Metamath Proof Explorer


Theorem perfectALTV

Description: The Euclid-Euler theorem, or Perfect Number theorem. A positive even integer N is a perfect number (that is, its divisor sum is 2 N ) if and only if it is of the form 2 ^ ( p - 1 ) x. ( 2 ^ p - 1 ) , where 2 ^ p - 1 is prime (a Mersenne prime). (It follows from this that p is also prime.) This is Metamath 100 proof #70. (Contributed by Mario Carneiro, 17-May-2016) (Revised by AV, 1-Jul-2020) (Proof modification is discouraged.)

Ref Expression
Assertion perfectALTV ⊢ N ∈ ℕ ∧ N ∈ Even → 1 σ N = 2 ⋅ N ↔ ∃ p ∈ ℤ 2 p − 1 ∈ ℙ ∧ N = 2 p − 1 ⁢ 2 p − 1

Proof

Step Hyp Ref Expression
1 2dvdseven ⊢ N ∈ Even → 2 ∥ N
2 1 ad2antlr ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → 2 ∥ N
3 2prm ⊢ 2 ∈ ℙ
4 simpll ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → N ∈ ℕ
5 pcelnn ⊢ 2 ∈ ℙ ∧ N ∈ ℕ → 2 pCnt N ∈ ℕ ↔ 2 ∥ N
6 3 4 5 sylancr ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → 2 pCnt N ∈ ℕ ↔ 2 ∥ N
7 2 6 mpbird ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → 2 pCnt N ∈ ℕ
8 7 nnzd ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → 2 pCnt N ∈ ℤ
9 8 peano2zd ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → 2 pCnt N + 1 ∈ ℤ
10 pcdvds ⊢ 2 ∈ ℙ ∧ N ∈ ℕ → 2 2 pCnt N ∥ N
11 3 4 10 sylancr ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → 2 2 pCnt N ∥ N
12 2nn ⊢ 2 ∈ ℕ
13 7 nnnn0d ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → 2 pCnt N ∈ ℕ 0
14 nnexpcl ⊢ 2 ∈ ℕ ∧ 2 pCnt N ∈ ℕ 0 → 2 2 pCnt N ∈ ℕ
15 12 13 14 sylancr ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → 2 2 pCnt N ∈ ℕ
16 nndivdvds ⊢ N ∈ ℕ ∧ 2 2 pCnt N ∈ ℕ → 2 2 pCnt N ∥ N ↔ N 2 2 pCnt N ∈ ℕ
17 4 15 16 syl2anc ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → 2 2 pCnt N ∥ N ↔ N 2 2 pCnt N ∈ ℕ
18 11 17 mpbid ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → N 2 2 pCnt N ∈ ℕ
19 18 nnzd ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → N 2 2 pCnt N ∈ ℤ
20 pcndvds2 ⊢ 2 ∈ ℙ ∧ N ∈ ℕ → ¬ 2 ∥ N 2 2 pCnt N
21 3 4 20 sylancr ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → ¬ 2 ∥ N 2 2 pCnt N
22 isodd3 ⊢ N 2 2 pCnt N ∈ Odd ↔ N 2 2 pCnt N ∈ ℤ ∧ ¬ 2 ∥ N 2 2 pCnt N
23 19 21 22 sylanbrc ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → N 2 2 pCnt N ∈ Odd
