let p be Prime; :: thesis:
let a, m be Nat; :: thesis:
let f be FinSequence of NAT ; :: thesis:
assume that
A1: p > 2 and
A2: a gcd p = 1 and
A3: f = a * (idseq ((p -' 1) div 2)) and
A4: m = card { k where k is Element of NAT : ( k in rng (f mod p) & k > p / 2 ) } ; :: thesis:
set f1 = f mod p;
A5: len (f mod p) = len f by EULER_2:def 2;
set X = { k where k is Element of NAT : ( k in rng (f mod p) & k > p / 2 ) } ;
for x being object st x in { k where k is Element of NAT : ( k in rng (f mod p) & k > p / 2 ) } holds
x in rng (f mod p)

proof

let x be object ; :: thesis:
assume x in { k where k is Element of NAT : ( k in rng (f mod p) & k > p / 2 ) } ; :: thesis:
then ex k being Element of NAT st
( x = k & k in rng (f mod p) & k > p / 2 ) ;
hence x in rng (f mod p) ; :: thesis:

end;

then A6: { k where k is Element of NAT : ( k in rng (f mod p) & k > p / 2 ) } c= rng (f mod p) ;
then reconsider X = { k where k is Element of NAT : ( k in rng (f mod p) & k > p / 2 ) } as finite set ;
reconsider X = X as finite Subset of NAT by A6, XBOOLE_1:1;
card X is Element of NAT ;
then reconsider m = m as Element of NAT by A4;
A7: (rng (f mod p)) \ X c= rng (f mod p) by XBOOLE_1:36;
reconsider Y = (rng (f mod p)) \ X as finite Subset of NAT ;
A8: a |^ ((p -' 1) div 2), Lege (a,p) are_congruent_mod p by A1, A2, Th28, INT_1:14;
set f2 = Sgm (rng (f mod p));
(Product (f mod p)) mod p = (Product f) mod p by EULER_2:11;
then A9: Product (f mod p), Product f are_congruent_mod p by NAT_D:64;
A10: p > 1 by INT_2:def 4;
then A11: p -' 1 = p - 1 by XREAL_1:233;
then A12: p -' 1 > 0 by A10, XREAL_1:50;
set p9 = (p -' 1) div 2;
rng (idseq ((p -' 1) div 2)) = Seg ((p -' 1) div 2) ;
then reconsider I = idseq ((p -' 1) div 2) as FinSequence of NAT by FINSEQ_1:def 4;
dom f = dom I by A3, VALUED_1:def 5;
then A13: len f = len I by FINSEQ_3:29
.= (p -' 1) div 2 by CARD_1:def 7 ;
p >= 2 + 1 by A1, NAT_1:13;
then p - 1 >= 3 - 1 by XREAL_1:9;
then f mod p <> {} by A13, A11, A5, NAT_2:13;
then rng (f mod p) is non empty Subset of NAT ;
then consider n1 being Element of NAT such that
A14: rng (f mod p) c= (Seg n1) \/ {0} by HEYTING3:1;
I is Element of ((p -' 1) div 2) -tuples_on REAL by FINSEQ_2:109, NUMBERS:19;
then A15: Product f = (Product (((p -' 1) div 2) |-> a)) * (Product I) by A3, RVSUM_1:108
.= (a |^ ((p -' 1) div 2)) * (Product I) by NEWTON:def 1 ;
p is odd by A1, PEPIN:17;
then A16: p -' 1 is even by A11, HILBERT3:2;
then A17: (p -' 1) div 2 = ((p -' 1) + 1) div 2 by NAT_2:26
.= p div 2 by A10, XREAL_1:235 ;
2 divides p -' 1 by A16, PEPIN:22;
then A18: p -' 1 = 2 * ((p -' 1) div 2) ;
then (p -' 1) div 2 divides p -' 1 ;
then (p -' 1) div 2 <= p -' 1 by A12, NAT_D:7;
then A19: (p -' 1) div 2 < p by A11, XREAL_1:146, XXREAL_0:2;
for d being Nat st d in dom I holds
(I . d) gcd p = 1

proof

let d be Nat; :: thesis:
assume d in dom I ; :: thesis:
then A20: d in Seg (len I) by FINSEQ_1:def 3;
then A21: d in Seg ((p -' 1) div 2) by CARD_1:def 7;
then A22: I . d = d by FINSEQ_2:49;
d <= (p -' 1) div 2 by A21, FINSEQ_1:1;
then A23: d < p by A19, XXREAL_0:2;
d >= 1 by A20, FINSEQ_1:1;
then d,p are_coprime by A23, EULER_1:2;
hence (I . d) gcd p = 1 by A22, INT_2:def 3; :: thesis:

end;

then A24: (Product I) gcd p = 1 by WSIERP_1:36;
A25: for d being Nat st d in dom f holds
f . d = a * d

proof

let d be Nat; :: thesis:
assume A26: d in dom f ; :: thesis:
then d in dom I by A3, VALUED_1:def 5;
then d in Seg (len I) by FINSEQ_1:def 3;
then A27: d is Element of Seg ((p -' 1) div 2) by CARD_1:def 7;
thus f . d = a * (I . d) by A3, A26, VALUED_1:def 5
.= a * d by A27 ; :: thesis:

end;

