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```
f:R→R f((x-y) 2 )= (f(x)) 2 -2xf(y) +y 2 then f(x)=?

```
11 years ago

```							Setting x = y = 0, we get f(0) = f2(0) which means f(0) = 0 or f(0) = 1

Swapping x and y we get

f[(y-x)2] = f[(x-y)2] = f2(y) - 2y f(x) + x2

Hence we get f2(x) - 2x f(y) + y2 = f2(y) - 2y f(x) + x2

Now putting y = 0, we get f2(x) - 2x f(0)  = f2(0) - x2

If f(0) = 0, then f2(x) = x2 and we get f(x) = x after discarding f(x) = -x as extraneous

Likewise, if f(0) = 1, then f(x) = x+1 satisfies the given equation
```
11 years ago
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