3

Find all functions, $f:\mathbb{R}\to \mathbb{R}$ such that $$f(xf(y)+y)=f(xy)+f(y)$$

What i have got till now is

Let $$P(x,y): f(xf(y)+y)=f(xy)+f(y)$$Then \begin{align*} &P(0,0): f(0)=f(0)+f(0)\implies f(0)=0\\ &P\Bigg(\frac{y}{y-f(y)},y\Bigg): f(y)=0 \end{align*} Hence $f(x)=0$ is a solution.

But we can see $f(x)=x$ is also a solution but I can't find that.

Soham Chatterjee
  • 1,236

1 Answers1

4

For all $y$ such that $f(y) \ne y$, replace $(x,y) = (\frac{y}{y-f(y)},y)$ in the equation, we have that $f(y) = 0$. Indeed, $$f(\frac{y}{y-f(y)}f(y)+y) = f(\frac{y}{y-f(y)}y)+f(y)$$ $$\iff f(\frac{y^2}{y-f(y)}) = f(\frac{y}{y-f(y)}y)+f(y)$$ $$\iff 0 = f(y)$$

So, we have for all $x$, $f(x) = 0$ or $f(x) =x$.

If $\exists y_0 \ne 0$ such that $f(y_0) = 0$

Then we have, for all $x$ $$f(xf(y_0)+y_0)=f(xy_0)+f(y_0)$$ $$\iff f(x.0+y_0)=f(xy_0)+0$$ $$\iff 0=f(xy_0)$$ Hence, for all $x\in \mathbb{R}$: $f(x)=0$

If $\not\exists y_0 \ne 0$ such that $f(y_0) = 0$

Then for all $x\ne0$, $f(x) \ne 0$. Because $f(x)$ can only receive value in $\{0,x \}$, then $f(x) = x$ for all $x$

Conclusion

There are 2 solutions: $f(x) = 0$ or $f(x) = x$.

NN2
  • 15,892