គណិតវិទ្យា

លំហាត់ដកស្រងពី​​ AoPs:
Let \,{\mathbb{R}}\, denote the set of all real numbers. Find all functions \,f: {\mathbb{R}}\rightarrow {\mathbb{R}}\, such that f\left( x^{2}+f(y)\right) =y+\left( f(x)\right) ^{2}\hspace{0.2in}\text{for all}\,x,y\in \mathbb{R}.

                                Solution
The first step is to establish that f(0) = 0. Putting x = y = 0, and f(0) = t, we get f(t) = t^2.
Also, f(x^2+t) = f(x)^2, and f(f(x)) = x + t^2.
We now evaluate f(t^2+f(1)^2) two ways.
First, it is f(f(1)^2 + f(t)) = t + f(f(1))^2 = t + (1 + t^2)^2 = 1 + t + 2t^2 + t^4.
Second, it is f(t^2 + f(1 + t)) = 1 + t + f(t)^2 = 1 + t + t^4. So t = 0, as required.

It follows immediately that f(f(x)) = x, and f(x^2) = f(x)^2.
Given any y, let z = f(y). Then y = f(z), so f(x^2 + y) = z + f(x)^2 = f(y) + f(x)^2.
Now given any positive x, take z so that x = z^2.
Then f(x + y) = f(z^2 + y) = f(y) + f(z)^2 = f(y) + f(z^2) = f(x) + f(y).
Putting y = -x, we get 0 = f(0) = f(x + -x) = f(x) + f(-x). Hence f(-x) = - f(x).
It follows that f(x + y) = f(x) + f(y) and f(x - y) = f(x) - f(y) hold for all x, y.

Take any x.
Let f(x) = y. If y > x, then let z = y - x. f(z) = f(y - x) = f(y) - f(x) = x - y = -z.
If y < x, then let z = x - y and f(z) = f(x - y) = f(x) - f(y) = y - x.
In either case we get some z > 0 with f(z) = -z < 0.
But now take w so that w^2 = z, then f(z) = f(w^2) = f(w)^2 \ge 0.
Contradiction. So we must have f(x) = x 
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