An algebra problem by Rahil Sehgal

We have

$f(x+y) = f(x) f(y)$

Setting $x=a$ and $y = \{1,2,3,\cdots,n\}$ we find that

$\begin{aligned} f(a+1) &= f(a) f(1) &= 2 f(a) \\ f(a+2) &= f(a+1) f(1) &= 4 f(a) \\ f(a+3) &= f(a+2) f(1) &= 8 f(a) \\ \vdots \\ f(a+k) &= f(a+k-1) f(1) &= 2^k f(a) \end{aligned}$

Thus

$\begin{aligned} \displaystyle \sum_{k=1}^n f(a+k) &= f(a) \sum_{k=1}^n 2^k \\ &= 2 f(a) \left( 2^n - 1 \right) \end{aligned}$

Hence

$2 f(a) \left( 2^n - 1 \right) = 16 \left( 2^n - 1 \right) \implies f(a) = 8 = 2^2 \times 2 = 2^2 f(1)$

It was shown that

$f(a+k) = 2^k f(a)$

Setting $k=2$ and $a=1$ gives

$f(3) = 2^2 f(1) = 8$

Hence $a=\boxed{3}$ .