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Does the equation $$f(x)+g(y)=x^2+xy+y^2 \mbox{ } \forall x,y \in \mathbb{R}$$ have solutions in real functions $f$ and $g$?

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    $\begingroup$ @RaziehNoori Your argument requires, e.g., that $f$ is differentiable. $\endgroup$ – Travis Willse Nov 27 '15 at 18:21
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Assume that such functions exist. From the given condition we have

$$f(0) + g(0) = 0$$ $$f(x) + g(0) = x^2 \rightarrow f(x) = x^2 + f(0)$$ $$f(0) + g(y) = y^2 \rightarrow g(y) = y^2 + g(0)$$

Assert this into the first identity we conclue $f(0) + g(0) = xy, \forall x, y \in \mathbb{R}$. The LHS is a constant, while the RHS can be an arbitrary real number. This leads to a contradiction, hence no such functions exist.

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No. Assume for contradiction that such $f(x), g(y)$ exist and let $P(x,y)$ be the statement $f(x)+g(y)=x^2+xy+y^2$. Then

$P(0,0)\implies g(0)=-f(0)$.

$P(x,0)\implies f(x)=x^2+f(0)$.

Similarly, $g(y)=y^2+g(0)$. But then $f(x)+g(y)=x^2+(f(0)+g(0))+y^2$ and $P(x,y)$ gives $f(0)+g(0)=xy,\, \forall x,y\in\Bbb R$, contradiction.

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