How to find the all continuous functions $f\colon \mathbb{R}\to\mathbb{R}$ such that $f(x)+f(y)+f(z)+f(x+y+z)=f(x+y)+f(y+z)+f(z+x)$
The functional equation $f(x)+f(y)+f(z)+f(x+y+z)=f(x+y)+f(y+z)+f(z+x)$
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calculus
functional-equations
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4It's easy to see that $f(0)=0$. The solutions form a linear space, and include $f(x)=x$ and $f(x) = x^2$ but no other powers. I would guess that $f(x) = ax + bx^2$ is the general solution. – 2012-08-23
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3A more descriptive title would be useful: as it stands, it is almost unconnected to the question! – 2012-08-23
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3Setting $x=y=z$ and differentiating the equality many times and evaluating at zero shows Robert's solutions are all the entire ones. – 2012-08-23
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0@MarianoSuárez-Alvarez Is it obvious then any continuous function satisfying that equality is differentiable? – 2012-08-23
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0If it were obvious I would have said «Robert's solutions are all solutions» and not what I wrote! – 2012-08-23