Metamath Proof Explorer


Theorem lgslem1

Description: When a is coprime to the prime p , a ^ ( ( p - 1 ) / 2 ) is equivalent mod p to 1 or -u 1 , and so adding 1 makes it equivalent to 0 or 2 . (Contributed by Mario Carneiro, 4-Feb-2015)

Ref Expression
Assertion lgslem1 A P 2 ¬ P A A P 1 2 + 1 mod P 0 2

Proof

Step Hyp Ref Expression
1 eldifi P 2 P
2 1 3ad2ant2 A P 2 ¬ P A P
3 prmnn P P
4 2 3 syl A P 2 ¬ P A P
5 simp1 A P 2 ¬ P A A
6 prmz P P
7 2 6 syl A P 2 ¬ P A P
8 5 7 gcdcomd A P 2 ¬ P A A gcd P = P gcd A
9 simp3 A P 2 ¬ P A ¬ P A
10 coprm P A ¬ P A P gcd A = 1
11 2 5 10 syl2anc A P 2 ¬ P A ¬ P A P gcd A = 1
12 9 11 mpbid A P 2 ¬ P A P gcd A = 1
13 8 12 eqtrd A P 2 ¬ P A A gcd P = 1
14 eulerth P A A gcd P = 1 A ϕ P mod P = 1 mod P
15 4 5 13 14 syl3anc A P 2 ¬ P A A ϕ P mod P = 1 mod P
16 phiprm P ϕ P = P 1
17 2 16 syl A P 2 ¬ P A ϕ P = P 1
18 nnm1nn0 P P 1 0
19 4 18 syl A P 2 ¬ P A P 1 0
20 17 19 eqeltrd A P 2 ¬ P A ϕ P 0
21 zexpcl A ϕ P 0 A ϕ P
22 5 20 21 syl2anc A P 2 ¬ P A A ϕ P
23 1zzd A P 2 ¬ P A 1
24 moddvds P A ϕ P 1 A ϕ P mod P = 1 mod P P A ϕ P 1
25 4 22 23 24 syl3anc A P 2 ¬ P A A ϕ P mod P = 1 mod P P A ϕ P 1
26 15 25 mpbid A P 2 ¬ P A P A ϕ P 1
27 19 nn0cnd A P 2 ¬ P A P 1
28 2cnd A P 2 ¬ P A 2
29 2ne0 2 0
30 29 a1i A P 2 ¬ P A 2 0
31 27 28 30 divcan1d A P 2 ¬ P A P 1 2 2 = P 1
32 17 31 eqtr4d A P 2 ¬ P A ϕ P = P 1 2 2
33 32 oveq2d A P 2 ¬ P A A ϕ P = A P 1 2 2
34 5 zcnd A P 2 ¬ P A A
35 2nn0 2 0
36 35 a1i A P 2 ¬ P A 2 0
37 oddprm P 2 P 1 2
38 37 3ad2ant2 A P 2 ¬ P A P 1 2
39 38 nnnn0d A P 2 ¬ P A P 1 2 0
40 34 36 39 expmuld A P 2 ¬ P A A P 1 2 2 = A P 1 2 2
41 33 40 eqtrd A P 2 ¬ P A A ϕ P = A P 1 2 2
42 41 oveq1d A P 2 ¬ P A A ϕ P 1 = A P 1 2 2 1
43 sq1 1 2 = 1
44 43 oveq2i A P 1 2 2 1 2 = A P 1 2 2 1
45 42 44 eqtr4di A P 2 ¬ P A A ϕ P 1 = A P 1 2 2 1 2
46 zexpcl A P 1 2 0 A P 1 2
47 5 39 46 syl2anc A P 2 ¬ P A A P 1 2
48 47 zcnd A P 2 ¬ P A A P 1 2
49 ax-1cn 1
50 subsq A P 1 2 1 A P 1 2 2 1 2 = A P 1 2 + 1 A P 1 2 1