A28: for d, e being Nat st 1 <= d & d < e & e <= len (f mod p) holds
(f mod p) . d <> (f mod p) . e

proof

let d, e be Nat; :: thesis:
assume that
A29: 1 <= d and
A30: d < e and
A31: e <= len (f mod p) ; :: thesis:
A32: e <= len f by A31, EULER_2:def 2;
1 <= e by A29, A30, XXREAL_0:2;
then A33: e in dom f by A32, FINSEQ_3:25;
then A34: (f mod p) . e = (f . e) mod p by EULER_2:def 2;
d < len f by A30, A32, XXREAL_0:2;
then A35: d in dom f by A29, FINSEQ_3:25;
then A36: (f mod p) . d = (f . d) mod p by EULER_2:def 2;

now :: thesis:

assume (f mod p) . d = (f mod p) . e ; :: thesis:
then f . e,f . d are_congruent_mod p by A36, A34, NAT_D:64;
then p divides (a * e) - (f . d) by A25, A33;
then p divides (a * e) - (a * d) by A25, A35;
then A37: p divides a * (e - d) ;
A38: ((p -' 1) div 2) - 1 < p by A19, XREAL_1:147;
reconsider dd = e - d as Element of NAT by A30, NAT_1:21;
A39: |.p.| = p by ABSVALUE:def 1;
A40: |.dd.| = dd by ABSVALUE:def 1;
A41: dd <= ((p -' 1) div 2) - 1 by A13, A5, A29, A31, XREAL_1:13;
dd <> 0 by A30;
then p <= dd by A2, A37, A39, A40, INT_4:6, WSIERP_1:29;
hence contradiction by A41, A38, XXREAL_0:2; :: thesis:

end;

hence (f mod p) . d <> (f mod p) . e ; :: thesis:

end;

then A42: len (f mod p) = card (rng (f mod p)) by GRAPH_5:7;
then A43: f mod p is one-to-one by FINSEQ_4:62;
A44: dom (f mod p) = dom f by A5, FINSEQ_3:29;
not 0 in rng (f mod p)

proof

reconsider a = a as Element of NAT by ORDINAL1:def 12;
assume 0 in rng (f mod p) ; :: thesis:
then consider n being Nat such that
A45: n in dom (f mod p) and
A46: (f mod p) . n = 0 by FINSEQ_2:10;
0 = (f . n) mod p by A44, A45, A46, EULER_2:def 2
.= (a * n) mod p by A25, A44, A45 ;
then A47: p divides a * n by PEPIN:6;
n >= 1 by A45, FINSEQ_3:25;
then A48: p <= n by A2, A47, NAT_D:7, WSIERP_1:30;
n <= (p -' 1) div 2 by A13, A5, A45, FINSEQ_3:25;
hence contradiction by A19, A48, XXREAL_0:2; :: thesis:

end;

proof

let k, l be Nat; :: thesis:
assume that
A59: k in Y and
A60: l in X ; :: thesis:
A61: not k in X by A59, XBOOLE_0:def 5;
A62: ex l1 being Element of NAT st
( l1 = l & l1 in rng (f mod p) & l1 > p / 2 ) by A60;
k in rng (f mod p) by A59, XBOOLE_0:def 5;
then k <= p / 2 by A61;
hence k < l by A62, XXREAL_0:2; :: thesis:

end;

then Sgm (Y \/ X) = (Sgm Y) ^ (Sgm X) by a52, a53, FINSEQ_3:42;
then Sgm ((rng (f mod p)) \/ X) = (Sgm Y) ^ (Sgm X) by XBOOLE_1:39;
then A63: Sgm (rng (f mod p)) = (Sgm Y) ^ (Sgm X) by A6, XBOOLE_1:12;
A64: for d being Nat st d in dom (((len ((Sgm (rng (f mod p))) /^ n)) |-> p) - ((Sgm (rng (f mod p))) /^ n)) holds
(((len ((Sgm (rng (f mod p))) /^ n)) |-> p) - ((Sgm (rng (f mod p))) /^ n)) . d = p - (((Sgm (rng (f mod p))) /^ n) . d)

proof

let d be Nat; :: thesis:
assume A65: d in dom (((len ((Sgm (rng (f mod p))) /^ n)) |-> p) - ((Sgm (rng (f mod p))) /^ n)) ; :: thesis:
then d in Seg (len ((Sgm (rng (f mod p))) /^ n)) by A57, FINSEQ_1:def 3;
then ((len ((Sgm (rng (f mod p))) /^ n)) |-> p) . d = p by FINSEQ_2:57;
hence (((len ((Sgm (rng (f mod p))) /^ n)) |-> p) - ((Sgm (rng (f mod p))) /^ n)) . d = p - (((Sgm (rng (f mod p))) /^ n) . d) by A65, VALUED_1:13; :: thesis:

end;

