1. Argument by Contradiction

April 2025
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Problem 1: Determine all functions $f : \mathbb{N}\rightarrww \mathbb{N}$ satisfying

$xf(y)+yf(x) = (x +y)f(x^2 +y^2)$

for all positive integers $x$ and $y$ .

idea and solution: We could attempt some functions. We find that constant function $f(x)=k$ works, $f(x)=x$ does not work. $f(x)=1/x$ does not work. This prompt us, maybe the only possible solution is $f(x)=k$ ? This is because the left and right hand side has different order, so only a function with zero order works.

We could do some operation. For $x=a$ , $y=b$ , we get $f(a^2+b^2)=\frac{af(b)+bf(a)}{a+b}$ . This expression could be simplified a lot if $f(b)=f(a)$ . So let’s just assume on is greater, let’s say $f(b)$ , then

$f(a)<f(a^2+b^2)<f(b)$

We know it is impossible since we cannot add infinitely many natural numbers in $[f(a),f(b)]$ . Q.E.D.

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