# find all the functions $f:\mathbb{R}→\mathbb{R}$ such that $f(1)=1$ and $f(xy+f(x))=xf(y)+f(x)$

here is what i did i found that $$f(f(0))=f(0)$$

i need to prove that f is injective so that $$f(0)$$ will be equal to $$0$$

Hence $$f(f(x))=xf(0)+f(x)$$ so if $$f(0)=0$$ and $$f$$ is injective $$f(x)=x$$ please don't give me the solution , just give me a hint to how to prove that $$f$$ is injective

Also , i found that $$f(x+f(x))=x+f(x)$$ i don't if that relation means that $$f$$ is injective

SORRY , GUYS! I FOUND A SOLUTION WITHOUT INJECTIVITY

$$x=1$$ and $$y=0$$ yields

$$f(f(1))=f(0)+f(1)$$

Use $$f(1)=1$$.

Added Setting $$y=0$$ you get $$f(f(x))=f(x)$$

Next, set $$y=\frac{1}{x}$$ and use $$f(f(x))=f(x)$$ and $$f(x+1)=f(x)+1$$.

• I think this still doesn't address injectivity does it? – DanielV Oct 10 '18 at 20:08
• yeah but now i have $f(x)=f(f(x))$ so i have to prove injectivity – user600785 Oct 10 '18 at 20:11
• and i also have $f(y+1)=f(y)+1$ – user600785 Oct 10 '18 at 20:13
• @user600785 Check the addition. – N. S. Oct 10 '18 at 20:52