Find a solution for the equation $a(4n) = {a(n)\over 4}$

Question

I was given the equation $a(4n) = {a(n)\over 4}$ where $a(1) = 1$. I know that ${1\over \sqrt n }$ solves this equation, but I don't know how I would find this solution by hand if I didn't know about it.
Any hints on this matter are greatly appreciated.
EDIT: oh snap I made a typo. It is $a(4n) = {a(n)\over 2}$. I apologize for my mistake.

I'm not sure why my question was put on hold. To clarify, I was asking how one would systematically solve the recursive equation given above, as I was only able to see the solution, but not how one would find it.
Anyway, my question has been answered by eranreches (do I need to something else apart from selecting the best answer to mark the question as answered? -- if I have to, I'm sorry, it is my first time posting here...)

Answer 1

If you want a more systematic way, define

$$b_{n}\equiv a_{4^{n}}$$

Then

$$b_{0}=a_{1}=1$$

$$b_{n+1}=a_{4^{n+1}}=a_{4\cdot4^{n}}=\frac{a_{4^{n}}}{2}=\frac{b_{n}}{2}$$

with a solution $b_{n}=\frac{1}{2^{n}}$. Now reverse to get

$$a_{n}=b_{\log_{4}n}=\frac{1}{2^{\log_{4}n}}=\frac{1}{4^{\frac{1}{2}\log_{4}n}}=\frac{1}{n^{\frac{1}{2}}}=\frac{1}{\sqrt{n}}$$

as wanted.

Answer 2

Let $n=4^k$, so that $4n=4^{k+1}$, and let $b(k):=a(2^k)=a(n)$.

Now

$$b(k+1)=\frac{b(k)}2,$$ which is an ordinary recurrence, with the particular solution (fulfilling $b(0)=1$)

$$b(k)=\frac 1{2^k}=\frac 1{\sqrt n}=a(n).$$