Metamath Proof Explorer


Theorem 4001lem3

Description: Lemma for 4001prm . Calculate a power mod. In decimal, we calculate 2 ^ 1 0 0 0 = 2 ^ 8 0 0 x. 2 ^ 2 0 0 == 2 3 1 1 x. 9 0 2 = 5 2 1 N + 1 and finally 2 ^ ( N - 1 ) = ( 2 ^ 1 0 0 0 ) ^ 4 == 1 ^ 4 = 1 . (Contributed by Mario Carneiro, 3-Mar-2014) (Revised by Mario Carneiro, 20-Apr-2015) (Proof shortened by AV, 16-Sep-2021)

Ref Expression
Hypothesis 4001prm.1 ⊢ N = 4001
Assertion 4001lem3 ⊢ 2 N − 1 mod N = 1 mod N

Proof

Step Hyp Ref Expression
1 4001prm.1 ⊢ N = 4001
2 4nn0 ⊢ 4 ∈ ℕ 0
3 0nn0 ⊢ 0 ∈ ℕ 0
4 2 3 deccl ⊢ 40 ∈ ℕ 0
5 4 3 deccl ⊢ 400 ∈ ℕ 0
6 1nn ⊢ 1 ∈ ℕ
7 5 6 decnncl ⊢ 4001 ∈ ℕ
8 1 7 eqeltri ⊢ N ∈ ℕ
9 2nn ⊢ 2 ∈ ℕ
10 2nn0 ⊢ 2 ∈ ℕ 0
11 10 3 deccl ⊢ 20 ∈ ℕ 0
12 11 3 deccl ⊢ 200 ∈ ℕ 0
13 12 3 deccl ⊢ 2000 ∈ ℕ 0
14 0z ⊢ 0 ∈ ℤ
15 1nn0 ⊢ 1 ∈ ℕ 0
16 10nn0 ⊢ 10 ∈ ℕ 0
17 16 3 deccl ⊢ 100 ∈ ℕ 0
18 17 3 deccl ⊢ 1000 ∈ ℕ 0
19 8nn0 ⊢ 8 ∈ ℕ 0
20 19 3 deccl ⊢ 80 ∈ ℕ 0
21 20 3 deccl ⊢ 800 ∈ ℕ 0
22 5nn0 ⊢ 5 ∈ ℕ 0
23 22 10 deccl ⊢ 52 ∈ ℕ 0
24 23 15 deccl ⊢ 521 ∈ ℕ 0
25 24 nn0zi ⊢ 521 ∈ ℤ
26 3nn0 ⊢ 3 ∈ ℕ 0
27 10 26 deccl ⊢ 23 ∈ ℕ 0
28 27 15 deccl ⊢ 231 ∈ ℕ 0
29 28 15 deccl ⊢ 2311 ∈ ℕ 0
30 9nn0 ⊢ 9 ∈ ℕ 0
31 30 3 deccl ⊢ 90 ∈ ℕ 0
32 31 10 deccl ⊢ 902 ∈ ℕ 0
33 1 4001lem2 ⊢ 2 800 mod N = 2311 mod N
34 1 4001lem1 ⊢ 2 200 mod N = 902 mod N
35 eqid ⊢ 800 = 800
36 eqid ⊢ 200 = 200
37 eqid ⊢ 80 = 80
38 eqid ⊢ 20 = 20
39 8p2e10 ⊢ 8 + 2 = 10
40 00id ⊢ 0 + 0 = 0
41 19 3 10 3 37 38 39 40 decadd ⊢ 80 + 20 = 100
42 20 3 11 3 35 36 41 40 decadd ⊢ 800 + 200 = 1000
43 15 dec0h ⊢ 1 = 01
44 eqid ⊢ 400 = 400
45 23 nn0cni ⊢ 52 ∈ ℂ
46 45 addlidi ⊢ 0 + 52 = 52
47 eqid ⊢ 40 = 40
48 5cn ⊢ 5 ∈ ℂ
49 48 addridi ⊢ 5 + 0 = 5
50 22 dec0h ⊢ 5 = 05
51 49 50 eqtri ⊢ 5 + 0 = 05
52 40 3 eqeltri ⊢ 0 + 0 ∈ ℕ 0
53 eqid ⊢ 521 = 521
54 eqid ⊢ 52 = 52
55 5t4e20 ⊢ 5 ⋅ 4 = 20
