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Question

Find all functions f:RR such that f(x+y)f(xy)=(f(x)+f(y))24x2f(y)..........(1)
for all x,yR, where R denotes the set of all real numbers.

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Solution

Put x=y=0 ; we get f(0)2=4f(0)2 and hence f(0)=0.
Put x=y; we get 4f(x)24x2f(x)=0 for all x.
Hence for each x, either f(x)=0 or f(x)=x2.
Suppose f(x)0.
Then we can find x00 such that f(x0)0.
Then f(x0)=x200.
Assume that there exists some y00 such that f(y0)=0.
Then f(x0+y0)f(x0y0)=f(x0)2.
Now f(x0+y0)f(x0y0)=0 or f(x0+y0)f(x0y0)=(x0+y0)2(x0y0)2.
If f(x0+y0)f(x0y0)=0, then f(x0)=0, a contradiction.
Hence it must be the later so that (x20y20)2=x40.
This reduces to y20(y202x20)=0. Since y00, we get y0=±2x0.
Suppose y0=2x0.
Put x=2x0andy=x0 in (1);
we get f((2+1)x0)f((21)x0)=(f(2x0)+f(x0))24(2x20)f(x0).
But f(2x0)=f(y0)=0.
Thus we get f((2+1)x0)f((21)x0)=f(x0)28x20f(x0)
=x408x40=7x40.
Now if LHS is equal to 0, we get x0=0, a contradiction.
Otherwise LHS is equal to (2+1)2(21)2x40 which reduces to x40.
x40=7x40 and this forces again x0=0.
Hence there is no y0 such that f(y)=0.
f(x)=x2 for all x.
Thus there are two solutions : f(x)=0 for all x or f(x)=x2, for all x.

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