Metamath Proof Explorer


Theorem flt4lem5

Description: In the context of the lemmas of pythagtrip , M and N are coprime. (Contributed by SN, 23-Aug-2024)

Ref Expression
Hypotheses flt4lem5.1 M = C + B + C B 2
flt4lem5.2 N = C + B C B 2
Assertion flt4lem5 A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A M gcd N = 1

Proof

Step Hyp Ref Expression
1 flt4lem5.1 M = C + B + C B 2
2 flt4lem5.2 N = C + B C B 2
3 simp3l A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A A gcd B = 1
4 simp11 A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A A
5 simp12 A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A B
6 coprmgcdb A B i i A i B i = 1 A gcd B = 1
7 4 5 6 syl2anc A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i A i B i = 1 A gcd B = 1
8 3 7 mpbird A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i A i B i = 1
9 simplr A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N i
10 9 nnzd A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N i
11 1 pythagtriplem11 A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A M
12 11 ad2antrr A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N M
13 12 nnsqcld A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N M 2
14 13 nnzd A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N M 2
15 2 pythagtriplem13 A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A N
16 15 ad2antrr A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N N
17 16 nnsqcld A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N N 2
18 17 nnzd A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N N 2
19 simprl A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N i M
20 12 nnzd A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N M
21 2nn 2
22 21 a1i A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N 2
23 dvdsexp2im i M 2 i M i M 2
24 10 20 22 23 syl3anc A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N i M i M 2
25 19 24 mpd A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N i M 2
26 simprr A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N i N
27 16 nnzd A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N N
28 dvdsexp2im i N 2 i N i N 2
29 10 27 22 28 syl3anc A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N i N i N 2
30 26 29 mpd A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N i N 2
31 10 14 18 25 30 dvds2subd A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N i M 2 N 2
32 1 2 pythagtriplem15 A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A A = M 2 N 2
33 32 ad2antrr A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N A = M 2 N 2
34 31 33 breqtrrd A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N i A
35 2z 2
36 35 a1i A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N 2
37 12 16 nnmulcld A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N M N
38 37 nnzd A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N M N
39 10 20 27 26 dvdsmultr2d A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N i M N
40 10 36 38 39 dvdsmultr2d A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N i 2 M N
41 1 2 pythagtriplem16 A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A B = 2 M N
42 41 ad2antrr A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N B = 2 M N
43 40 42 breqtrrd A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N i B
44 34 43 jca A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N i A i B
45 44 ex A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N i A i B
46 45 imim1d A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i A i B i = 1 i M i N i = 1
47 46 ralimdva A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i A i B i = 1 i i M i N i = 1
48 8 47 mpd A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N i = 1
49 coprmgcdb M N i i M i N i = 1 M gcd N = 1
50 11 15 49 syl2anc A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A i i M i N i = 1 M gcd N = 1
51 48 50 mpbid A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A M gcd N = 1