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If f(x) is a polynomial satisfying $2 + f(x)f(y)=f(x)+f(y)+f(xy)$, find $f(f(2))$, given $f(2)=5.$

ATTEMPT:- $f(f(2))=f(5)$, We can find $f(0)$,$f(1)$ and $f(1/2)$ to be $1,2$ and $5/4$ respectively.

we can change the function to the form $g(x)*g(y)=g(xy)$ by basic transformation.

but how to get $f(5)$ without using transformations.

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  • $\begingroup$ How is $g(x)$ defined? And what is $g(2)$? $\endgroup$ – Hans Engler May 7 '15 at 0:25
  • $\begingroup$ Have you tried using the fact that you are given $f$ as a polynomial? $\endgroup$ – MCT May 7 '15 at 0:33
  • $\begingroup$ Why would you care about whether you use a transformation or not? $\endgroup$ – Thomas Andrews May 7 '15 at 0:33
  • $\begingroup$ #HansEngler we can change the given polynomial function to the form g(x)∗g(y)=g(xy) where g(x)=f(x)-1, g(y)=f(y)-1 and g(xy)=f(xy)-1 $\endgroup$ – yasir May 7 '15 at 9:44
  • $\begingroup$ #Thomas we can use transformation but there is always some other way round, may be even better than transformations. $\endgroup$ – yasir May 7 '15 at 9:45
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Hint: $(f(x)-1)(f(y)-1)=f(xy)-1$ so letting $h(x)=f(x)-1$ we have $h(x)h(y)=h(xy)$. Use that $h$ is a polynomial to show that $h(x)=x^n$ for some $n$.

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Notice that you can get values for $f(2n) = (2n)^2 + 1$. Observing this, I suppose that $f$ is a quadratic polynomial with form $f(t) = at^2 + bt + 1$ ($+1$ since we know that $f(0) = 1$).

Writing the functional equation as $(f(x) - 1)(f(y) - 1) = f(xy) - 1$, we get $(ax^2 + bx)(ay^2 + by) = (ax^2 y^2 + bxy)$, so comparing coefficients we get the following system:

$a^2 = a$

$ab = 0$

$b^2 = b$

So this gives answers of $(1, 0), (0,1),$ and $(0,0)$. This gives answers of $f(x) = 1, f(x) = x+1, f(x) = x^2 + 1$. Using our initial condition $f(2) = 5$, we know that only the last equation can be correct.

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Substitute $y = 2$ to get $2+f(x)f(2) = f(x)+f(2)+f(2x)$, which simplifies to $f(2x)-1 = 4(f(x)-1)$ for all reals $x$.

Now, suppose that $x_0 \neq 0$ is a zero of $f(x)-1$. Then $f(2x_0)-1 = 4(f(x_0)-1) = 4 \cdot 0 = 0$, and so, $2x_0$ is also a zero of $f(x)-1$. Continue this process indefinitely to get that $2^kx_0$ is a zero of $f(x)-1$ for all non-negative integers $k$. But $f(x)-1$ is a non-zero polynomial, and thus, can only have finitely many zeros, a contradiction.

Therefore, the only zero of $f(x) - 1$ is $0$, i.e. $f(x)-1 = Cx^n$ for some constant $C$ and some non-negative integer $n$.

Now, use the fact that $f(2x)-1 = 4(f(x)-1)$ to determine $n$, and then use the fact that $f(2) = 5$ to determine $C$.

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