Problem:
For any positive integer a,σ(a) denotes the sum of the positive integer divisors of a. Let n be the least positive integer such that σ(an)−1 is divisible by 2021 for all positive integers a. Find the sum of the prime factors in the prime factorization of n.
Solution:
If a has prime factorization p1α1p2α2⋯, then σ(a)=σ(p1α1)σ(p2α2)⋯ and hence σ(an)=σ(p1nα1)σ(p2nα2)⋯. Therefore it suffices to find the least positive integer n such that σ(pnα)≡1(mod2021) for all prime powers pα. Because 2021=43⋅47, by the Chinese Remainder Theorem, it is sufficient that σ(pnα)≡1(mod43) and σ(pnα)≡1(mod47) for all prime powers pα.
Assume that n satisfies the required condition. In particular, for all p and α,n must satisfy
σ(pnα)=1+p+p2+⋯+pnα≡1(modq),
where q=43 or q=47.
- If p=q, this sum will always be congruent to 1(modq).
- If p≡1(modq), then each term in the sum is 1(modq), so
σ(pnα)≡nα+1≡1(modq)
Thus the required n must satisfy q∣nα for all α, so q∣n.
Note that 43⋅4+1=173 and 47⋅6+1=283 are both prime numbers, so such p exist for both q=43 and q=47.
- If p is a prime such that p=q and gcd(p−1,q)=1, then
1+p+p2+⋯+pnα=p−1pnα+1−1≡1(modq)
which, after clearing the denominators and canceling a factor of p, reduces to
pnα≡1(modq).
By Fermat's Little Theorem, it is sufficient to have q−1∣n. However, for both q=43 and q=47 there exists a prime p such that p is a primitive root modulo q. For example, p=5 is a primitive root modulo both 43 and 47. Therefore the condition that q−1∣n is also necessary.
It follows that n must be divisible by 42,43,46, and 47 . The requested sum is 2+3+7+23+43+47=125.