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I've started studying functions and I am having trouble with the following question:

Find all solutions to the functional equation $f(x) +f(x+y)=y+2 $

Using the substitution technique when $y=0$ I have $f(x)=1$.

This implies that also $f(x+y)=1 $ and since $f(x)+f(x+y)=y+2$ , I am left with the conclusion that there are not solutions for the above functional equation.

Is this correct ?

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You are correct. Setting $y = 0$ gives us $$ \forall x \in \mathbb{R} : f(x) = 1$$ In particular $$ \forall y \in \mathbb{R} : 1 + 1 = y + 2 \iff y = 0$$ which clearly is a contradiction!

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  • $\begingroup$ Thanks for the check up.It follows that is a contradiction if we consider as a solution a function which satisfy the above condition for all pairs $(x,y)$, right ? $\endgroup$ – Mr. Y Dec 22 '15 at 12:55
  • $\begingroup$ Yes. In full rigor the idea is that if we assume there exists some solution $g(x)$ to the functional equation, we know that it must have the property $g(x) = 1$ for all x. It then follows that this equation does not satisfy the functional equation which is a contradiction, sicne we assumed that it does! $\endgroup$ – Kayle of the Creeks Dec 22 '15 at 12:57
  • $\begingroup$ Thank you.That cleared all my doubts.I will accept your answer in six minutes (reputation system). $\endgroup$ – Mr. Y Dec 22 '15 at 12:58
  • $\begingroup$ If you are really pedantic, the answer is that there is no solution over the reals. For example, the identity over the field of two elements $F_2$ does satisfy the equation. $\endgroup$ – Klaus Draeger Dec 22 '15 at 13:02

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