56 2t4e8 ⊢ 2 ⋅ 4 = 8
57 2 22 10 54 55 56 decmul1 ⊢ 52 ⋅ 4 = 208
58 4cn ⊢ 4 ∈ ℂ
59 58 mullidi ⊢ 1 ⋅ 4 = 4
60 59 40 oveq12i ⊢ 1 ⋅ 4 + 0 + 0 = 4 + 0
61 58 addridi ⊢ 4 + 0 = 4
62 60 61 eqtri ⊢ 1 ⋅ 4 + 0 + 0 = 4
63 23 15 52 53 2 57 62 decrmanc ⊢ 521 ⋅ 4 + 0 + 0 = 2084
64 24 nn0cni ⊢ 521 ∈ ℂ
65 64 mul01i ⊢ 521 ⋅ 0 = 0
66 65 oveq1i ⊢ 521 ⋅ 0 + 5 = 0 + 5
67 48 addlidi ⊢ 0 + 5 = 5
68 66 67 50 3eqtri ⊢ 521 ⋅ 0 + 5 = 05
69 2 3 3 22 47 51 24 22 3 63 68 decma2c ⊢ 521 ⋅ 40 + 5 + 0 = 20845
70 65 oveq1i ⊢ 521 ⋅ 0 + 2 = 0 + 2
71 2cn ⊢ 2 ∈ ℂ
72 71 addlidi ⊢ 0 + 2 = 2
73 10 dec0h ⊢ 2 = 02
74 70 72 73 3eqtri ⊢ 521 ⋅ 0 + 2 = 02
75 4 3 22 10 44 46 24 10 3 69 74 decma2c ⊢ 521 ⋅ 400 + 0 + 52 = 208452
76 45 mulridi ⊢ 52 ⋅ 1 = 52
77 ax-1cn ⊢ 1 ∈ ℂ
78 77 mullidi ⊢ 1 ⋅ 1 = 1
79 78 oveq1i ⊢ 1 ⋅ 1 + 1 = 1 + 1
80 1p1e2 ⊢ 1 + 1 = 2
81 79 80 eqtri ⊢ 1 ⋅ 1 + 1 = 2
82 23 15 15 53 15 76 81 decrmanc ⊢ 521 ⋅ 1 + 1 = 522
83 5 15 3 15 1 43 24 10 23 75 82 decma2c ⊢ 521 ⋅ N + 1 = 2084522
84 eqid ⊢ 902 = 902
85 6nn0 ⊢ 6 ∈ ℕ 0
86 2 85 deccl ⊢ 46 ∈ ℕ 0
87 86 10 deccl ⊢ 462 ∈ ℕ 0
88 eqid ⊢ 90 = 90
89 eqid ⊢ 462 = 462
90 eqid ⊢ 2311 = 2311
91 86 nn0cni ⊢ 46 ∈ ℂ
92 91 addridi ⊢ 46 + 0 = 46
93 4p1e5 ⊢ 4 + 1 = 5
94 93 22 eqeltri ⊢ 4 + 1 ∈ ℕ 0
95 eqid ⊢ 231 = 231
96 eqid ⊢ 23 = 23
97 9cn ⊢ 9 ∈ ℂ
98 9t2e18 ⊢ 9 ⋅ 2 = 18
99 97 71 98 mulcomli ⊢ 2 ⋅ 9 = 18
100 15 19 10 99 80 39 decaddci2 ⊢ 2 ⋅ 9 + 2 = 20
101 7nn0 ⊢ 7 ∈ ℕ 0
102 7p1e8 ⊢ 7 + 1 = 8
103 3cn ⊢ 3 ∈ ℂ
104 9t3e27 ⊢ 9 ⋅ 3 = 27
105 97 103 104 mulcomli ⊢ 3 ⋅ 9 = 27
106 10 101 102 105 decsuc ⊢ 3 ⋅ 9 + 1 = 28
107 10 26 15 96 30 19 10 100 106 decrmac ⊢ 23 ⋅ 9 + 1 = 208
108 97 mullidi ⊢ 1 ⋅ 9 = 9
109 108 93 oveq12i ⊢ 1 ⋅ 9 + 4 + 1 = 9 + 5
110 9p5e14 ⊢ 9 + 5 = 14
111 109 110 eqtri ⊢ 1 ⋅ 9 + 4 + 1 = 14
112 27 15 94 95 30 2 15 107 111 decrmac ⊢ 231 ⋅ 9 + 4 + 1 = 2084
