How to find the all functions $f$ :$ \mathbb{R}\longrightarrow\mathbb{R}$ such that $f(2011)=2012$,for every $x,y\in\mathbb{R}$ then: $$f(4xy)=2yf(x+y)+f(x-y)$$
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7$\begingroup$ Plugging in $y=0$ gives $f(0)=f(x)$... $\endgroup$– anonCommented Jul 13, 2012 at 11:28
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$\begingroup$ Then f(0)=2012 when taking x=2011 => f(x) = 2012 for every x. $\endgroup$– superMCommented Jul 13, 2012 at 11:31
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1 Answer
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If such a function exists, then $f(x)=f(0) \ne 0$ for every $x$, therefore $f(0)=2yf(0)+f(0)$ for every $y$, i.e. $y=0$ for every $y$. Thus such a function does not exist.
answered Jul 13, 2012 at 11:35