Metamath Proof Explorer


Theorem phiprmpw

Description: Value of the Euler phi function at a prime power. Theorem 2.5(a) in ApostolNT p. 28. (Contributed by Mario Carneiro, 24-Feb-2014)

Ref Expression
Assertion phiprmpw ⊢ P ∈ ℙ ∧ K ∈ ℕ → ϕ ⁡ P K = P K − 1 ⁢ P − 1

Proof

Step Hyp Ref Expression
1 prmnn ⊢ P ∈ ℙ → P ∈ ℕ
2 nnnn0 ⊢ K ∈ ℕ → K ∈ ℕ 0
3 nnexpcl ⊢ P ∈ ℕ ∧ K ∈ ℕ 0 → P K ∈ ℕ
4 1 2 3 syl2an ⊢ P ∈ ℙ ∧ K ∈ ℕ → P K ∈ ℕ
5 phival ⊢ P K ∈ ℕ → ϕ ⁡ P K = x ∈ 1 … P K | x gcd P K = 1
6 4 5 syl ⊢ P ∈ ℙ ∧ K ∈ ℕ → ϕ ⁡ P K = x ∈ 1 … P K | x gcd P K = 1
7 nnm1nn0 ⊢ K ∈ ℕ → K − 1 ∈ ℕ 0
8 nnexpcl ⊢ P ∈ ℕ ∧ K − 1 ∈ ℕ 0 → P K − 1 ∈ ℕ
9 1 7 8 syl2an ⊢ P ∈ ℙ ∧ K ∈ ℕ → P K − 1 ∈ ℕ
10 9 nncnd ⊢ P ∈ ℙ ∧ K ∈ ℕ → P K − 1 ∈ ℂ
11 1 nncnd ⊢ P ∈ ℙ → P ∈ ℂ
12 11 adantr ⊢ P ∈ ℙ ∧ K ∈ ℕ → P ∈ ℂ
13 ax-1cn ⊢ 1 ∈ ℂ
14 subdi ⊢ P K − 1 ∈ ℂ ∧ P ∈ ℂ ∧ 1 ∈ ℂ → P K − 1 ⁢ P − 1 = P K − 1 ⁢ P − P K − 1 ⋅ 1
15 13 14 mp3an3 ⊢ P K − 1 ∈ ℂ ∧ P ∈ ℂ → P K − 1 ⁢ P − 1 = P K − 1 ⁢ P − P K − 1 ⋅ 1
16 10 12 15 syl2anc ⊢ P ∈ ℙ ∧ K ∈ ℕ → P K − 1 ⁢ P − 1 = P K − 1 ⁢ P − P K − 1 ⋅ 1
17 10 mulridd ⊢ P ∈ ℙ ∧ K ∈ ℕ → P K − 1 ⋅ 1 = P K − 1
18 17 oveq2d ⊢ P ∈ ℙ ∧ K ∈ ℕ → P K − 1 ⁢ P − P K − 1 ⋅ 1 = P K − 1 ⁢ P − P K − 1
19 fzfi ⊢ 1 … P K ∈ Fin
20 ssrab2 ⊢ x ∈ 1 … P K | x gcd P K = 1 ⊆ 1 … P K
21 ssfi ⊢ 1 … P K ∈ Fin ∧ x ∈ 1 … P K | x gcd P K = 1 ⊆ 1 … P K → x ∈ 1 … P K | x gcd P K = 1 ∈ Fin
22 19 20 21 mp2an ⊢ x ∈ 1 … P K | x gcd P K = 1 ∈ Fin
23 ssrab2 ⊢ x ∈ 1 … P K | P ∥ x − 0 ⊆ 1 … P K
24 ssfi ⊢ 1 … P K ∈ Fin ∧ x ∈ 1 … P K | P ∥ x − 0 ⊆ 1 … P K → x ∈ 1 … P K | P ∥ x − 0 ∈ Fin
