RSA cracking: The same message is sent to two different people problem
$c_1 \equiv m^{e_1} \bmod n$

$c_2 \equiv m^{e_2} \bmod n$

if $\gcd(e_1,e_2)=1$ then $\exists a,b\in\mathbb{Z} : e_1\cdot a + e_2\cdot b = 1$ ($a$ and $b$ can be found by extended euclidean algorithm)

And,

$m\equiv c_1^a\cdot c_2^b \bmod n$

Note: In practice, either $a$ or $b$ will be negative. WLOG, let $b$ be negative. This leads to problems in the above equation. To get around this, use the following computation.

Let $i\equiv c_2^{-1} \bmod n$ (i.e., the modular inverse of $c_2$)

$m\equiv c_1^a\cdot i^{-b} \bmod n$

Update to show details of $m\equiv c_1^a\cdot c_2^b\bmod n$. All math below done modulo $n$.

$$
c_1^a\cdot c_2^b\\
(m^{e_1})^a\cdot (m^{e_2})^b\\
m^{e_1a}\cdot m^{e_2b}\\
m^{e_1a+e_2b}\\
m^1\\
m
$$