Problems from Ukrainian Authors

Let us denote the equation given in the statement of a problem by $(\ast )$ and substitute $y=-x$ into it. We obtain $$f^2(x)+f^2(-x)+2f(-x^2)=f^2(0)\text{{ (**).}}$$ Substituting $x=x+y,y=-x$ into $(\ast )$, we get $$f^2(y)=f^2(x+y)+f^2(-x)+2f(-x^2-xy).$$ Adding $(\ast )$ and $(\ast \ast )$ to the last-mentioned equality, we obtain $$f(xy)+f(-x^2-xy)=f(-x^2)-f^2(0)/2.$$ Let us denote $g(x)=f(x)+f^2(0)/2$. Let $a=-x^2\le 0,b=xy\in \mathbb{R}$ when $a\ne 0$. The last-mentioned equality assumes the form $$g(a)+g(b)=g(a+b)\text{{ (***)}.}$$ If $a,b<0$, then we obtain the Cauchy's equation. Let us prove that $g(x)$ is bounded above when $x<0$. Then from the Cauchy's equation we get that $g(x)=kx$ for arbitrary negative number $x$ and real number $k$. From $(\ast \ast )$ we get

Thus, $g(x)=kx\forall x<0$. Let us assume that $x\ge 0$. Substituting $y=-1-x$ into $(\ast \ast \ast )$, we obtain $g(x)=g(-1)-g(-1-x)=kx$ for every positive number $x$. Thus, we proved that $$f(x)=g(x)-f^2(0)/2=kx-f^2(0)/2=kx+c$$ for every real $x$. Substituting the explicit form of the function $f$ into $(\ast )$, we obtain: $$\begin{array}{rl}{k}^2(x+y{)}^2+2kc(x+y)+c^2& =k^2{x}^2+2kcx+c^2+2kxy+2c+k^2{y}^2+2kcy+c^2,\\ (k^2-k)xy& =c^2+2c.\end{array}$$ Considering that $xy$ is able to possess arbitrary real value, from the last-mentioned equality we get that $k\in \{0,1\}$, $c\in \{0,-2\}$. Thus, we obtain that the condition can be fulfilled only by the functions $f(x)=0$, $f(x)=-2$, $f(x)=x$, and $f(x)=x-2$. We verify that all these functions fulfill the condition of the problem by checking.