MathLabs

Problem 2

Find all functions f:R→Rf:\mathbb R\to\mathbb R such that f(x2+f(y))=y+f(x)2f(x^2+f(y))=y+f(x)^2 for all real x,yx,y.
Step 2 of 5: Force f(0)=0f(0)=0
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In plain words

Force f(0)=0f(0)=0

f(t2+f(1)2)=1+t+2t2+t4=1+t+t4f(t^2+f(1)^2)=1+t+2t^2+t^4=1+t+t^4
Detailed analysis

The first identity gives f(t2+f(1)2)=t+(1+t2)2=1+t+2t2+t4f(t^2+f(1)^2)=t+(1+t^2)^2=1+t+2t^2+t^4. On the other hand, apply the original equation with x=t,y=1+tx=t,y=1+t and use f(1+t)=f(1)2f(1+t)=f(1)^2 and f(t)=t2f(t)=t^2 to get f(t2+f(1)2)=1+t+t4f(t^2+f(1)^2)=1+t+t^4. Hence 2t2=02t^2=0, so t=0t=0.