If $f(x) $ be a polynomial function satisfying $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$ and $f(4) =65$ then find $f(6)$

Question

Problem :

If $f(x) $ be a polynomial function satisfying $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$ and $f(4) =65$ then find $f(6)$

Solution :

$f(x) f(\frac{1}{x})-f(x) =f(\frac{1}{x})$

$\Rightarrow f(x) =\frac{f(1/x)}{f(1/x)-1}$.....(i)

Also $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$

$\Rightarrow f(\frac{1}{x})=\frac{f(x)}{f(x)-1}$ ......(ii)

On multiplying (i) and (ii) , we get

$f(x) .f(\frac{1}{x})=\frac{f(1/x).f(x)}{(f(1/x)-1) ((f)(x)-1)}$

$\Rightarrow (f(\frac{1}{x}) -1)(f(x)-1)=1$

Please suggest how to proceed here since $f(x)-1 ; \& f(\frac{1}{x}-1)$ are reciprocal to each other Thanks

Answer 1

Let $f$ be an integer polynomial with $f(x)f(1/x) = f(x) + f(1/x)$. There are four solutions, $f = 0$, $f = 2$, and $f = 1 \pm x^n$.

Clearly $f = 0$ is a solution.

As in the question, we have:

$$ (f(x) - 1)(f(1/x) - 1) = 1 $$

Let $n$ be the degree of $f \neq 0$. Multiplying both sides of the above by $x^n$ gives:

$$ (f(x) - 1)(x^n f(1/x) - x^n) = x^n $$

Now both terms are now polynomials in $\mathbb Z [x]$ with degree $n$. Therefore, by unique prime factorization one of the factors on the left must be $x^n$ (up to units), and the other factor is some unit. This gives two possibilities, either:

$$ f(x) - 1 = \pm x^n $$ $$ f(x) = 1 \pm x^n $$

The other possibility:

$$ x^n f(1/x) - x^n = \pm x^n$$ $$ f(1/x) = 0, 2 \implies f(x) = 2 \text{ since $f \neq 0$}$$

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Applying this to your question, we get two possibilities for $f(x)$. Either:

$$ 4^n + 1 = 65 $$ $$ 4^n - 1 = 65 $$

Since $n$ has to be a natural number, the only solution is $n=3$, giving the polynomial $f(x) = x^3 + 1 $.

Answer 2

The equations are satisfied by $f(x) = x^3 + 1$. For: $$ f(x) f\left(\frac{1}{x}\right) = (x^3 + 1) \left(\frac{1}{x^3} + 1\right) = 1 + x^3 + \frac{1}{x^3} + 1 = f(x) + f \left(\frac{1}{x}\right) $$ and $f(4) = 4^3 + 1 = 65$. So $f(6) = 6^3 + 1 = 217$.

Answer 3

Let $f(x)=\sum_{k=0}^na_kx^k,f\neq0$, where $n$ is the degree of $f$, i.e. $a_n\neq0$. If $n=0$, then your equation is equivalent to $a_0^2=2a_0$, which implies $a_0=2$. So let's assume that $n\geq1$. Multiply both sides of your equation by $x^n$ and use the Cauchy product formula for finite sums. This yields after some simplifications \begin{align*} \sum_{j=0}^{n-k}a_ja_{j+k}&=a_k,\qquad\forall k\in\{1,\ldots,n\}, \\ \sum_{j=0}^na_j^2&=2a_0. \end{align*} For $k=n$ we get $a_0a_n=a_n$, and since $a_n\neq0$ we have $a_0=1$. For $k=n-1$ we get $a_0a_{n-1}+a_1a_n=a_{n-1}$, which is equivalent to $a_1a_n=0$ and hence $a_1=0$. For $k=n-2$ we obtain $a_0a_{n-2}+a_1a_{n-1}+a_2a_n=a_{n-2}$, thus $a_2a_n=0$ and finally $a_2=0$. Continuing this process gives $a_k=0$ for each $k\in\{1,\ldots,n-1\}$. From the last equality we then have that $1+a_n^2=2$, which implies $a_n=\pm1$. Consequently, $f$ is of the form $f(x)=1\pm x^n$.

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Using $f(4)=65$, we conclude $f(x)=x^3+1$ (see George V. Williams' answer).

Answer 4

An easy inspection shows that $f$ is nonconstant. Let $f(x)=x^ng(x)$, with $n\geq0$ and $g(0)\ne0$. Then $f(x)f(1/x)=g(x)g(1/x)$ and $f(x)+f(1/x)=x^ng(x)+\frac{g(1/x)}{x^n}$, hence $x^ng(x)g(1/x)=x^{2n}g(x)+g(1/x)$. Let $m$ be the degree of $g$, and consider the reciprocal polynomial of $g$, that is $x^mg(1/x)$, who also has degree $m$. We have

$$x^ng(x)x^mg(1/x)=x^{2n+m}g(x)+x^mg(1/x)\,.$$

Note that the LHS has degree $n+2m$, and the summands on the RHS have degrees $2n+2m$ and $m$ respectively. Now $2n+2m=m$ implies $n=0$, and $2n+2m>m$ implies that the degree of the RHS is equal to $2n+2m$, so $2n+2m=n+2m$ and again $n=0$. Thus $f(x)$ has the form $f(x)=a_0+a_1x+\cdots+a_mx^m$, with $m>0$ and $a_0a_m\ne0$, and the equality above becomes $f(x)h(x)=x^mf(x)+h(x)\,,$ where $h(x)=x^mf(1/x)=a_m+a_{m-1}x+\cdots+a_0x^m$. We can rewrite this equality as

$$\bigl(f(x)-1\bigr)h(x)=x^mf(x)\,.$$

Since $0$ is not a root of $h(x)$ then $x^m$ divides $f(x)-1$, and since both $x^m$ and $f(x)-1$ have degree $m$, it follows that $f(x)-1=cx^m$ for some $c\ne0$. Replacing again yields $c(x^m+c)=cx^m+1$, which implies $c=\pm1$. Finally $64=f(4)-1=4^mc$, which in turn forces $c=1$ and $m=3$. We conclude then that $f(x)=x^3+1$.