25 19 23 24 mp2an ⊢ x ∈ 1 … P K | P ∥ x − 0 ∈ Fin
26 inrab ⊢ x ∈ 1 … P K | x gcd P K = 1 ∩ x ∈ 1 … P K | P ∥ x − 0 = x ∈ 1 … P K | x gcd P K = 1 ∧ P ∥ x − 0
27 elfzelz ⊢ x ∈ 1 … P K → x ∈ ℤ
28 prmz ⊢ P ∈ ℙ → P ∈ ℤ
29 rpexp ⊢ P ∈ ℤ ∧ x ∈ ℤ ∧ K ∈ ℕ → P K gcd x = 1 ↔ P gcd x = 1
30 28 29 syl3an1 ⊢ P ∈ ℙ ∧ x ∈ ℤ ∧ K ∈ ℕ → P K gcd x = 1 ↔ P gcd x = 1
31 30 3expa ⊢ P ∈ ℙ ∧ x ∈ ℤ ∧ K ∈ ℕ → P K gcd x = 1 ↔ P gcd x = 1
32 31 an32s ⊢ P ∈ ℙ ∧ K ∈ ℕ ∧ x ∈ ℤ → P K gcd x = 1 ↔ P gcd x = 1
33 simpr ⊢ P ∈ ℙ ∧ K ∈ ℕ ∧ x ∈ ℤ → x ∈ ℤ
34 zexpcl ⊢ P ∈ ℤ ∧ K ∈ ℕ 0 → P K ∈ ℤ
35 28 2 34 syl2an ⊢ P ∈ ℙ ∧ K ∈ ℕ → P K ∈ ℤ
36 35 adantr ⊢ P ∈ ℙ ∧ K ∈ ℕ ∧ x ∈ ℤ → P K ∈ ℤ
37 33 36 gcdcomd ⊢ P ∈ ℙ ∧ K ∈ ℕ ∧ x ∈ ℤ → x gcd P K = P K gcd x
38 37 eqeq1d ⊢ P ∈ ℙ ∧ K ∈ ℕ ∧ x ∈ ℤ → x gcd P K = 1 ↔ P K gcd x = 1
39 coprm ⊢ P ∈ ℙ ∧ x ∈ ℤ → ¬ P ∥ x ↔ P gcd x = 1
40 39 adantlr ⊢ P ∈ ℙ ∧ K ∈ ℕ ∧ x ∈ ℤ → ¬ P ∥ x ↔ P gcd x = 1
41 32 38 40 3bitr4d ⊢ P ∈ ℙ ∧ K ∈ ℕ ∧ x ∈ ℤ → x gcd P K = 1 ↔ ¬ P ∥ x
42 zcn ⊢ x ∈ ℤ → x ∈ ℂ
43 42 adantl ⊢ P ∈ ℙ ∧ K ∈ ℕ ∧ x ∈ ℤ → x ∈ ℂ
44 43 subid1d ⊢ P ∈ ℙ ∧ K ∈ ℕ ∧ x ∈ ℤ → x − 0 = x
45 44 breq2d ⊢ P ∈ ℙ ∧ K ∈ ℕ ∧ x ∈ ℤ → P ∥ x − 0 ↔ P ∥ x
46 45 notbid ⊢ P ∈ ℙ ∧ K ∈ ℕ ∧ x ∈ ℤ → ¬ P ∥ x − 0 ↔ ¬ P ∥ x
47 41 46 bitr4d ⊢ P ∈ ℙ ∧ K ∈ ℕ ∧ x ∈ ℤ → x gcd P K = 1 ↔ ¬ P ∥ x − 0
48 27 47 sylan2 ⊢ P ∈ ℙ ∧ K ∈ ℕ ∧ x ∈ 1 … P K → x gcd P K = 1 ↔ ¬ P ∥ x − 0
49 48 biimpd ⊢ P ∈ ℙ ∧ K ∈ ℕ ∧ x ∈ 1 … P K → x gcd P K = 1 → ¬ P ∥ x − 0
50 imnan ⊢ x gcd P K = 1 → ¬ P ∥ x − 0 ↔ ¬ x gcd P K = 1 ∧ P ∥ x − 0
51 49 50 sylib ⊢ P ∈ ℙ ∧ K ∈ ℕ ∧ x ∈ 1 … P K → ¬ x gcd P K = 1 ∧ P ∥ x − 0
