SinaQane wrote: For a polynomial $P(x,y)$ with real coefficients, there exists a function $f(x)$ such that for all real numbers $x, y$ we have: $$ P(x, y) = f(x + y) - f(x) - f(y). $$Prove that $f(x)$ is a polynomial as well. Trivially wrong since "$P(x,y)=f(x+y)-f(x)-f(y)$" implies "$P(x,y)=g(x+y)-g(x)-g(y)$", for any $g(x)=f(x)+$any non linear additive function And obviously both $f(x)$ and $g(x)$ can not be polynomial at the same time

I have attached the Russian document maybe it's: "Prove that a polynomial $q$ can be subtracted form $f$."

Attachments:

@above edited it.

SinaQane wrote: For a polynomial $P(x,y)$ with real coefficients, there exists a function $f(x)$ such that for all real numbers $x, y$ we have: $$ P(x, y) = f(x + y) - f(x) - f(y). $$Prove that a polynomial $q$ can be subtracted form $f$. The last line isn't really intelligible. The correct translation is as follows: Given a polynomial $P(x,y)$ with real coefficients, suppose that some real function $f:\mathbb R \to \mathbb R$ satisfies $$P(x,y) = f(x+y)-f(x)-f(y)$$for all $x,y\in\mathbb R$. Show that some polynomial $q$ satisfies $$P(x,y) = q(x+y)-q(x)-q(y)$$

@above Thanks a lot. I'll edit it right away.

matinyousefi wrote: I have attached the Russian document maybe it's: "Prove that a polynomial $q$ can be subtracted form $f$." Could you share the book please (I know russian, no need to look for english resource)

CROWmatician wrote: matinyousefi wrote: I have attached the Russian document maybe it's: "Prove that a polynomial $q$ can be subtracted form $f$." Could you share the book please (I know russian, no need to look for english resource) But what I have attached is Russian.

Here is the correct translation: Quote: Given a polynomial $P(x,y)$ with real coefficients, suppose that some real function $f:\mathbb R \to \mathbb R$ satisfies $$P(x,y) = f(x+y)-f(x)-f(y)$$for all $x,y\in\mathbb R$. Show that there exists some polynomial $q$ such that $f(x) = q(x)$ for infinitely many x.

It was posted (and solved) here.