Perfect numbers among polynomial values

Transcription

1 Perfect numbers among polynomial values Oleksiy Klurman Université de Montréal June 8, 2015

2 Outline Classical results Radical of perfect number and the ABC-conjecture Idea of the proof

3 Let σ(n) = d n d. Perfect number: σ(n) = 2n.

4 Let σ(n) = d n d. Perfect number: σ(n) = 2n. Question: Do there exist infinitely many odd and even perfect numbers?

5 Let σ(n) = d n d. Perfect number: σ(n) = 2n. Question: Do there exist infinitely many odd and even perfect numbers? Euler:n is even perfect number iff and 2 p 1 is prime. n = 2 p 1 (2 p 1)

6 Euler: If n is odd perfect number, then n = q α m 2, where q = 1(mod4) and (q, m) = 1.

7 Euler: If n is odd perfect number, then n = q α m 2, where q = 1(mod4) and (q, m) = 1. Sylvester: If n is odd perfect then ω(n) 5.

8 Euler: If n is odd perfect number, then n = q α m 2, where q = 1(mod4) and (q, m) = 1. Sylvester: If n is odd perfect then ω(n) 5. Pomerance: If n is odd perfect then ω(n) 7. Nielsen: If n is odd perfect then ω(n) 9.

9 Heuristic: there exists only finitely many odd perfect numbers. Goal: There exists only finitely many perfect numbers of the special form.

10 Heuristic: there exists only finitely many odd perfect numbers. Goal: There exists only finitely many perfect numbers of the special form. Luca: perfect numbers among Fibonacci and Lucas sequences. Pollack: perfect numbers with identical digits.

11 Perfect numbers and the ABC-conjecture The ABC-conjecture: suppose that a, b, c N, (a, b, c) = 1 and a + b = c. Then rad(abc) 1+ε ε c Luca, Pomerance: Assume the ABC-conjecture. Then for any odd k N, x y = k has finitely many solutions in perfect numbers.

12 Perfect numbers and the ABC-conjecture The ABC-conjecture: suppose that a, b, c N, (a, b, c) = 1 and a + b = c. Then rad(abc) 1+ε ε c Luca, Pomerance: Assume the ABC-conjecture. Then for any odd k N, x y = k has finitely many solutions in perfect numbers. Luca, Pomerance: Assume the ABC-conjecture. Then for any k N, a n+2 a n k has finitely many solutions.

13 Key idea of the proof Luca, Pomerance: if n is perfect, then rad(n) n Acquaah and Konyagin: if n is perfect, then rad(n) n 2 3.

14 Key idea of the proof Luca, Pomerance: if n is perfect, then rad(n) n Acquaah and Konyagin: if n is perfect, then rad(n) n 2 3. Conjecture: if n is perfect, then rad(n) n 1 2. Note: if n = 2 p 1 (2 p 1) then rad(n) 2(2 p 1) 4 n

15 Ellia and Ochem, Rao: if rad(n) > n then the special prime q > if 3 n and q 223 otherwise. K.: if n is perfect, then rad(n) n 5 8.

16 Applications K.: Assume the ABC-conjecture. Let P(x) Z[x] of degree 3 has no repeated factors. Then, P(n) = m, has finitely many solutions for m perfect. K.: Assume that the ABC- conjecture. Let f (x, y) Z[x] be a homogeneous form of degree d 5 without repeated linear factors. Then, f (n, k) = m, has finitely many solutions for m perfect.

17 Recall: n is even perfect number iff n = 2 p 1 (2 p 1). K.: Suppose P(x) Z[x] of degree 2. Then there exist only finitely many integers n, such that P(n) is an even perfect number unless P(x) = Q2 (x) 1, 8 where Q(x) = 2 k (a 0 x + b 0 ) m 1 for some k, m, a, b Z.

18 Proofs Elkies, Langevin: Assume the ABC- conjecture. Suppose that f (x) Z[x] has no repeated roots. Then p f (m) p ε m deg f 1 ε.

19 Proofs Elkies, Langevin: Assume the ABC- conjecture. Suppose that f (x) Z[x] has no repeated roots. Then p f (m) p ε m deg f 1 ε. Observe, n 9 14 d rad(p(n)) n d 1 ε.

20 Proofs Elkies, Langevin: Assume the ABC- conjecture. Suppose that f (x) Z[x] has no repeated roots. Then p f (m) p ε m deg f 1 ε. Observe, n 9 14 d rad(p(n)) n d 1 ε. Thus, or d ε < d d 1 ε

21 Start with P(n) = 2 p 1 (2 p 1) = 2 2p 1 2 p 1.

22 Start with P(n) = 2 p 1 (2 p 1) = 2 2p 1 2 p 1. 8P(n) + 1 = (2 p+1 1) 2.

23 Start with P(n) = 2 p 1 (2 p 1) = 2 2p 1 2 p 1. 8P(n) + 1 = (2 p+1 1) 2. Applying Faltings theorem, we get Iterating, we end up with Q(n) + 1 = 2 p+1 = (2 p+1 2 ) 2 an 2 + bn + c = 2 m. This is Ramanujan-Nagell type equation. Results of Shorey and Stewart applies.

24 Bounding radical Suppose rad(n) = q p 1 p 2... p k n β

25 Bounding radical Suppose rad(n) = q p 1 p 2... p k n β Recall, q k i=1 2α i β 1 1 β pi 2q k i=1 p 2α i i = (q + 1) k (1 + p i + + p 2α i i ). i=1

26 Bounding radical Suppose rad(n) = q p 1 p 2... p k n β Recall, q k i=1 2α i β 1 1 β pi 2q k i=1 p 2α i i = (q + 1) k (1 + p i + + p 2α i i ). i=1 WLOG, α 1 = 1 and q 1 + p 1 + p 2 1.

27 After some arguments we arrive at the following cases p2 2 q + 1, p 2 q + 1 and p p p 2α 3 3, p p p 2α 3 3, p p p 2α 3 3 and p p p 2α 4 4.

28 Thank you!