I am thinking that this is a variation of the Chinese Remainder Theorem as the iff qualifies that this set of equations is not exactly the definition of the Chinese Remainder Theorem, leading me to believe that this will indeed hold. However, I am not exactly sure how to start this off.
Again, the question is formalized as:
Given the gcd($a$, $b$) = 1, prove that
$$ x \equiv y\pmod a $$ $$ x \equiv y\pmod b $$
iff
$$ x \equiv y \space mod \space (ab) $$
Let's prove the case two variables $a$ and $b$.If $a|(x-y)$, and $b|(x-y)$, and $(a,b) = 1$, then write: $am + bn = 1$ by Bezout's formula. and use: $x-y = as$, and $x-y = bt$, then: $x-y = as(am + bn) = sma^2 + snab$, and $x-y = bt(am + bn) = tmab + tnb^2$. So: $a |sma^2 + snab = tmab + tnb^2$. So: $a |tnb^2$. Here we need to prove that: $a|tn$, and we are done. Suppose not, then there is a prime $p$ such that: $p|a$ but $p \not| nt$, then $p \not| b^2$. So $p \not| ntb^2$, contradiction. So $a|nt$. So: $ab|tnb^2$, then $ab|tmab + tnb^2 = x-y$. The other direction is simple because: $a|ab$ and $b|ab$. So if$ab|x-y$ then $a|x-y$ and $b|x-y$.