52 51 ralrimiva ⊢ P ∈ ℙ ∧ K ∈ ℕ → ∀ x ∈ 1 … P K ¬ x gcd P K = 1 ∧ P ∥ x − 0
53 rabeq0 ⊢ x ∈ 1 … P K | x gcd P K = 1 ∧ P ∥ x − 0 = ∅ ↔ ∀ x ∈ 1 … P K ¬ x gcd P K = 1 ∧ P ∥ x − 0
54 52 53 sylibr ⊢ P ∈ ℙ ∧ K ∈ ℕ → x ∈ 1 … P K | x gcd P K = 1 ∧ P ∥ x − 0 = ∅
55 26 54 eqtrid ⊢ P ∈ ℙ ∧ K ∈ ℕ → x ∈ 1 … P K | x gcd P K = 1 ∩ x ∈ 1 … P K | P ∥ x − 0 = ∅
56 hashun ⊢ x ∈ 1 … P K | x gcd P K = 1 ∈ Fin ∧ x ∈ 1 … P K | P ∥ x − 0 ∈ Fin ∧ x ∈ 1 … P K | x gcd P K = 1 ∩ x ∈ 1 … P K | P ∥ x − 0 = ∅ → x ∈ 1 … P K | x gcd P K = 1 ∪ x ∈ 1 … P K | P ∥ x − 0 = x ∈ 1 … P K | x gcd P K = 1 + x ∈ 1 … P K | P ∥ x − 0
57 22 25 55 56 mp3an12i ⊢ P ∈ ℙ ∧ K ∈ ℕ → x ∈ 1 … P K | x gcd P K = 1 ∪ x ∈ 1 … P K | P ∥ x − 0 = x ∈ 1 … P K | x gcd P K = 1 + x ∈ 1 … P K | P ∥ x − 0
58 unrab ⊢ x ∈ 1 … P K | x gcd P K = 1 ∪ x ∈ 1 … P K | P ∥ x − 0 = x ∈ 1 … P K | x gcd P K = 1 ∨ P ∥ x − 0
59 48 biimprd ⊢ P ∈ ℙ ∧ K ∈ ℕ ∧ x ∈ 1 … P K → ¬ P ∥ x − 0 → x gcd P K = 1
60 59 con1d ⊢ P ∈ ℙ ∧ K ∈ ℕ ∧ x ∈ 1 … P K → ¬ x gcd P K = 1 → P ∥ x − 0
61 60 orrd ⊢ P ∈ ℙ ∧ K ∈ ℕ ∧ x ∈ 1 … P K → x gcd P K = 1 ∨ P ∥ x − 0
62 61 ralrimiva ⊢ P ∈ ℙ ∧ K ∈ ℕ → ∀ x ∈ 1 … P K x gcd P K = 1 ∨ P ∥ x − 0
63 rabid2 ⊢ 1 … P K = x ∈ 1 … P K | x gcd P K = 1 ∨ P ∥ x − 0 ↔ ∀ x ∈ 1 … P K x gcd P K = 1 ∨ P ∥ x − 0
64 62 63 sylibr ⊢ P ∈ ℙ ∧ K ∈ ℕ → 1 … P K = x ∈ 1 … P K | x gcd P K = 1 ∨ P ∥ x − 0
65 58 64 eqtr4id ⊢ P ∈ ℙ ∧ K ∈ ℕ → x ∈ 1 … P K | x gcd P K = 1 ∪ x ∈ 1 … P K | P ∥ x − 0 = 1 … P K
66 65 fveq2d ⊢ P ∈ ℙ ∧ K ∈ ℕ → x ∈ 1 … P K | x gcd P K = 1 ∪ x ∈ 1 … P K | P ∥ x − 0 = 1 … P K
