ELEMENTARY NUMBER THEORY - PART 2
CIPRIAN MANOLESCU
1. Some results
To a positive integer n we can associate the quantities:
τ (n) = the number of positive divisors of n,
σ(n) = the sum of positive divisors of n,
ϕ(n) = the number of integers m relatively prime to n, with 1 ≤ m ≤ n.
The function ϕ(n) is called Euler’s totient function.
If n = pk11 . . . pkmm is the prime factorization of n, we have the formulas:
τ (n) = (k1 + 1) · . . . · (km + 1),
σ(n) =
pkkm +1 − 1
pk11 +1 − 1
· ... ·
,
p1 − 1
pk − 1
ϕ(n) = pk11 −1 (p1 − 1) · . . . · pkmm −1 (pm − 1).
Recall that we write a ≡ b (mod m) if m divides a − b. By a (mod m) we mean the remainder
in the division of a by m.
Theorem 1.1 (Fermat’s Little Theorem). If a is a positive integer and p is a prime, then
ap ≡ a
(mod p).
Theorem 1.2 (Euler). If a and m are relatively prime positive integers, then
aϕ(m) ≡ 1 (mod m).
Example 1.3. What is 544 (mod 49)? We have ϕ(49) = 7 · 6 = 42. Since 5 and 49 are relatively
prime, we can apply Euler’s theorem to get 542 ≡ 1 (mod 49). Hence:
544 ≡ 52 · 542 ≡ 52 · 1 ≡ 25 (mod 49).
Theorem 1.4 (Wilson). If p is a prime, then
(p − 1)! ≡ −1 (mod p).
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CIPRIAN MANOLESCU
2. Problems
1. Calculate: τ (500), σ(500), ϕ(500).
2. Calculate: τ (280), σ(280), ϕ(280).
3. How many even positive divisors does 1000 have? What is their sum?
4. What is the product of all positive divisors of 500?
5. If n = pk11 . . . pkmm is the prime factorization of n, find a formula for the product of all positive
divisors of n.
6. Calculate 2123 (mod 11).
7. Calculate 2123 + 5123 (mod 99).
8. Show that ϕ(nm ) = nm−1 · ϕ(n), for all m, n ≥ 1.
9. Find the greatest common divisor of 100! + 1 and 101!.
10. Prove that for any even positive integer n, the number n2 − 1 divides 2n! − 1.
11. Determine the last three digits of the number
20032002
2001
.
12. Show that for any fixed positive integer n, the sequence
2
22
2, 22 , 22 , 22 , . . .
is eventually constant.
(mod n)