[ [ Methods for solving problems with parameters $\quad]$

Given the quadratic trinomial $f(x)=x^{2}+a x+b$. It is known that for any real $x$ there exists a real $y$ such that $f(y)=f(x)+y$. Find the greatest possible value of $a$.

#
From the condition, it follows that the quadratic equation $f(y)-y-f(x)=0$ is solvable with respect to $y$ for any $x$. Substituting $x=-\frac{a}{2}$, we get the equation
$y^{2}+(a-1) y+a^{2} / 4$, the discriminant of which is $(a-1)^{2}-a^{2}=1-2 a$, from which $a \leq 1 / 2$. On the other hand, if $a$ $=1 / 2$, then for any $x$ we can set $y=-x$: then $f(y)=x^{2}-1 / 2 x+b=\left(x^{2}+1 / 2 x+b\right)-x=f(x)+y$, as required.

## Answer

$a=1 / 2$.