Metamath Proof Explorer


Theorem dvdssq

Description: Two numbers are divisible iff their squares are. (Contributed by Scott Fenton, 18-Apr-2014) (Revised by Mario Carneiro, 19-Apr-2014)

Ref Expression
Assertion dvdssq MNMNM2N2

Proof

Step Hyp Ref Expression
1 breq1 M=0MN0N
2 sq0i M=0M2=0
3 2 breq1d M=0M2N20N2
4 1 3 bibi12d M=0MNM2N20N0N2
5 nnabscl MM0M
6 breq2 N=0MNM0
7 sq0i N=0N2=0
8 7 breq2d N=0M2N2M20
9 6 8 bibi12d N=0MNM2N2M0M20
10 nnabscl NN0N
11 dvdssqlem MNMNM2N2
12 10 11 sylan2 MNN0MNM2N2
13 nnz MM
14 simpl NN0N
15 dvdsabsb MNMNMN
16 13 14 15 syl2an MNN0MNMN
17 nnsqcl MM2
18 17 nnzd MM2
19 zsqcl NN2
20 19 adantr NN0N2
21 dvdsabsb M2N2M2N2M2N2
22 18 20 21 syl2an MNN0M2N2M2N2
23 zcn NN
24 23 adantr NN0N
25 abssq NN2=N2
26 24 25 syl NN0N2=N2
27 26 breq2d NN0M2N2M2N2
28 27 adantl MNN0M2N2M2N2
29 22 28 bitr4d MNN0M2N2M2N2
30 12 16 29 3bitr4d MNN0MNM2N2
31 30 anassrs MNN0MNM2N2
32 dvds0 MM0
33 zsqcl MM2
34 dvds0 M2M20
35 33 34 syl MM20
36 32 35 2thd MM0M20
37 13 36 syl MM0M20
38 37 adantr MNM0M20
39 9 31 38 pm2.61ne MNMNM2N2
40 5 39 sylan MM0NMNM2N2
41 absdvdsb MNMNMN
42 41 adantlr MM0NMNMN
43 zsqcl MM2
44 43 adantr MM0M2
45 absdvdsb M2N2M2N2M2N2
46 44 19 45 syl2an MM0NM2N2M2N2
47 zcn MM
48 abssq MM2=M2
49 47 48 syl MM2=M2
50 49 eqcomd MM2=M2
51 50 adantr MM0M2=M2
52 51 breq1d MM0M2N2M2N2
53 52 adantr MM0NM2N2M2N2
54 46 53 bitrd MM0NM2N2M2N2
55 40 42 54 3bitr4d MM0NMNM2N2
56 55 an32s MNM0MNM2N2
57 0dvds N0NN=0
58 sqeq0 NN2=0N=0
59 23 58 syl NN2=0N=0
60 57 59 bitr4d N0NN2=0
61 0dvds N20N2N2=0
62 19 61 syl N0N2N2=0
63 60 62 bitr4d N0N0N2
64 63 adantl MN0N0N2
65 4 56 64 pm2.61ne MNMNM2N2