Chinese Remainder Theorem

**SlipEternal** · September 6th 2012, 08:19 AM

I have a system of congruences modulo different powers of three which satisfies the conditions of the Chinese Remainder Theorem. I have $n+1$ variables (all positive integers) and I have $n+1$ equations. For each $k=1,2,...,n+1$ :

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

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

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

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

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