# Chinese Remainder Theorem

• Sep 6th 2012, 08:19 AM
SlipEternal
Chinese Remainder Theorem
I have a system of congruences modulo different powers of three which satisfies the conditions of the Chinese Remainder Theorem. I have $\displaystyle n+1$ variables (all positive integers) and I have $\displaystyle n+1$ equations. For each $\displaystyle k=1,2,...,n+1$:

$\displaystyle \displaymode 2^{a_0}\equiv 1+\sum_{i=1}^{k-1}\left(3^i\left(\frac{3^{k}+1}{2}\right)^{\sum_{j =1}^i {a_j}\right) (\text{mod }3^k)$

So, for example, if $\displaystyle n=2$, I have:

$\displaystyle \displaymode 2^{a_0}\equiv 1 (\text{mod }3)$
$\displaystyle \displaymode 2^{a_0}\equiv 1+3\cdot 5^{a_1} (\text{mod }9)$
$\displaystyle \displaymode 2^{a_0}\equiv 1+3\cdot 14^{a_1}+9\cdot 14^{a_1+a_2} (\text{mod }27)$

If I use the Chinese Remainder Theorem, I wind up with a system of nonlinear equations in $\displaystyle \mathbb{Z} / 3^{n+1}\mathbb{Z}$. Is this possible to solve for $\displaystyle a_0 (\text{mod }2\cdot 3^n)$?