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Can anyone give a proof to the following theorem:

For $n$, let $n-1$ $=$ $q^k*t$ with $q$ prime and $k > 0$. For some integer $1$ $<$ $a$ $<$ $n-1$ if $a^{n-1}$ $=$ $1$ $\pmod n$ and $\gcd(a^{(n-1)/q}-1$, $n$) $=$ $1$ each divisor of $n$ has the form $q^k*h+1$.

Let $S$ denote all the numbers following the theorem above for a specific $q^k$. Is there a special form that $S$ has besides $q^k*s$? For example,

each number $S$ that have ALL divisors congruent to $1$ $\pmod 3$ have the form $S$ $=$ $3a^2+b^2$.

each number $S$ that have ALL divisors congruent to $1$ $\pmod 4$ have the form $S$ $=$ $a^2+b^2$.

what about numbers $S$ that have ALL divisors congruent to $1$ $\pmod 5$? What is their special form?

Thanks in advance.

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