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

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 Lakshya Sinha
9 months, 2 weeks ago

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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 \(\boxed{f(x)=x}\). Shivam Jadhav · 9 months, 2 weeks ago

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@Shivam Jadhav 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. Shrihari B · 9 months ago

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@Shivam Jadhav nice but typo.. if you put y=0 in the first line you would get \[f(xf(x))=f(0)+x^2\] Aareyan Manzoor · 9 months, 2 weeks ago

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@Aareyan Manzoor Nice observations but it did not affect the solution greatly. Shivam Jadhav · 9 months, 2 weeks ago

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@Aareyan Manzoor Yeah just saw that now Lakshya Sinha · 9 months, 2 weeks ago

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