# Maths Competition problem!

Hi guys I am new here. So I found this problem I can not solve, can someone help me out? :))

http://prntscr.com/27jqyk

Would like an answer for both questions and please provide valid proofs, thanks very much! :D

Note by Igor Filkovski
4 years, 7 months ago

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Well, I can get the discussion started, but I don't have a full answer. It looks like $$f(x) = 1-x^2$$ satisfies the condition. (I rewrote $$f(y)$$ as $$a$$; the $$2xa$$ term suggested looking at squares.)

If $$f(y) = 0$$ for some $$y$$, then $$f(0) = 1$$ (plug into the equation and solve). And in fact, if the range of $$f$$ is all of $$\mathbb R$$, it's easy to show that $$f(x) = 1-x^2$$: for all $$x$$ find $$y$$ such that $$f(y) = x$$, then we get $$f(0) = f(x) + 2x^2 + f(x) - 1$$, and $$f(0) = 1$$, so $$f(x) = 1-x^2$$. (The same argument shows that regardless of the range of $$f$$, $$f(f(y)) = C - f(y)^2,$$, where $$C = (f(0)+1)/2.$$ )

I'm not sure what to do without the assumption on the range of $$f$$, though.

- 4 years, 7 months ago

Oh also just paste the link I put there, I could not get it work :((

- 4 years, 7 months ago

Hi Igor,

Welcome to Brilliant! I just edited your discussion to add a direct link to your problem. You can see what I did by clicking the "edit this discussion" button on your post. Further info on using markdown formatting can be found here.

Staff - 4 years, 7 months ago

Thank you very much Mr. Taylor! I was directed here by a friend of mine, I can see what he meant by saying the people are friendly :))

- 4 years, 7 months ago

Hi! I am new here .Can u plz teel me how to post a problem that we are not able to solve?

- 4 years, 7 months ago