Find $g$(3) if $g(x)g(y)=g(x)+g(y)+g(xy)-2$ and $g(2)=5$ 
If $g(x)$ is a polynomial function satisfying $g(x)g(y)=g(x)+g(y)+g(xy)-2$ for all real $x$ and $y$ and $g(2)=5$, then find $g$(3)

Putting $x=1,y=2$
$$
5g(1)=g(1)+5+5-2\implies \boxed{g(1)=2}
$$
And I think I can evaluate for $g(1/2)$ also.
Solution given in my reference is $g(3)=10$, but I think I am stuck with this.
 A: The point is that $g$ is a polynomial.
Thus write $g(x) = a_n x^n + \dotsc + a_0$, and you have an equality:
$$\sum_{i, j = 0}^n a_i a_j x^i y^j = (\sum_{i = 0}^n a_i (x^iy^i + x^i + y^i)) - 2.$$
Since this is true for all $x, y$, it means that it is an equality as polynomials in $x, y$.
By comparing coefficients, we see that $a_i a_j$ must be zero whenever $i > j > 0$ (because there is no $x^i y^j$ term on the right).
Therefore $g$ is of the form $ax^n + b$, and our equation becomes:
$$a^2x^ny^n + abx^n + aby^n + b^2 = ax^ny^n + b + ax^n + b + ay^n + b - 2,$$
from which we deduce that $a = 0$ or $a = 1$.
If $a = 0$, then $b^2 = 3b - 2$, so that $b = 1$ or $b = 2$: this contradicts $g(2) = 5$. 
The only possibility is then $a = 1$. Comparing coefficients before $x^n$, we have $ab = a$, which tells us $b = 1$.
It turns out that any polynomial $g$ of the form $x^n + 1$ will satisfy that equation for all $x, y$.
From $g(2) = 5$ we then conclude that $n = 2$, hence $g(3) = 3^2 + 1 = 10$.
A: Substitute $y=2$, then $$g(2x)=4g(x)-3.$$
Suppose $g(x)=x^2+1$; a result which is true for $x=1$ and $2$. 
Then $g(2x)=4g(x)-3=4x^2+1$. Therefore $g(x)=x^2+1$ for $x=1,2,4,8,16,..$.
Then the polynomial $g(x)-x^2-1=0$ has infinitely many roots and is therefore identically zero. 
Therefore $g(x)\equiv x^2+1$ and $g(3)=10.$
A: 
I'll compute not only $\,g(3)\,$ but simply whole $\,g.\,$
  Polynomial g is unique.
 

Substitute $\,y:=2\,$ into the given identity (remember that $\,g(2)=5).\,$ We obtain:
$$ g(2\cdot x)\,\,=\,\,4\cdot g(x) - 3 $$
hence
$$ g(0)=1\qquad\mbox{and}\qquad g(1)=2 $$
Next, apply the first two derivatives of the above identity (for
$\,x\,$ only):
$$ g'(2\cdot x)\,=\,2\cdot g'(x) $$
and
  $$ g''(2\cdot x)\,=\,g''(x) $$
This means that $\,g''\,$ is constant, i.e. $\, \deg(g)\le 2.\,$
Thus,
  $$ g(x)\,\,=\,\, a\cdot x^2\,+\,b\cdot x\,+\,c $$
for certain coefficients $\,a\,b\,c.$
From $\,g(0)=1\,$ we get $\,c=1\,$ hence
   $\,g(x)\,\,=\,\, a\cdot x^2\,+\,b\cdot x\,+\,1.$
Then $\, g(1)=2\,$ and $\,g(2)=5\,$ give
  $$ a+b=1 $$
and $\, 4\cdot a\,+\,2\cdot b = 4,\,$ i.e.
  $$ 2\cdot a\,+\,b = 2 $$
respectively. It follows that
  $$ a=1\qquad b=0\qquad c=1 $$
hence
$$ g(x)\,=\,x^2+1 $$
In particular
$$ g(3)\, =\, 10 $$
Great!

We can see value $\,a\,$ straight from the initial identity (in $x$ and $y$) that $\,a^2=a\,$ hence $\,a=1\,$ (compare the highest coefficients of the left and of the right side of the identity; I saw it from the start but... This $\,a=1\,$ would save me a bit of computation). 

A: If $f(x)=g(x)-1$ then we get $$f(x)f(y)=f(xy)$$ and $f(2)=4$. We are seeking for $f(3)+1$. 
We have $f(4)=16$ and $f(2)=f(2)f(1)$ so $f(1)=1$. We can see by induction now that $f(2^n) =4^n$, so $f(x)=x^2$ for infinitely $x$ so it is for all $x$ a squre and thus the result is $10$.
