Note : This problem has no specific source .
Let $n$ be a composite number of the form :
$n=p^{a_1}_1\cdot p^{a_2}_2 \ldots p^{a_k}_k $ , where $p_1,p_2 , \ldots p_k$ are distinct primes and $a_1,a_2,\ldots a_k >0$ ; $k>1$ .
Is it true that :
If $ ~p^{n-1}_i \equiv r_i \pmod n ~\text{and}~ 0 \leq r_i \leq n-1 ~\text{then}~ r_i >1$
My attempt :
a) suppose $r_i=1$
According to Euler Theorem :
If $~\gcd(p_i,n)=1~$ then $~p^{\varphi(n)}_i \equiv 1 \pmod n$
$\bullet $ if $n$ is a prime then $p^{n-1}_i \equiv 1 \pmod n$
$\bullet $ if $n$ is a pseudoprime or composite such that $\varphi(n) \mid n-1$ then $p^{n-1}_i \equiv 1 \pmod n$
By contraposition of Euler Theorem it follows :
if $~\gcd(p_i,n) \neq 1~$ then $~p^{n-1}_i \not\equiv 1 \pmod n$
So , since $~\gcd(p_i,n) \neq 1~$ it follows $r_i \neq 1$
b) suppose $r_i=0$
According to the contraposition of Chinese Remainder Theorem :
$\text{ iff } \begin{cases} p^{n-1}_i \not\equiv 0 \pmod {p^{a_1}_1} \\ p^{n-1}_i \not\equiv 0 \pmod {p^{a_2}_2} \\ \vdots \\ p^{n-1}_i \equiv 0 \pmod {p^{a_i}_i} \\ \vdots \\ p^{n-1}_i \not\equiv 0 \pmod {p^{a_k}_k} \end{cases} ~\text { then }~ p^{n-1}_i \not \equiv 0 \pmod {p^{a_1}_1\cdot p^{a_2}_2 \ldots p^{a_k}_k}$
hence $~p^{n-1}_i \not \equiv 0 \pmod n ~$ and therefore $r_i \neq 0$
So , since $r_i \neq 0 \text { and } r_i \neq 1$ it follows $r_i > 1$
Q.E.D.
Question : Is my proof correct and if it is not where is a mistake ?