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

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.

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## Comments

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TopNewestPut \(y=0\) 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 \(\boxed{f(x)=x}\).

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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.

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

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Nice observations but it did not affect the solution greatly.

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Yeah just saw that now

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