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

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 ?

## 1 Answer

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!

• 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 ? – Mr. Y Dec 22 '15 at 12:55
• 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! – Kayle of the Creeks Dec 22 '15 at 12:57
• Thank you.That cleared all my doubts.I will accept your answer in six minutes (reputation system). – Mr. Y Dec 22 '15 at 12:58
• 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. – Klaus Draeger Dec 22 '15 at 13:02