51 48 49 50 sylancl A P 2 ¬ P A A P 1 2 2 1 2 = A P 1 2 + 1 A P 1 2 1
52 45 51 eqtrd A P 2 ¬ P A A ϕ P 1 = A P 1 2 + 1 A P 1 2 1
53 26 52 breqtrd A P 2 ¬ P A P A P 1 2 + 1 A P 1 2 1
54 47 peano2zd A P 2 ¬ P A A P 1 2 + 1
55 peano2zm A P 1 2 A P 1 2 1
56 47 55 syl A P 2 ¬ P A A P 1 2 1
57 euclemma P A P 1 2 + 1 A P 1 2 1 P A P 1 2 + 1 A P 1 2 1 P A P 1 2 + 1 P A P 1 2 1
58 2 54 56 57 syl3anc A P 2 ¬ P A P A P 1 2 + 1 A P 1 2 1 P A P 1 2 + 1 P A P 1 2 1
59 53 58 mpbid A P 2 ¬ P A P A P 1 2 + 1 P A P 1 2 1
60 dvdsval3 P A P 1 2 + 1 P A P 1 2 + 1 A P 1 2 + 1 mod P = 0
61 4 54 60 syl2anc A P 2 ¬ P A P A P 1 2 + 1 A P 1 2 + 1 mod P = 0
62 2z 2
63 62 a1i A P 2 ¬ P A 2
64 moddvds P A P 1 2 + 1 2 A P 1 2 + 1 mod P = 2 mod P P A P 1 2 + 1 - 2
65 4 54 63 64 syl3anc A P 2 ¬ P A A P 1 2 + 1 mod P = 2 mod P P A P 1 2 + 1 - 2
66 2re 2
67 66 a1i A P 2 ¬ P A 2
68 4 nnrpd A P 2 ¬ P A P +
69 0le2 0 2
70 69 a1i A P 2 ¬ P A 0 2
71 4 nnred A P 2 ¬ P A P
72 prmuz2 P P 2
73 2 72 syl A P 2 ¬ P A P 2
74 eluzle P 2 2 P
75 73 74 syl A P 2 ¬ P A 2 P
76 eldifsni P 2 P 2
77 76 3ad2ant2 A P 2 ¬ P A P 2
78 67 71 75 77 leneltd A P 2 ¬ P A 2 < P
79 modid 2 P + 0 2 2 < P 2 mod P = 2
80 67 68 70 78 79 syl22anc A P 2 ¬ P A 2 mod P = 2
81 80 eqeq2d A P 2 ¬ P A A P 1 2 + 1 mod P = 2 mod P A P 1 2 + 1 mod P = 2
82 df-2 2 = 1 + 1
83 82 oveq2i A P 1 2 + 1 - 2 = A P 1 2 + 1 - 1 + 1
84 49 a1i A P 2 ¬ P A 1
85 48 84 84 pnpcan2d A P 2 ¬ P A A P 1 2 + 1 - 1 + 1 = A P 1 2 1
86 83 85 syl5eq A P 2 ¬ P A A P 1 2 + 1 - 2 = A P 1 2 1
87 86 breq2d A P 2 ¬ P A P A P 1 2 + 1 - 2 P A P 1 2 1
88 65 81 87 3bitr3rd A P 2 ¬ P A P A P 1 2 1 A P 1 2 + 1 mod P = 2
89 61 88 orbi12d A P 2 ¬ P A P A P 1 2 + 1 P A P 1 2 1 A P 1 2 + 1 mod P = 0 A P 1 2 + 1 mod P = 2
90 59 89 mpbid A P 2 ¬ P A A P 1 2 + 1 mod P = 0 A P 1 2 + 1 mod P = 2
91 ovex A P 1 2 + 1 mod P V
92 91 elpr A P 1 2 + 1 mod P 0 2 A P 1 2 + 1 mod P = 0 A P 1 2 + 1 mod P = 2
93 90 92 sylibr A P 2 ¬ P A A P 1 2 + 1 mod P 0 2