Find all solutions to the equation functional $f(x)+f(x+y)=y+2$
Letting $y=0$ one gets $2f(x) = 2 \Rightarrow f(x) = 1.$
I found this problem on aops and the proposed solution made the same observation and then stated
"Substituting this into the given functional equation yields $2=y+2$, which is not true for all $y$. Thus the given functional equation has no solutions."
However I don't see how they made the conclusion? Substituting this back gets
$1+f(x+y)=y+2$? If I would keep $y=0$ I would get $1+f(x) = 1+1 = 2$? What am I missing here?
We are trying to find functions ${f}$ that satisfy
$${f(x) + f(x+y) = y + 2}$$
for all ${x,y \in \mathbb{R}}$. Setting ${y=0}$ (which we are completely allowed to do as it should hold for any ${x,y \in \mathbb{R}}$) shows us, as you pointed out, that
$${f(x) = 1\ \forall\ x \in \mathbb{R}}$$
In particular, we will also then have that
$${f(x+y)=1\ \forall\ x,y \in \mathbb{R}}$$
So, you see that
$${f(x)+f(x+y)=y+2}$$
would imply that
$${1+1=y+2\Leftrightarrow 2=y+2}$$
... but there is an obvious problem with this. This equation is only satisfied for ${y=0}$, hence there is no function $f$ such that ${\forall\ x,y \in \mathbb{R}}$ we have
$${f(x)+f(x+y)=y+2}$$
It's not possible.