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

http://prntscr.com/27jqyk

or click here

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

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

http://prntscr.com/27jqyk

or click here

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

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TopNewestWell, 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. – Patrick Corn · 3 years, 8 months ago

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Oh also just paste the link I put there, I could not get it work :(( – Igor Filkovski · 3 years, 8 months ago

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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. – Peter Taylor Staff · 3 years, 8 months ago

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– Igor Filkovski · 3 years, 8 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 :))Log in to reply

– Anuva Agrawal · 3 years, 8 months ago

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