ACCCGS8 wrote: Find all functions $ f:\mathbb{R}\rightarrow\mathbb{R} $ such that $f(x^2 - y^2) = (x+y)(f(x) - f(y))$ for all real numbers $x$ and $y$. x=y=0 ===>f(0)=0 , if x=0 ==>xf(x)=f(x²) , if x=0 : f (y)=-f(-y) , f(1)=0 x=-y ==> f(2x²)=4xf(x)=4f(x²) & it's easy to prove that: f(nx)=n²f(x) so the fonctionnal equation become : f(x²-y²) = f(x²)-f(y²)+yx²f(1)-xy²f(1) ==> f(x²-y²) = x²-y² if we replace f(x²-y²) by x²-y² in the fonctionnal equation we get f(x)=ax for all a£R

there is also another solution is: f(x)=0 we get it if we repalce n by x in this : f(nx)=n²f(x)

ryuzaki wrote: x=-y ==> f(2x²)=4xf(x)=4f(x²) Here is an obvious mistake, taking $x=-y$ will yield $f((-y)^2-y^2)=(-y+y)(f(-y)-f(y))$, which is equivalent to $f(0)=0$.

ACCCGS8 wrote: Find all functions $ f:\mathbb{R}\rightarrow\mathbb{R} $ such that $f(x^2 - y^2) = (x+y)(f(x) - f(y))$ for all real numbers $x$ and $y$. It's very easy one , first just putting $x=y=0$ get $f(0)=0$ now putting $y=0$ we conclude $f$ is odd. Now so we've $(x+y)(f(x)-f(y))=(x-y)(f(x)+f(y))\implies xf(y)=yf(x)\implies f(x)=cx$

TheBottle wrote: ryuzaki wrote: x=-y ==> f(2x²)=4xf(x)=4f(x²) Here is an obvious mistake, taking $x=-y$ will yield $f((-y)^2-y^2)=(-y+y)(f(-y)-f(y))$, which is equivalent to $f(0)=0$. I do not pay attention to the square

subham1729 wrote: xf(y)=yf(x)\implies f(x)=cx$ Sorry but can someone explain how to prove this. Thanks

Goikoetxea wrote: subham1729 wrote: $xf(y)=yf(x)\implies f(x)=cx$ Sorry but can someone explain how to prove this. Thanks Because, that is same as $\dfrac {f(x)}{x}=\dfrac {f(y)}{y}$; putting $y=1$ we have $\dfrac {f(x)}{x}=f(1)=c$.

Although it is pretty easy, I've got another cool approach: As we can rapidly check, $f(0)=0$ Lets, then, define $g(x) = f(x)-f(1)x$. then $g$ satisfies exactly the same equation that $f$ does, but $g(1)=0$. So, taking $x=-1$ along with $y=0$, $g(-1)=0$. So, taking in the equation, successively, $y=1$ and $y=-1$, we have that $g(x^2-1) = (x+1)g(x) = (x-1)g(x) \iff g(x) \equiv 0$. So, we are done.

Can someone help me prove that $f(nx)=n^2f(x)$? [No, since it's false ... in fact $f(nx) = nf(x)$]

$P(0,0) : f(0) = 0$. $P(x,0) : f(x^2) = xf(x)$ $(1)$ $P(0,x) : f(-x^2) = -xf(x)$ $(2)$ From $(1),(2)$ we have $f$ is odd function. $P(x,-y) : f(x^2 - y^2) = (x-y)(f(x) - f(-y)) = (x-y)(f(x) + f(y)) = (x+y)(f(x) - f(y)) \implies xf(y) = yf(x)$. Let $y = 1$ so we have $xf(1) = f(x) \implies f(x) = ax$. Answers : $f(x) = ax$.