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Problem 5

Find all functions f:R→Rf:\mathbb R\to\mathbb R such that for all x,y,z∈Rx,y,z\in\mathbb R, f(f(x)+f(y)+f(z))=f(f(x)−f(y))+f(2xy+f(z))+2f(xz−yz)f(f(x)+f(y)+f(z))=f(f(x)-f(y))+f(2xy+f(z))+2f(xz-yz).
Step 7 of 8: Nonconstant case
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f(x)=x2f(x)=x^2
Detailed analysis

Putting x=yx=y and using f(0)=0f(0)=0 gives f(2f(x)+f(z))=f(2x2+f(z))f(2f(x)+f(z))=f(2x^2+f(z)). The fibre property and nonconstancy rule out the negative alternative, so f(x)=x2f(x)=x^2 for every xx.