Problem 4

Determine all pairs

(n,p)

of positive integers such that

p

is a prime,

n

not exceeded

2p

, and

(p-1)^n+1

is divisible by

n^{p-1}

Step 2 of 5: Identify the smallest prime divisor

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In plain words

The least-prime-divisor argument forces the same prime.

q=p

Detailed analysis

Let $q$ be the smallest prime divisor of $n$ . Using $(p-1)^n\equiv-1\pmod q$ and the least positive exponents giving residues $-1$ and $1$ , a Euclidean-division argument shows that the exponent $n$ is a multiple of the least $-1$ exponent, which must be $1$ . Thus $p-1\equiv-1\pmod q$ , and since $p,q$ are prime, $q=p$ .