A Cool Functional Equation

Let $f(x) = a_nx^n + ... + a_2x^2 + a_1x + a_0$ . Since $f(0) = 1$ , $\Rightarrow \color{#3D99F6}{a_0 = 1}$ , $\Rightarrow f(x) = a_nx^n + ... + a_2x^2 + a_1x + \color{#3D99F6}{1}$ .

$\begin{aligned} f(x) f(2x^2) & = f(2x^3+x) \\ (a_nx^n + ... + 1) (a_n(2^nx^{2n}) + ... + 1) & = a_n(2x^3+x)^n + ... + 1 \\ 2 \color{#3D99F6} {a_n^2} x^{3n} + ... + 1 & = 2\color{#3D99F6} {a_n} x^{3n} + ... + 1 \end{aligned}$

This implies that $a_n^2 = a_n$ , $\Rightarrow a_n = 1$ , $\Rightarrow f(x) = x^n + ... + a_2x^2 + a_1x + 1$ .

From $f(x) f(2x^2) = f(2x^3+x)$ ,

$\begin{aligned} \Rightarrow f(2x^2) & = \frac {f(2x^3+x)} {f(x)} \\ f(2(-x)^2) & = \frac {f(2[-x]^3+[-x])} {f(-x)} \\ f(2x^2) & = \frac {f(-[2x^3+x])} {f(-x)} \\ \Rightarrow \frac {f(2x^3+x)} {f(x)} & = \frac {f(-[2x^3+x])} {f(-x)} \end{aligned}$

This implies that $f(x)$ is even and $a_k = 0$ for $k$ is odd; $\Rightarrow f(x) = a_nx^n + ... + a_4x^4 + a_2x^2 + 1$ , where $n$ is even.

For $n=2$ , then $f(x) = x^2 +1$ and:

$\begin{aligned} \Rightarrow f(2) + f(3) & = (2^2+1) + (3^2+1) = 15 \ne 125 \quad \Rightarrow n \ne 2 \end{aligned}$

For $n=4$ , then $f(x) = x^4 + a_2x^2+1$ and:

$\begin{aligned} \Rightarrow f(2) + f(3) & = (2^4 + 2^2a_2+1) + (3^4+3^2a_2+1) = 99 + 13a_2 = 125 \\ \Rightarrow a_2 & = \frac{125-99} {13} = 2 \end{aligned}$

$\Rightarrow f(x) = x^4 + 2x^2 + 1 = (x^2+1)^2$ .

Now let us check if $f(x) f(2x^2) = f(2x^3+x)$ is satisfied.

$\begin{aligned} \Rightarrow LHS & = f(x) f(2x^2) \\ & = (x^2+1)^ 2 (4x^4+1)^2 \\ & = (4x^6 + 4x^4 + x^2 + 1)^2 \\ & = ([2x^3+x]^2+1)^2 \\ & = f(2x^3+x) = RHS \end{aligned}$

Therefore, $f(5) = (5^2+1)^2 = 26^2 = \boxed{676}$

Let $r$ be a real root of $f(x)$ . This means that $f(r) = 0$ and implies, because of the last condition, that $f(2r^3+r) = 0$ which means $2r^3+r$ is also a real root. However you can do this infinitely and since the polynomial has finite degree, there must be no real roots.

Now after experimenting, it seems that the usual values of $0, 1, ...$ don't seem to work, and we sigh because we are tired, and bored.

However, then we think of trying nonreal numbers, more specifically, $\pm i$ !

Plugging in $\pm i$ gets us:

$f(i) f(-2) = f(-i)$

$f(-i)f(-2) = f(i)$

Substituting gives us that $f(i) = f(-i) = 0$ since $f(x)$ has no real roots.

We can rewrite $f$ as $f(x) = (x^2+1)Q(x)$

Now we substitute this into $f(x) f(2x^2) = f(2x^3+x)$ and we get.... the exact same identity!

This implies we can keep on squeezing factors of $i$ and $-i$ out until we get $f(x) = (x^2+1)^n$ for some integer $n$ .

We see that the $f(0) = 1$ condition is fulfilled.

We now use the $f(2) + f(3) = 125$ condition to get that $5^n + 10^n = 125$

By looking at the equation, it is easy to see the solution: $n = 2$ .

Finally we have our function $f$ :

$f(x) = (x^2 + 1)^2 \implies f(5) = 26^2 = \boxed{676}$