24 simpr ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → 1 σ N = 2 ⋅ N
25 nncn ⊢ N ∈ ℕ → N ∈ ℂ
26 25 ad2antrr ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → N ∈ ℂ
27 15 nncnd ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → 2 2 pCnt N ∈ ℂ
28 15 nnne0d ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → 2 2 pCnt N ≠ 0
29 26 27 28 divcan2d ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → 2 2 pCnt N ⁢ N 2 2 pCnt N = N
30 29 oveq2d ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → 1 σ 2 2 pCnt N ⁢ N 2 2 pCnt N = 1 σ N
31 29 oveq2d ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → 2 ⁢ 2 2 pCnt N ⁢ N 2 2 pCnt N = 2 ⋅ N
32 24 30 31 3eqtr4d ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → 1 σ 2 2 pCnt N ⁢ N 2 2 pCnt N = 2 ⁢ 2 2 pCnt N ⁢ N 2 2 pCnt N
33 7 18 23 32 perfectALTVlem2 ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → N 2 2 pCnt N ∈ ℙ ∧ N 2 2 pCnt N = 2 2 pCnt N + 1 − 1
34 33 simprd ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → N 2 2 pCnt N = 2 2 pCnt N + 1 − 1
35 33 simpld ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → N 2 2 pCnt N ∈ ℙ
36 34 35 eqeltrrd ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → 2 2 pCnt N + 1 − 1 ∈ ℙ
37 7 nncnd ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → 2 pCnt N ∈ ℂ
38 ax-1cn ⊢ 1 ∈ ℂ
39 pncan ⊢ 2 pCnt N ∈ ℂ ∧ 1 ∈ ℂ → 2 pCnt N + 1 - 1 = 2 pCnt N
40 37 38 39 sylancl ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → 2 pCnt N + 1 - 1 = 2 pCnt N
41 40 eqcomd ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → 2 pCnt N = 2 pCnt N + 1 - 1
42 41 oveq2d ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → 2 2 pCnt N = 2 2 pCnt N + 1 - 1
43 42 34 oveq12d ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → 2 2 pCnt N ⁢ N 2 2 pCnt N = 2 2 pCnt N + 1 - 1 ⁢ 2 2 pCnt N + 1 − 1
44 29 43 eqtr3d ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → N = 2 2 pCnt N + 1 - 1 ⁢ 2 2 pCnt N + 1 − 1
45 oveq2 ⊢ p = 2 pCnt N + 1 → 2 p = 2 2 pCnt N + 1
46 45 oveq1d ⊢ p = 2 pCnt N + 1 → 2 p − 1 = 2 2 pCnt N + 1 − 1
47 46 eleq1d ⊢ p = 2 pCnt N + 1 → 2 p − 1 ∈ ℙ ↔ 2 2 pCnt N + 1 − 1 ∈ ℙ
48 oveq1 ⊢ p = 2 pCnt N + 1 → p − 1 = 2 pCnt N + 1 - 1
49 48 oveq2d ⊢ p = 2 pCnt N + 1 → 2 p − 1 = 2 2 pCnt N + 1 - 1
50 49 46 oveq12d ⊢ p = 2 pCnt N + 1 → 2 p − 1 ⁢ 2 p − 1 = 2 2 pCnt N + 1 - 1 ⁢ 2 2 pCnt N + 1 − 1
