$f:\mathbb{R}\rightarrow \mathbb{R}$ satisfies the equation \[f(x)f(y)-af(xy)=x+y\] , for every real numbers $x,y$. Find all possible real values of $a$.
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Tags: algebra, polynomial
r31415
18.01.2013 21:27
Letting x,y=0 gives $f(0)=a$ or $f(0)=0$. Plugging in x=0, y=y gives $f(0)f(y)-af(0)=y$. Therefore, we can see that if f(0)=0, it cannot satisfy all y. Substituting $f(0)=a$ gives $f(y)=\frac {a^2+y} {a} $, and substituting this into the original equation gives $(\frac {a^2+x} {a} )(\frac {a^2+y} {a} )-a (\frac {a^2+xy} {a}) =x+y$, and expressing this as a polynomial in a gives
\[(x+y)a^2-(xy+x+y)^2+xy=0\]
\[a=\frac {xy+x+y \pm \sqrt {(xy+x+y)^2-4xy(x+y)}} {2x+2y}\]
\[a= \frac {xy+x+y \pm (x+y-xy)} {2x+2y}\]
Therefore, a=2 or $\frac {xy} {x+y}$, and the latter is clearly impossible. $\boxed {a=2} $.
Mr.A
07.05.2016 13:39
Letting x,y=0 gives $f(0)=a$ or $f(0)=0$. Plugging in x=0, y=y gives $f(0)f(y)-af(0)=y$. Therefore, we can see that if f(0)=0, it cannot satisfy all y. Substituting $f(0)=a$ gives $f(y)=\frac {a^2+y} {a} $, and substituting this into the original equation gives $(\frac {a^2+x} {a} )(\frac {a^2+y} {a} )-a (\frac {a^2+xy} {a}) =x+y$, $\Rightarrow$ $(1-a)(\frac{xy}{a}-x-y)=0$ Plugging $x=1,y=0$ $(a-1)=0$ Thus a=1 Answer:a=1
tugra_ozbey_eratli
07.03.2024 22:25
$P(0,0)$ gives $f(0)^2-af(0)=0=f(0)(f(0)-a)=>f(0)=0$ or $f(0)=a$ $P(x,0)$ gives $f(x)f(0)-af(0)=x$ if $f(0)=0$ this gives a contradiction so $f(0)=a$ $f(0)=a=>f(x)f(0)-af(0)=x=af(x)-a^2=>f(x)=a+\frac x{a}$ $P(a,a)$ gives $f(a)^2-af(a^2)=2a=(a+1)^2-2a^2=2a=>a^2=1=>a=1,-1$ if $a=1=>f(x)=x+1$ if $a=-1=>f(x)=-x-1$ Clearly, these functions works so Answer: $\boxed {a=1,-1}$