F.E. from Fahim Vai

[Phlembac Adib Hasan]

Find all functions $f:\mathbb {R}^+\to \mathbb {R}$ such that
1.$\exists a\in \mathbb {R}^+$ such that $f(a)=1$
2.$f(x)f(y)+f\left (\frac {a}{x}\right )f\left (\frac {a}{y}\right )=2f(xy)$

[Phlembac Adib Hasan]

Solution:(Somebody check it, please!)

Let $P(x,y)\Rightarrow f(x)f(y)+f\left (\frac {a}{x}\right )f\left (\frac {a}{y}\right )=2f(xy)$
Take $P(1,1)$ to show $f(1)=1$.So $P(x,1)\Rightarrow f(x)=f(\frac {a}{x})$.Then $P(x,y)\Rightarrow f(xy)=f(x)f(y)$
So $f(1)=f(x\cdot \frac {a}{x})=f(x)f(\frac a x)=f(x)^2$.
$f(x)=-1\forall x\in \mathbb R^+$ is not possible due to $f(a)=1$.So all the functions are:
1.$f(x)=1\forall x\in \mathbb {R}^+$
2.$f(1)=f(a)=1\; and\; f(x)=1\; or\; -1\forall x\in \mathbb {R}^+\backslash \{1,a\}$
It's easy to see the first is a correct solution, but I am not sure about the second.(Although I found nothing wrong about it.)