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Metamath Proof Explorer

Theorem pythagtriplem8

Description: Lemma for pythagtrip . Show that ( sqrt( C - B ) ) is a positive integer. (Contributed by Scott Fenton, 17-Apr-2014) (Revised by Mario Carneiro, 19-Apr-2014)

Ref Expression
Assertion pythagtriplem8 
|- ( ( ( A e. NN /\ B e. NN /\ C e. NN ) /\ ( ( A ^ 2 ) + ( B ^ 2 ) ) = ( C ^ 2 ) /\ ( ( A gcd B ) = 1 /\ -. 2 || A ) ) -> ( sqrt ` ( C - B ) ) e. NN )

Proof

Step Hyp Ref Expression
1 pythagtriplem6 
 |-  ( ( ( A e. NN /\ B e. NN /\ C e. NN ) /\ ( ( A ^ 2 ) + ( B ^ 2 ) ) = ( C ^ 2 ) /\ ( ( A gcd B ) = 1 /\ -. 2 || A ) ) -> ( sqrt ` ( C - B ) ) = ( ( C - B ) gcd A ) )
2 nnz 
 |-  ( C e. NN -> C e. ZZ )
3 nnz 
 |-  ( B e. NN -> B e. ZZ )
4 zsubcl 
 |-  ( ( C e. ZZ /\ B e. ZZ ) -> ( C - B ) e. ZZ )
5 2 3 4 syl2anr 
 |-  ( ( B e. NN /\ C e. NN ) -> ( C - B ) e. ZZ )
6 5 3adant1 
 |-  ( ( A e. NN /\ B e. NN /\ C e. NN ) -> ( C - B ) e. ZZ )
7 nnz 
 |-  ( A e. NN -> A e. ZZ )
8 7 3ad2ant1 
 |-  ( ( A e. NN /\ B e. NN /\ C e. NN ) -> A e. ZZ )
9 nnne0 
 |-  ( A e. NN -> A =/= 0 )
10 9 neneqd 
 |-  ( A e. NN -> -. A = 0 )
11 10 intnand 
 |-  ( A e. NN -> -. ( ( C - B ) = 0 /\ A = 0 ) )
12 11 3ad2ant1 
 |-  ( ( A e. NN /\ B e. NN /\ C e. NN ) -> -. ( ( C - B ) = 0 /\ A = 0 ) )
13 gcdn0cl 
 |-  ( ( ( ( C - B ) e. ZZ /\ A e. ZZ ) /\ -. ( ( C - B ) = 0 /\ A = 0 ) ) -> ( ( C - B ) gcd A ) e. NN )
14 6 8 12 13 syl21anc 
 |-  ( ( A e. NN /\ B e. NN /\ C e. NN ) -> ( ( C - B ) gcd A ) e. NN )
15 14 3ad2ant1 
 |-  ( ( ( A e. NN /\ B e. NN /\ C e. NN ) /\ ( ( A ^ 2 ) + ( B ^ 2 ) ) = ( C ^ 2 ) /\ ( ( A gcd B ) = 1 /\ -. 2 || A ) ) -> ( ( C - B ) gcd A ) e. NN )
16 1 15 eqeltrd 
 |-  ( ( ( A e. NN /\ B e. NN /\ C e. NN ) /\ ( ( A ^ 2 ) + ( B ^ 2 ) ) = ( C ^ 2 ) /\ ( ( A gcd B ) = 1 /\ -. 2 || A ) ) -> ( sqrt ` ( C - B ) ) e. NN )