Is it mainstream?

Let $f(x) = \displaystyle \sum_{i=0}^n a_ix^i$ where $a_n \neq 0$ . In this form the condition above looks like $\displaystyle \sum_{i=0}^n a_ix^i + \displaystyle \sum_{i=0}^n a_ix^{-i} = (\displaystyle \sum_{i=0}^n a_ix^i) \cdot (\displaystyle \sum_{i=0}^n a_ix^{-i})$ .

Clearly f is not a constant function, otherwise we would have $f(x) = f(7) = 2402$ and $2 \cdot 2402 = 2402^2$ from the condition, which is a contradiction. This means that $n \geq 1$ .

Observe that on the left side for every $i \geq 0$ the coefficient of $x^i$ is $a_i$ . If we evaluate the product on the right side we can see that for an $i \geq 0$ the exponent of x in the result will be $i$ if and only if we multiply together the terms in the form $a_{k+i}x^{k+i}$ and $a_kx^{-k}$ for $0 \leq k \leq n-i$ . Thus for every $i \geq 0$ the coefficient of $x^i$ will be $\displaystyle \sum_{k=0}^{n-i} a_{k+i}a_k$ . Two polynomials are equal if and only if all the respective coefficients are equal, hence we obtain $a_i = \displaystyle \sum_{k=0}^{n-i} a_{k+i}a_k$ for every $0 \leq i \leq n$ . (1)

Considering the last equation with $i = n$ we get $a_n = a_na_0$ , which implies $a_0 = 1$ . Next we are going to show by induction that $a_i = 0$ for every $0 < i < n$ . Using equation (1) for $i = n-1$ we obtain $a_{n-1} = a_na_1 + a_{n-1}a_0$ . From $a_0 = 1$ we can deduce that $a_na_1 = 0$ hence $a_1 = 0$ since $a_n \neq 0$ . Assume by induction that our claim is true for every $0 < i < j$ . Then applying (1) to $n-j$ leads us to $a_{n-j} = \displaystyle \sum_{k=0}^{j} a_{k+n-j}a_k$ where our induction hypothesis yields the terms for $0 < k < j$ to be 0, so the equation can be simplified into the form $a_{n-j} = a_na_j + a_{n-j}$ which implies $a_j = 0$ in view of $a_n \neq 0$ . Henceforth our induction is complete.

We have shown that $a_0 = 1$ and $a_i = 0$ for $0 < i < n$ so our function is in the form $a_nx^n+1$ . Then from the original condition we obtain $a_nx^n+1 + a_nx^{-n}+1 = (a_nx^n+1)(a_nx^{-n}+1)$ $\Rightarrow$ $a_nx^n + a_nx^{-n} + 2 = a_n^2 + a_nx^n + a_nx^{-n} + 1$ $\Rightarrow$ $a_n^2 = 1$ $\Rightarrow$ $a_n = \pm 1$

Hence $f(x) = \pm x^n + 1$ . Assuming $f(x) = -x^n + 1$ we would get $2402 = f(7) = -7^n + 1 \leq -6 < 2402$ by $n \geq 1$ so $f(x) = x^n + 1$ . For x = 7 we have $7^n + 1 = f(7) = 2042$ which implies that n = 4 so $f(x) = x^4 + 1$ . Putting in x = 11 we obtain the final answer to be 14642.

[This is an almost complete solution. Looking over Sandor's solution, it is equivalent to his, so you can view the details there.]

Let $g(x) = f(x) - 1$ . We have

$1 = f(x) f( \frac{1}{x}) - f(x) - f( \frac{1}{x} ) + 1 = [ f(x) - 1 ] [ f( \frac{1}{x} ) -1 ] = g(x) g( \frac{1}{x} ).$

Claim: The only polynomial that satisfies this condition is $g (x) = \pm x^n$ .

Proof: Let $g(x) = \sum_{i=0}^n a_i x^i$ with $a_n \neq 0$ .
We have $( \sum a_i x^i ) \times ( \sum a_i x^{-i} ) = 1 \times x^0$ .
Compare the coefficients $x^n, x^{n-1} , \ldots , x^0$ .
Conclude that $a_0, a_1, a_2, \ldots, a_{n-1} = 0$ and $a_n = \pm1$ .

By inspection, we can see the following type of polynomial functions satisfy the given condition. $f(x)=1+x^k$ Now, $f(7)=1+7^k$ $\implies 7^k=2401$ $\implies k=4$ Therefore, $f(11)=11^4+1=14642$

[P.S. I'm not sure about the uniqueness of this solution. If you find other functions that work as well, please let me know.]

We note that if $\alpha$ is a zero of $f(x)$ then so is $\frac{1}{\alpha}.$ Therefore $f(x)$ is a reciprocal polynomial (A polynomial is called reciprocal if $a_i=a_{n-i}$ where $a_i$ are coefficients of the polynomial and $n$ is the degree.) whose degree is $2n$ as it should have even number of terms. Now every reciprocal polynomial can be written as $f(x)=x^n g(x+\frac{1}{x}),$ putting this in the given relation and after reducing it we get $g(x+\frac{1}{x})\{g(x+\frac{1}{x})- x^n-\frac{1}{x^n}\}=0.$ Since $f(7)=2402$ , hence $g(x+\frac{1}{x})= x^n+\frac{1}{x^n}.$ Therefore $f(x)=x^{2n}+1$ , and using $f(7)=2402$ we get $f(x)=x^4+1.$ Thus $f(11)=14642.$