1. Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ that satisfy the conditions
a) $f(x+f(y))=f(x+y)+1$ for all $x, y \in \mathbb{R}$,

b) $f$ is a strictly increasing function.
Solution. From condition a), it follows that for any real numbers $x$ and $y$,

$$
f(x+f(y))=f(x+y)+1=f(y+x)+1=f(y+f(x))
$$

Furthermore, according to b), the function is strictly increasing, and since every strictly increasing function is injective, from (1) it follows that $x+f(y)=y+f(x)$. Specifically, if we substitute $y=0$ in the last equation, we get $f(x)=x+f(0)$. Again, from condition a), we obtain $x+f(y)+f(0)=x+y+f(0)+1$, which simplifies to $f(y)=y+1$ for any real number $y$. Clearly, the last function satisfies both conditions of the problem.