Function in function

Play around with the given information and we find

$\begin{array}{ccc} & & f(1) = 4 \\ f(1 + f(1)) = 4f(1) & f(1 + 4) = 4\cdot 4 & f(5) = 16 \\ f(5 + f(5)) = 4f(5) & f(5 + 16) = 4\cdot 16 & f(21) = \boxed{64} \end{array}$

and that's all there is to it...

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In the same way we can show that $f\left(\frac{4^n - 1}3 \right) = f(1 + 4 + 4^2 + \cdots + 4^{n-1}) = 4^n;$ this leads to the conjecture that $f\left(\frac{y-1}3\right) = y;$ $f(x) = 3x + 1$

In that case we find indeed that $f(x + f(x)) = f(4x + 1) = 12x + 4 = 4(3x+1) = 4f(x).$

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Alternatively, we can assume that $f(x)$ is a polynomial function. Suppose its degree is one, i.e. $f(x) = ax + b$ .

From $f(1) = 4$ we have $a + b = 4$ .

Also $f(x + f(x)) = a(x + ax + b) + b = (a^2 + a)x + (ab + b);\ \ \ \ \ \ 4f(x) = 4ax + 4b.$ Equate the coefficients and substitute $b = 4-a$ : $a^2 + a = 4a\ \ \ \ \therefore\ \ \ \ a = 0 \ \ \ \text{or}\ \ \ \ a = 3.$ $ab + b = 4b\ \ \ \ \therefore\ \ \ \ b = 0 \ \ \ \text{or}\ \ \ \ a = 3.$

Since $a + b = 4$ , we must rule out $b = 0$ and are left with $a = 3$ , $b = 1$ .

Note: There may be other, non-linear functions that satisfy the given information! In fact, there are many non-continuous functions that satisfy. The only thing we really know about this function is that $f(x) = 3x + 1\ \ \ \ \ \ \ \text{for}\ x = 4^n,\ \ n\ \text{an integer}.$

$f(1+f(1)) = 4f(1) \Rightarrow f(5) = 16$ $f(5+f(5)) = 4f(5) \Rightarrow f(21) = 64$