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

or click here

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

#OlympiadMath #Competitions #MathProblem #Math

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
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Comments

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

Igor Filkovski - 7 years, 6 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.

Peter Taylor Staff - 7 years, 6 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 :))

Igor Filkovski - 7 years, 6 months ago

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

Anuva Agrawal - 7 years, 6 months ago

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.

Patrick Corn - 7 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$