Metamath Proof Explorer


Theorem fmtnoprmfac1

Description: Divisor of Fermat number (special form of Euler's result, see fmtnofac1 ): Let F_n be a Fermat number. Let p be a prime divisor of F_n. Then p is in the form: k*2^(n+1)+1 where k is a positive integer. (Contributed by AV, 25-Jul-2021)

Ref Expression
Assertion fmtnoprmfac1 N P P FermatNo N k P = k 2 N + 1 + 1

Proof

Step Hyp Ref Expression
1 breq1 P = 2 P FermatNo N 2 FermatNo N
2 1 adantr P = 2 N P FermatNo N 2 FermatNo N
3 nnnn0 N N 0
4 fmtnoodd N 0 ¬ 2 FermatNo N
5 3 4 syl N ¬ 2 FermatNo N
6 5 adantl P = 2 N ¬ 2 FermatNo N
7 6 pm2.21d P = 2 N 2 FermatNo N k P = k 2 N + 1 + 1
8 2 7 sylbid P = 2 N P FermatNo N k P = k 2 N + 1 + 1
9 8 a1d P = 2 N P P FermatNo N k P = k 2 N + 1 + 1
10 9 ex P = 2 N P P FermatNo N k P = k 2 N + 1 + 1
11 10 3impd P = 2 N P P FermatNo N k P = k 2 N + 1 + 1
12 simpr1 ¬ P = 2 N P P FermatNo N N
13 neqne ¬ P = 2 P 2
14 13 anim2i P ¬ P = 2 P P 2
15 eldifsn P 2 P P 2
16 14 15 sylibr P ¬ P = 2 P 2
17 16 ex P ¬ P = 2 P 2
18 17 3ad2ant2 N P P FermatNo N ¬ P = 2 P 2
19 18 impcom ¬ P = 2 N P P FermatNo N P 2
20 simpr3 ¬ P = 2 N P P FermatNo N P FermatNo N
21 fmtnoprmfac1lem N P 2 P FermatNo N od P 2 = 2 N + 1
22 12 19 20 21 syl3anc ¬ P = 2 N P P FermatNo N od P 2 = 2 N + 1
23 prmnn P P
24 23 ad2antll ¬ P = 2 N P P
25 2z 2
26 25 a1i ¬ P = 2 N P 2
27 13 necomd ¬ P = 2 2 P
28 27 adantr ¬ P = 2 N P 2 P
29 2prm 2
30 29 a1i N 2
31 30 anim1i N P 2 P
32 31 adantl ¬ P = 2 N P 2 P
33 prmrp 2 P 2 gcd P = 1 2 P
34 32 33 syl ¬ P = 2 N P 2 gcd P = 1 2 P
35 28 34 mpbird ¬ P = 2 N P 2 gcd P = 1
36 odzphi P 2 2 gcd P = 1 od P 2 ϕ P
37 24 26 35 36 syl3anc ¬ P = 2 N P od P 2 ϕ P
38 phiprm P ϕ P = P 1
39 38 ad2antll ¬ P = 2 N P ϕ P = P 1
40 39 breq2d ¬ P = 2 N P od P 2 ϕ P od P 2 P 1
41 breq1 od P 2 = 2 N + 1 od P 2 P 1 2 N + 1 P 1
42 41 adantl ¬ P = 2 N P od P 2 = 2 N + 1 od P 2 P 1 2 N + 1 P 1
43 2nn 2
44 43 a1i N 2
45 peano2nn N N + 1
46 45 nnnn0d N N + 1 0
47 44 46 nnexpcld N 2 N + 1
48 23 nnnn0d P P 0
49 prmuz2 P P 2
50 eluzle P 2 2 P
51 49 50 syl P 2 P
52 nn0ge2m1nn P 0 2 P P 1
53 48 51 52 syl2anc P P 1
54 47 53 anim12i N P 2 N + 1 P 1
55 54 adantl ¬ P = 2 N P 2 N + 1 P 1
56 nndivides 2 N + 1 P 1 2 N + 1 P 1 k k 2 N + 1 = P 1
57 55 56 syl ¬ P = 2 N P 2 N + 1 P 1 k k 2 N + 1 = P 1
58 eqcom k 2 N + 1 = P 1 P 1 = k 2 N + 1
59 58 a1i ¬ P = 2 N P k k 2 N + 1 = P 1 P 1 = k 2 N + 1
60 23 nncnd P P
61 60 adantl N P P
62 61 adantr N P k P
63 1cnd N P k 1
64 nncn k k
65 64 adantl N P k k
66 peano2nn0 N 0 N + 1 0
67 3 66 syl N N + 1 0
68 44 67 nnexpcld N 2 N + 1
69 68 nncnd N 2 N + 1
70 69 adantr N P 2 N + 1
71 70 adantr N P k 2 N + 1
72 65 71 mulcld N P k k 2 N + 1
73 62 63 72 subadd2d N P k P 1 = k 2 N + 1 k 2 N + 1 + 1 = P
74 73 adantll ¬ P = 2 N P k P 1 = k 2 N + 1 k 2 N + 1 + 1 = P
75 eqcom k 2 N + 1 + 1 = P P = k 2 N + 1 + 1
76 75 a1i ¬ P = 2 N P k k 2 N + 1 + 1 = P P = k 2 N + 1 + 1
77 59 74 76 3bitrd ¬ P = 2 N P k k 2 N + 1 = P 1 P = k 2 N + 1 + 1
78 77 rexbidva ¬ P = 2 N P k k 2 N + 1 = P 1 k P = k 2 N + 1 + 1
79 78 biimpd ¬ P = 2 N P k k 2 N + 1 = P 1 k P = k 2 N + 1 + 1
80 57 79 sylbid ¬ P = 2 N P 2 N + 1 P 1 k P = k 2 N + 1 + 1
81 80 adantr ¬ P = 2 N P od P 2 = 2 N + 1 2 N + 1 P 1 k P = k 2 N + 1 + 1
82 42 81 sylbid ¬ P = 2 N P od P 2 = 2 N + 1 od P 2 P 1 k P = k 2 N + 1 + 1
83 82 ex ¬ P = 2 N P od P 2 = 2 N + 1 od P 2 P 1 k P = k 2 N + 1 + 1
84 83 com23 ¬ P = 2 N P od P 2 P 1 od P 2 = 2 N + 1 k P = k 2 N + 1 + 1
85 40 84 sylbid ¬ P = 2 N P od P 2 ϕ P od P 2 = 2 N + 1 k P = k 2 N + 1 + 1
86 37 85 mpd ¬ P = 2 N P od P 2 = 2 N + 1 k P = k 2 N + 1 + 1
87 86 3adantr3 ¬ P = 2 N P P FermatNo N od P 2 = 2 N + 1 k P = k 2 N + 1 + 1
88 22 87 mpd ¬ P = 2 N P P FermatNo N k P = k 2 N + 1 + 1
89 88 ex ¬ P = 2 N P P FermatNo N k P = k 2 N + 1 + 1
90 11 89 pm2.61i N P P FermatNo N k P = k 2 N + 1 + 1