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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.

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  • $\begingroup$ There is a bunch of duplicates on AoPS. One of them: artofproblemsolving.com/community/c6h1502528p8871300 $\endgroup$
    – VIVID
    Commented Oct 19, 2020 at 15:31
  • $\begingroup$ The link requires the function to be continuous while this question does not. $\endgroup$
    – cr001
    Commented Oct 19, 2020 at 15:33

1 Answer 1

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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.

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