51 50 eqeq2d ⊢ p = 2 pCnt N + 1 → N = 2 p − 1 ⁢ 2 p − 1 ↔ N = 2 2 pCnt N + 1 - 1 ⁢ 2 2 pCnt N + 1 − 1
52 47 51 anbi12d ⊢ p = 2 pCnt N + 1 → 2 p − 1 ∈ ℙ ∧ N = 2 p − 1 ⁢ 2 p − 1 ↔ 2 2 pCnt N + 1 − 1 ∈ ℙ ∧ N = 2 2 pCnt N + 1 - 1 ⁢ 2 2 pCnt N + 1 − 1
53 52 rspcev ⊢ 2 pCnt N + 1 ∈ ℤ ∧ 2 2 pCnt N + 1 − 1 ∈ ℙ ∧ N = 2 2 pCnt N + 1 - 1 ⁢ 2 2 pCnt N + 1 − 1 → ∃ p ∈ ℤ 2 p − 1 ∈ ℙ ∧ N = 2 p − 1 ⁢ 2 p − 1
54 9 36 44 53 syl12anc ⊢ N ∈ ℕ ∧ N ∈ Even ∧ 1 σ N = 2 ⋅ N → ∃ p ∈ ℤ 2 p − 1 ∈ ℙ ∧ N = 2 p − 1 ⁢ 2 p − 1
55 54 ex ⊢ N ∈ ℕ ∧ N ∈ Even → 1 σ N = 2 ⋅ N → ∃ p ∈ ℤ 2 p − 1 ∈ ℙ ∧ N = 2 p − 1 ⁢ 2 p − 1
56 perfect1 ⊢ p ∈ ℤ ∧ 2 p − 1 ∈ ℙ → 1 σ 2 p − 1 ⁢ 2 p − 1 = 2 p ⁢ 2 p − 1
57 2cn ⊢ 2 ∈ ℂ
58 mersenne ⊢ p ∈ ℤ ∧ 2 p − 1 ∈ ℙ → p ∈ ℙ
59 prmnn ⊢ p ∈ ℙ → p ∈ ℕ
60 58 59 syl ⊢ p ∈ ℤ ∧ 2 p − 1 ∈ ℙ → p ∈ ℕ
61 expm1t ⊢ 2 ∈ ℂ ∧ p ∈ ℕ → 2 p = 2 p − 1 ⋅ 2
62 57 60 61 sylancr ⊢ p ∈ ℤ ∧ 2 p − 1 ∈ ℙ → 2 p = 2 p − 1 ⋅ 2
63 nnm1nn0 ⊢ p ∈ ℕ → p − 1 ∈ ℕ 0
64 60 63 syl ⊢ p ∈ ℤ ∧ 2 p − 1 ∈ ℙ → p − 1 ∈ ℕ 0
65 expcl ⊢ 2 ∈ ℂ ∧ p − 1 ∈ ℕ 0 → 2 p − 1 ∈ ℂ
66 57 64 65 sylancr ⊢ p ∈ ℤ ∧ 2 p − 1 ∈ ℙ → 2 p − 1 ∈ ℂ
67 mulcom ⊢ 2 p − 1 ∈ ℂ ∧ 2 ∈ ℂ → 2 p − 1 ⋅ 2 = 2 ⁢ 2 p − 1
68 66 57 67 sylancl ⊢ p ∈ ℤ ∧ 2 p − 1 ∈ ℙ → 2 p − 1 ⋅ 2 = 2 ⁢ 2 p − 1
69 62 68 eqtrd ⊢ p ∈ ℤ ∧ 2 p − 1 ∈ ℙ → 2 p = 2 ⁢ 2 p − 1
70 69 oveq1d ⊢ p ∈ ℤ ∧ 2 p − 1 ∈ ℙ → 2 p ⁢ 2 p − 1 = 2 ⁢ 2 p − 1 ⁢ 2 p − 1
71 2cnd ⊢ p ∈ ℤ ∧ 2 p − 1 ∈ ℙ → 2 ∈ ℂ
72 prmnn ⊢ 2 p − 1 ∈ ℙ → 2 p − 1 ∈ ℕ
73 72 adantl ⊢ p ∈ ℤ ∧ 2 p − 1 ∈ ℙ → 2 p − 1 ∈ ℕ
74 73 nncnd ⊢ p ∈ ℤ ∧ 2 p − 1 ∈ ℙ → 2 p − 1 ∈ ℂ
75 71 66 74 mulassd ⊢ p ∈ ℤ ∧ 2 p − 1 ∈ ℙ → 2 ⁢ 2 p − 1 ⁢ 2 p − 1 = 2 ⁢ 2 p − 1 ⁢ 2 p − 1
76 56 70 75 3eqtrd ⊢ p ∈ ℤ ∧ 2 p − 1 ∈ ℙ → 1 σ 2 p − 1 ⁢ 2 p − 1 = 2 ⁢ 2 p − 1 ⁢ 2 p − 1
77 oveq2 ⊢ N = 2 p − 1 ⁢ 2 p − 1 → 1 σ N = 1 σ 2 p − 1 ⁢ 2 p − 1
78 oveq2 ⊢ N = 2 p − 1 ⁢ 2 p − 1 → 2 ⋅ N = 2 ⁢ 2 p − 1 ⁢ 2 p − 1
79 77 78 eqeq12d ⊢ N = 2 p − 1 ⁢ 2 p − 1 → 1 σ N = 2 ⋅ N ↔ 1 σ 2 p − 1 ⁢ 2 p − 1 = 2 ⁢ 2 p − 1 ⁢ 2 p − 1
80 76 79 syl5ibrcom ⊢ p ∈ ℤ ∧ 2 p − 1 ∈ ℙ → N = 2 p − 1 ⁢ 2 p − 1 → 1 σ N = 2 ⋅ N
81 80 impr ⊢ p ∈ ℤ ∧ 2 p − 1 ∈ ℙ ∧ N = 2 p − 1 ⁢ 2 p − 1 → 1 σ N = 2 ⋅ N
82 81 rexlimiva ⊢ ∃ p ∈ ℤ 2 p − 1 ∈ ℙ ∧ N = 2 p − 1 ⁢ 2 p − 1 → 1 σ N = 2 ⋅ N
83 55 82 impbid1 ⊢ N ∈ ℕ ∧ N ∈ Even → 1 σ N = 2 ⋅ N ↔ ∃ p ∈ ℤ 2 p − 1 ∈ ℙ ∧ N = 2 p − 1 ⁢ 2 p − 1