113 108 oveq1i ⊢ 1 ⋅ 9 + 6 = 9 + 6
114 9p6e15 ⊢ 9 + 6 = 15
115 113 114 eqtri ⊢ 1 ⋅ 9 + 6 = 15
116 28 15 2 85 90 92 30 22 15 112 115 decmac ⊢ 2311 ⋅ 9 + 46 + 0 = 20845
117 29 nn0cni ⊢ 2311 ∈ ℂ
118 117 mul01i ⊢ 2311 ⋅ 0 = 0
119 118 oveq1i ⊢ 2311 ⋅ 0 + 2 = 0 + 2
120 119 72 73 3eqtri ⊢ 2311 ⋅ 0 + 2 = 02
121 30 3 86 10 88 89 29 10 3 116 120 decma2c ⊢ 2311 ⋅ 90 + 462 = 208452
122 2t2e4 ⊢ 2 ⋅ 2 = 4
123 3t2e6 ⊢ 3 ⋅ 2 = 6
124 10 10 26 96 122 123 decmul1 ⊢ 23 ⋅ 2 = 46
125 71 mullidi ⊢ 1 ⋅ 2 = 2
126 10 27 15 95 124 125 decmul1 ⊢ 231 ⋅ 2 = 462
127 10 28 15 90 126 125 decmul1 ⊢ 2311 ⋅ 2 = 4622
128 29 31 10 84 10 87 121 127 decmul2c ⊢ 2311 ⋅ 902 = 2084522
129 83 128 eqtr4i ⊢ 521 ⋅ N + 1 = 2311 ⋅ 902
130 8 9 21 25 29 15 12 32 33 34 42 129 modxai ⊢ 2 1000 mod N = 1 mod N
131 18 nn0cni ⊢ 1000 ∈ ℂ
132 eqid ⊢ 1000 = 1000
133 eqid ⊢ 100 = 100
134 10 dec0u ⊢ 10 ⋅ 2 = 20
135 71 mul02i ⊢ 0 ⋅ 2 = 0
136 10 16 3 133 134 135 decmul1 ⊢ 100 ⋅ 2 = 200
137 10 17 3 132 136 135 decmul1 ⊢ 1000 ⋅ 2 = 2000
138 131 71 137 mulcomli ⊢ 2 ⋅ 1000 = 2000
139 8 nncni ⊢ N ∈ ℂ
140 139 mul02i ⊢ 0 ⋅ N = 0
141 140 oveq1i ⊢ 0 ⋅ N + 1 = 0 + 1
142 77 addlidi ⊢ 0 + 1 = 1
143 78 142 eqtr4i ⊢ 1 ⋅ 1 = 0 + 1
144 141 143 eqtr4i ⊢ 0 ⋅ N + 1 = 1 ⋅ 1
145 8 9 18 14 15 15 130 138 144 mod2xi ⊢ 2 2000 mod N = 1 mod N
146 13 nn0cni ⊢ 2000 ∈ ℂ
147 eqid ⊢ 2000 = 2000
148 10 10 3 38 122 135 decmul1 ⊢ 20 ⋅ 2 = 40
149 10 11 3 36 148 135 decmul1 ⊢ 200 ⋅ 2 = 400
150 10 12 3 147 149 135 decmul1 ⊢ 2000 ⋅ 2 = 4000
151 146 71 150 mulcomli ⊢ 2 ⋅ 2000 = 4000
152 5 3 deccl ⊢ 4000 ∈ ℕ 0
153 152 nn0cni ⊢ 4000 ∈ ℂ
154 eqid ⊢ 4000 = 4000
155 5 3 142 154 decsuc ⊢ 4000 + 1 = 4001
156 1 155 eqtr4i ⊢ N = 4000 + 1
157 153 77 156 mvrraddi ⊢ N − 1 = 4000
158 151 157 eqtr4i ⊢ 2 ⋅ 2000 = N − 1
159 8 9 13 14 15 15 145 158 144 mod2xi ⊢ 2 N − 1 mod N = 1 mod N