Chinese Remainder Theorem Confusion I need to use the Chinese Remainder Theorem to find
six consecutive integers, each divisible by a prime squared
greater 5.
So I chose to solve these 
$$x\equiv 0 \pmod {49}$$
$$x+1\equiv 0 \pmod {121}$$
$$x+2\equiv 0 \pmod {169}$$
$$x+3\equiv 0 \pmod {289}$$
$$x+4\equiv 0 \pmod {361}$$
$$x+5\equiv 0 \pmod {529}$$
I wrote a program to try and solve it but wasn't able to find a solution, could someone help get me started on how to use the CRT to solve this problem?
I'm not quite sure how to apply it to even tackle this problem..
 A: Let's solve just one pair of those congruences:
$\begin{align}
x+4&\equiv 0 \bmod {361} &\implies x&\equiv -4 \bmod {361}\\
x+5&\equiv 0 \bmod {529} &\implies x&\equiv -5 \bmod {529}\\
\end{align}$
We can jump into the extended Euclidean algorithm to find a combination of $361$ and $529$ that solves to their GCD of $1$:
$\begin{array}{|c|c|} \hline
\quad n \quad & \quad s \quad & \quad t \quad & \quad q \quad \\\hline
529 & 1 & 0 &  \\
361 & 0 & 1 & 1 \\
168 & 1 & -1 & 2 \\
25 & -2 & 3 & 6 \\
18 & 13 & -19 & 1 \\
7 & -15 & 22 & 2 \\
4 & 43 & -63 & 1 \\
3 & -58 & 85 & 1 \\
1 & 101 & -148 &  \\ \hline
\end{array}$
where each line solves $n=529s+361t$ and $q$ is a multiplier used to get to the reduced $n$ on the following line. From the last line we see Bézout's identity of $101\cdot 529 -148\cdot 361 =1$, giving $-148\cdot 361 \equiv 1 \bmod 529$.
Then $x=361k-4 \implies 361k+1\equiv 0\bmod 529  \implies k-148 \equiv 0\bmod 529 $ using the result from the Bézout identity, so $k=148$ is a result consistent with the $\bmod 529$ equivalence , and we get 
$$ x \equiv 361\cdot 148 -4 \equiv 53424 \bmod 190969 (=361\cdot529)$$
Then this result can be combined with one of the other equivalences, and so on to a (very large) solution.
