Atrocious primes

I'm considering $n\geq1$ .

First, we notice, $p^n$

$\equiv ((p-1)+1)^n$ $(mod$ $(p-1))$

$\equiv 1^n$ $(mod$ $(p-1))$ [With the help of Binomial Theorem ]

$\equiv 1$ $(mod$ $(p-1))$ .

So, $p^n \equiv 1$ $(mod$ $(p-1))$ $\cdots{ } \cdots{ } \cdots{ }(1)$ .

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Again, $7^{p^n}+1$

$\equiv 7^{p^n \mod (p-1)}+1$ $(\mod p)$ [By Euler's Theorem ]

$\equiv 7^1+1$ $(\mod p)$ [By (1) ]

$\equiv 8$ $(\mod p)$

Hence, $7^{p^n}+1 \equiv 2^3$ $(\mod p)$ $\cdots{ } \cdots{ } \cdots{ }(2)$

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Now, $p^n \mid (7^{p^n}+1)$ only if $p \mid (7^{p^n}+1)$ , since $p \not \mid (7^{p^n}+1) \implies$ $p^n \not \mid (7^{p^n}+1)$ , when $p$ is a prime.

From (2) , $p \mid (7^{p^n}+1) \iff p \mid 2^3 \iff p=2$ . So, $p^n$ with $p\neq2$ can't be Atrocious .

We're left only with integers $2^n$ .

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For $n=1$ , $2^n=2$ , which is an atrocious number [By (2) ]

For $n=2,3,4$ , you can calculate to observe $(7^{2^n}+1) \equiv 2 (\mod 2^n)$ . Then the reasonable conjecture can be drawn: " $(7^{2^n}+1) \equiv 2 (\mod 2^n)$ , for every $n\geq2$ ; which can be verified by a simple proof by induction on $n$ .

So, $2$ is the only 'atrocious' number. So, the answer is $\boxed{2}$ .

Due to the LFT $7^{{p^{n}-p^{n-1}}} \equiv 1 (mod p^n)$ because $\phi(p^n)= p^n(1-\frac{1}{p})$ = $p^{n}-p^{n-1}$

But $7^{p^n}$ = $7^{p^n} * \frac{7^{p^{n-1}}}{7^{p^{n-1}}}$ = $7^{p^{n}-p^{n-1}} * 7^{p^{n-1}} \equiv 7^{p^{n-1}} (mod p^{n})$ So if $p^n | 7^{p^n} +1$ then $p^n | 7^{p^{n-1}} +1$ . If we apply the theorem k times we get that $p^n | 7^{p^{n-k}} +1$ , in particular by using it n times we get that $p^n|7^{p^0}+1 \Rightarrow p^n|8$ now we can easly conclude that the only possible values of p and n are respectively 2 and 1 that is $p^n = 2^1=2$

Since 7 to the power anything is odd, so the right hand side is even. So the LHS must also be even. This is only possible in the case of p=2. {Since it is the only even prime, and even to the power anything is also even.}.

Given: $7^{p^{n}} + 1 \equiv 0 \pmod {p^{n}}$

But $7^{p^{n}} -1 \equiv 0 \pmod {p^{n}}$

Thus

$2 \equiv 0 \pmod {p^{n}}$

And hence

$p^{n}= 2$