)=x+1$.
ACGNmath
26.06.2022 16:05
Let $x$ be any positive real. Substituting $y=x+1$ into the second assertion yields \[(-f(x)+x+1)(g(x+1)+x)=0\]and since $g(x+1)+x>0$, it follows that $f(x)=x+1$. Now substitute this into the first assertion to get $(x+y)(g(y)+x-1)=(x+y)^2$. Thus $g(y)=y+1$. This means $f(x)=x+1, g(y)=y+1$ is the only answer, which is clearly seen to work.
rananjay
26.06.2022 22:16
Substituting x = y = 0 into both equations, we get (f(0) - 1)(g(0) - 1) = 0 and f(0)g(0) = 1. This means that f(0) = g(0) = 1. Plugging y = 0 into the first equation, we get f(x) = x + 1. Using the same process on g, we get g(x) = x + 1.
ZETA_in_olympiad
26.06.2022 22:29
rananjay wrote: Substituting x = y = 0 into both equations, we get (f(0) - 1)(g(0) - 1) = 0 and f(0)g(0) = 1. This means that f(0) = g(0) = 1. Plugging y = 0 into the first equation, we get f(x) = x + 1. Using the same process on g, we get g(x) = x + 1. You cannot use zero. The problem is for $x,y>0.$
Setting $x=y=1$ in the first yields $f(1)g(1)=4.$ Setting $x=y=1$ in the second yields $f(1)=g(1).$ So they both are $2.$ Setting $y=1$ in the first gives $f(x)=x+1.$ And setting $x=1$ gives $g(y)=y+1.$ This works.
SpecialMaths
09.07.2022 12:30
Here is a link to a video solution https://youtu.be/Mg-3s5uz3JU
ahizounemustapha
08.07.2023 17:46
here is my solution ; https://youtu.be/ljx_kEvmlKA