# For IMO

$\large{ f(xf(x + y)) = f(yf(x)) + x^2}$

Find all functions $$f$$ from the set of real numbers into the set of real numbers which satisfy for all real numbers $$x,y$$ and the equation above.

Note by Department 8
2 years, 8 months ago

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## Comments

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Put $$y=0$$ in the given functional equation.

we get $\boxed{f(xf(x))=f(0)+x^{2}}...........(1)$ Now put $$x=-x$$ in the given functional equation .

We get $\boxed{f(-xf(-x)=f(0)+x^{2}}...........(2)$

Equating $$(1),(2)$$ we get

$f(-xf(-x))=f(xf(x))$ $-xf(-x)=xf(x)$ $\boxed{f(x)+f(-x)=0}..........(3)$

Now by putting $$x=0$$ we get $$\boxed{f(0)=0}........(4)$$. Now equations $$(1),(2)$$ transform into

$\boxed{f(xf(x))=x^{2}}...........(I)$

$\boxed{f(xf(x))=x^{2}}...........(II)$

Our main functional equation is $$f(xf(x+y))=f(yf(x))+x^{2}$$

This implies that degree of $$f(yf(x)),f(xf(x+y))$$ is $$2$$.

This implies that degree of $$f(x),f(x+y)=1$$.

Therefore we can write $$f(x)=ax+b$$ where $$a,b$$ are real numbers.

But $$f(0)=0$$ therefore $$b=0$$ . therefore $$f(x)=ax$$ . But for $$f(x)=ax$$ to satisfy $$(I)$$ $$a=1$$ is a must condition .

Therefore our final function reduces to $$f(x)=x$$.

Therefore the required function is $$\boxed{f(x)=x}$$.

- 2 years, 8 months ago

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How can u say that in the line above (3) ? I dont think that it is obvious that the given functional equation is injective. If yes then please prove it Also I dont see how do u get f(0)=0. Also arguments like degree of so and so .... are not allowed. What is the function is not a polynomial ? If it is something like exponential then argument fails. Please check the solution again.

- 2 years, 7 months ago

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nice but typo.. if you put y=0 in the first line you would get $f(xf(x))=f(0)+x^2$

- 2 years, 8 months ago

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Nice observations but it did not affect the solution greatly.

- 2 years, 8 months ago

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Yeah just saw that now

- 2 years, 8 months ago

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