67 4 nnnn0d ⊢ P ∈ ℙ ∧ K ∈ ℕ → P K ∈ ℕ 0
68 hashfz1 ⊢ P K ∈ ℕ 0 → 1 … P K = P K
69 67 68 syl ⊢ P ∈ ℙ ∧ K ∈ ℕ → 1 … P K = P K
70 expm1t ⊢ P ∈ ℂ ∧ K ∈ ℕ → P K = P K − 1 ⁢ P
71 11 70 sylan ⊢ P ∈ ℙ ∧ K ∈ ℕ → P K = P K − 1 ⁢ P
72 66 69 71 3eqtrd ⊢ P ∈ ℙ ∧ K ∈ ℕ → x ∈ 1 … P K | x gcd P K = 1 ∪ x ∈ 1 … P K | P ∥ x − 0 = P K − 1 ⁢ P
73 1 adantr ⊢ P ∈ ℙ ∧ K ∈ ℕ → P ∈ ℕ
74 1zzd ⊢ P ∈ ℙ ∧ K ∈ ℕ → 1 ∈ ℤ
75 nn0uz ⊢ ℕ 0 = ℤ ≥ 0
76 1m1e0 ⊢ 1 − 1 = 0
77 76 fveq2i ⊢ ℤ ≥ 1 − 1 = ℤ ≥ 0
78 75 77 eqtr4i ⊢ ℕ 0 = ℤ ≥ 1 − 1
79 67 78 eleqtrdi ⊢ P ∈ ℙ ∧ K ∈ ℕ → P K ∈ ℤ ≥ 1 − 1
80 0zd ⊢ P ∈ ℙ ∧ K ∈ ℕ → 0 ∈ ℤ
81 73 74 79 80 hashdvds ⊢ P ∈ ℙ ∧ K ∈ ℕ → x ∈ 1 … P K | P ∥ x − 0 = P K − 0 P − 1 - 1 - 0 P
82 4 nncnd ⊢ P ∈ ℙ ∧ K ∈ ℕ → P K ∈ ℂ
83 82 subid1d ⊢ P ∈ ℙ ∧ K ∈ ℕ → P K − 0 = P K
84 83 oveq1d ⊢ P ∈ ℙ ∧ K ∈ ℕ → P K − 0 P = P K P
85 73 nnne0d ⊢ P ∈ ℙ ∧ K ∈ ℕ → P ≠ 0
86 nnz ⊢ K ∈ ℕ → K ∈ ℤ
87 86 adantl ⊢ P ∈ ℙ ∧ K ∈ ℕ → K ∈ ℤ
88 12 85 87 expm1d ⊢ P ∈ ℙ ∧ K ∈ ℕ → P K − 1 = P K P
89 84 88 eqtr4d ⊢ P ∈ ℙ ∧ K ∈ ℕ → P K − 0 P = P K − 1
90 89 fveq2d ⊢ P ∈ ℙ ∧ K ∈ ℕ → P K − 0 P = P K − 1
91 9 nnzd ⊢ P ∈ ℙ ∧ K ∈ ℕ → P K − 1 ∈ ℤ
92 flid ⊢ P K − 1 ∈ ℤ → P K − 1 = P K − 1
93 91 92 syl ⊢ P ∈ ℙ ∧ K ∈ ℕ → P K − 1 = P K − 1
94 90 93 eqtrd ⊢ P ∈ ℙ ∧ K ∈ ℕ → P K − 0 P = P K − 1
95 76 oveq1i ⊢ 1 - 1 - 0 = 0 − 0
96 0m0e0 ⊢ 0 − 0 = 0
97 95 96 eqtri ⊢ 1 - 1 - 0 = 0
98 97 oveq1i ⊢ 1 - 1 - 0 P = 0 P
99 12 85 div0d ⊢ P ∈ ℙ ∧ K ∈ ℕ → 0 P = 0
100 98 99 eqtrid ⊢ P ∈ ℙ ∧ K ∈ ℕ → 1 - 1 - 0 P = 0
101 100 fveq2d ⊢ P ∈ ℙ ∧ K ∈ ℕ → 1 - 1 - 0 P = 0
102 0z ⊢ 0 ∈ ℤ
103 flid ⊢ 0 ∈ ℤ → 0 = 0
104 102 103 ax-mp ⊢ 0 = 0
