# Find all functions $f : \mathbb R \to \mathbb R$ such that $f \big( x + f ( y ) \big) = y + f ( x + 1 )$.

Find all functions $$f : \mathbb R \to \mathbb R$$ such that $$f \big( x + f ( y ) \big) = y + f ( x + 1 ) \text .$$

I managed to prove this function is injective by a contradiction. Then, putting $$y = 0$$: $$f \big( x + f ( 0 ) \big) = f ( x + 1 )$$ $$x + f ( 0 ) = x + 1$$ $$f ( 0 ) = 1$$ I also tried to check the cases when $$x = 0$$ and when both $$x$$ and $$y$$ are $$0$$, but I didn't get anything useful from that.

• There is a bunch of duplicates on AoPS. One of them: artofproblemsolving.com/community/c6h1502528p8871300 Commented Oct 19, 2020 at 15:31
• The link requires the function to be continuous while this question does not. Commented Oct 19, 2020 at 15:33

You can show that the functions $$f : \mathbb R \to \mathbb R$$ satisfying $$f \big( x + f ( y ) \big) = y + f ( x + 1 ) \tag 0 \label 0$$ for all $$x , y \in \mathbb R$$ are exactly those of the form $$f ( x ) = A ( x ) + 1$$, where $$A$$ is an additive involution, i.e. $$A ( x + y ) = A ( x ) + A ( y )$$ and $$A \big( A ( x ) \big) = x$$ for all $$x , y \in \mathbb R$$. If further regularity conditions like continuity, local boundedness, integrability or measurability are assumed for $$f$$, then $$A$$ will be regular, too. The only regular additive functions are those of the form $$A ( x ) = c x$$ for some constant $$c \in \mathbb R$$ (see here). Thus the only regular additive involutions are $$A ( x ) = x$$ and $$A ( x ) = - x$$, as we must have $$A \big( A ( 1 ) \big) = c ^ 2 = 1$$. Therefore, the only regular solutions to \eqref{0} are $$f ( x ) = x + 1$$ and $$f ( x ) = - x + 1$$. Without any further conditions on $$f$$, one can use the axiom of choice to show that there are nonregular solutions, too (see Example 5.6 in this PDF file).
It's straightforward to check that if $$f$$ is of the form $$f ( x ) = A ( x ) + 1$$ for some additive involution $$A$$, then it satisfies \eqref{0}. We try to prove the converse.
Letting $$x = 0$$ in \eqref{0} we get $$f \big( f ( y ) \big) = y + f ( 1 ) \text . \tag 1 \label 1$$ In particular, \eqref{1} shows that if $$f ( x ) = f ( y )$$ then $$x = y$$. Letting $$y = 0$$ in \eqref{1} shows that $$f \big( f ( 0 ) \big) = f ( 1 )$$, and thus by injectivity, $$f ( 0 ) = 1$$. Hence, putting $$x = - 1$$ in \eqref{0} we have $$f \big( f ( y ) - 1 \big) = y + 1 \text . \tag 2 \label 2$$ \eqref{2} shows that if we substitute $$x - 1$$ for $$x$$ and $$f ( y ) - 1$$ for $$y$$ in \eqref{0}, we get $$f ( x + y ) = f ( x ) + f ( y ) - 1 \text . \tag 3 \label 3$$ Now, if we define $$A : \mathbb R \to \mathbb R$$ with $$A ( x ) = f ( x ) - 1$$, then by \eqref{3} we can see that $$A$$ is additive, and by \eqref{2} we can see that $$A$$ is an involution. This proves what was desired.