Show that if the encryption exponent $3$ is used for the RSA cryptosystem by three different people with different moduli, a plaintext message $P$ encrypted using each of their keys can be recovered from these resulting three ciphertext messages.
Proof. We have $x_1\equiv P^3\pmod{n_1},$ $x_2\equiv P^3\pmod{n_2},$ $x_3\equiv P^3\pmod{n_3},$ and using the Chinese Remainder Theorem, we can get $P^3\equiv x_1(n_2n_3)(n_2n_3)^{-1}+x_2(n_1n_3)(n_1n_3)^{-1}+x_3(n_1n_2)(n_1n_2)^{-1}\pmod{n_1n_2n_3}.$ Therefore, $P$ can be found by testing all cubes that satisfy this congruence. $\square$
Does this proof seem rigorous enough, or should I step it up one more notch and compute for $P$? Or is perhaps the exercise asking for something else? What do you guys think? Thanks in advance!