105 101 104 eqtrdi ⊢ P ∈ ℙ ∧ K ∈ ℕ → 1 - 1 - 0 P = 0
106 94 105 oveq12d ⊢ P ∈ ℙ ∧ K ∈ ℕ → P K − 0 P − 1 - 1 - 0 P = P K − 1 − 0
107 10 subid1d ⊢ P ∈ ℙ ∧ K ∈ ℕ → P K − 1 − 0 = P K − 1
108 81 106 107 3eqtrd ⊢ P ∈ ℙ ∧ K ∈ ℕ → x ∈ 1 … P K | P ∥ x − 0 = P K − 1
109 108 oveq2d ⊢ P ∈ ℙ ∧ K ∈ ℕ → x ∈ 1 … P K | x gcd P K = 1 + x ∈ 1 … P K | P ∥ x − 0 = x ∈ 1 … P K | x gcd P K = 1 + P K − 1
110 hashcl ⊢ x ∈ 1 … P K | x gcd P K = 1 ∈ Fin → x ∈ 1 … P K | x gcd P K = 1 ∈ ℕ 0
111 22 110 ax-mp ⊢ x ∈ 1 … P K | x gcd P K = 1 ∈ ℕ 0
112 111 nn0cni ⊢ x ∈ 1 … P K | x gcd P K = 1 ∈ ℂ
113 addcom ⊢ x ∈ 1 … P K | x gcd P K = 1 ∈ ℂ ∧ P K − 1 ∈ ℂ → x ∈ 1 … P K | x gcd P K = 1 + P K − 1 = P K − 1 + x ∈ 1 … P K | x gcd P K = 1
114 112 10 113 sylancr ⊢ P ∈ ℙ ∧ K ∈ ℕ → x ∈ 1 … P K | x gcd P K = 1 + P K − 1 = P K − 1 + x ∈ 1 … P K | x gcd P K = 1
115 109 114 eqtrd ⊢ P ∈ ℙ ∧ K ∈ ℕ → x ∈ 1 … P K | x gcd P K = 1 + x ∈ 1 … P K | P ∥ x − 0 = P K − 1 + x ∈ 1 … P K | x gcd P K = 1
116 57 72 115 3eqtr3rd ⊢ P ∈ ℙ ∧ K ∈ ℕ → P K − 1 + x ∈ 1 … P K | x gcd P K = 1 = P K − 1 ⁢ P
117 10 12 mulcld ⊢ P ∈ ℙ ∧ K ∈ ℕ → P K − 1 ⁢ P ∈ ℂ
118 112 a1i ⊢ P ∈ ℙ ∧ K ∈ ℕ → x ∈ 1 … P K | x gcd P K = 1 ∈ ℂ
119 117 10 118 subaddd ⊢ P ∈ ℙ ∧ K ∈ ℕ → P K − 1 ⁢ P − P K − 1 = x ∈ 1 … P K | x gcd P K = 1 ↔ P K − 1 + x ∈ 1 … P K | x gcd P K = 1 = P K − 1 ⁢ P
120 116 119 mpbird ⊢ P ∈ ℙ ∧ K ∈ ℕ → P K − 1 ⁢ P − P K − 1 = x ∈ 1 … P K | x gcd P K = 1
121 16 18 120 3eqtrrd ⊢ P ∈ ℙ ∧ K ∈ ℕ → x ∈ 1 … P K | x gcd P K = 1 = P K − 1 ⁢ P − 1
122 6 121 eqtrd ⊢ P ∈ ℙ ∧ K ∈ ℕ → ϕ ⁡ P K = P K − 1 ⁢ P − 1