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1. Argument by Contradiction

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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