A66: len (Sgm Y) = card Y by a53, FINSEQ_3:39
.= ((p -' 1) div 2) - m by A4, A13, A5, A6, A42, CARD_2:44 ;
then A67: (Sgm (rng (f mod p))) /^ n = Sgm X by A63, FINSEQ_5:37;
A68: for d being Nat st d in dom (((len ((Sgm (rng (f mod p))) /^ n)) |-> p) - ((Sgm (rng (f mod p))) /^ n)) holds
( (((len ((Sgm (rng (f mod p))) /^ n)) |-> p) - ((Sgm (rng (f mod p))) /^ n)) . d > 0 & (((len ((Sgm (rng (f mod p))) /^ n)) |-> p) - ((Sgm (rng (f mod p))) /^ n)) . d <= (p -' 1) div 2 )

proof

let d be Nat; :: thesis:
reconsider w = (((len ((Sgm (rng (f mod p))) /^ n)) |-> p) - ((Sgm (rng (f mod p))) /^ n)) . d as Element of INT by INT_1:def 2;
assume A69: d in dom (((len ((Sgm (rng (f mod p))) /^ n)) |-> p) - ((Sgm (rng (f mod p))) /^ n)) ; :: thesis:
then (Sgm X) . d in rng (Sgm X) by A67, A57, FUNCT_1:3;
then (Sgm X) . d in X by a52, FINSEQ_1:def 14;
then A70: ex ll being Element of NAT st
( ll = (Sgm X) . d & ll in rng (f mod p) & ll > p / 2 ) ;
then consider e being Nat such that
A71: e in dom (f mod p) and
A72: (f mod p) . e = ((Sgm (rng (f mod p))) /^ n) . d by A67, FINSEQ_2:10;
((Sgm (rng (f mod p))) /^ n) . d = (f . e) mod p by A44, A71, A72, EULER_2:def 2;
then A73: ((Sgm (rng (f mod p))) /^ n) . d < p by NAT_D:1;
A74: (((len ((Sgm (rng (f mod p))) /^ n)) |-> p) - ((Sgm (rng (f mod p))) /^ n)) . d = p - (((Sgm (rng (f mod p))) /^ n) . d) by A64, A69;
then w < p - (p / 2) by A67, A70, XREAL_1:10;
hence ( (((len ((Sgm (rng (f mod p))) /^ n)) |-> p) - ((Sgm (rng (f mod p))) /^ n)) . d > 0 & (((len ((Sgm (rng (f mod p))) /^ n)) |-> p) - ((Sgm (rng (f mod p))) /^ n)) . d <= (p -' 1) div 2 ) by A17, A74, A73, INT_1:54, XREAL_1:50; :: thesis:

end;

for d being Nat st d in dom (((len ((Sgm (rng (f mod p))) /^ n)) |-> p) - ((Sgm (rng (f mod p))) /^ n)) holds
(((len ((Sgm (rng (f mod p))) /^ n)) |-> p) - ((Sgm (rng (f mod p))) /^ n)) . d in NAT

proof

let d be Nat; :: thesis:
assume A76: d in dom (((len ((Sgm (rng (f mod p))) /^ n)) |-> p) - ((Sgm (rng (f mod p))) /^ n)) ; :: thesis:
(((len ((Sgm (rng (f mod p))) /^ n)) |-> p) - ((Sgm (rng (f mod p))) /^ n)) . d > 0 by A68, A76;
hence (((len ((Sgm (rng (f mod p))) /^ n)) |-> p) - ((Sgm (rng (f mod p))) /^ n)) . d in NAT by INT_1:3; :: thesis:

end;

then reconsider f3 = ((len ((Sgm (rng (f mod p))) /^ n)) |-> p) - ((Sgm (rng (f mod p))) /^ n) as FinSequence of NAT by FINSEQ_2:12;
|.((- 1) |^ m).| = 1 by SERIES_2:1;
then A77: ( (- 1) |^ m = 1 or - ((- 1) |^ m) = 1 ) by ABSVALUE:1;
f3 is FinSequence of NAT ;
then reconsider f4 = ((Sgm (rng (f mod p))) | n) ^ (((len ((Sgm (rng (f mod p))) /^ n)) |-> p) - ((Sgm (rng (f mod p))) /^ n)) as FinSequence of NAT by FINSEQ_1:75;
A78: (Sgm (rng (f mod p))) | n = Sgm Y by A63, A66, FINSEQ_3:113, FINSEQ_6:10;
A79: for d being Nat st d in dom f4 holds
( f4 . d > 0 & f4 . d <= (p -' 1) div 2 )

proof

let d be Nat; :: thesis:
assume A80: d in dom f4 ; :: thesis:

per cases by A80, FINSEQ_1:25;

suppose A81: d in dom ((Sgm (rng (f mod p))) | n) ; :: thesis:

reconsider d = d as Element of NAT by ORDINAL1:def 12;
((Sgm (rng (f mod p))) | n) . d in rng (Sgm Y) by A78, A81, FUNCT_1:3;
then A82: ((Sgm (rng (f mod p))) | n) . d in Y by a53, FINSEQ_1:def 14;
then A83: ((Sgm (rng (f mod p))) | n) . d in rng (f mod p) by XBOOLE_0:def 5;
not ((Sgm (rng (f mod p))) | n) . d in X by A82, XBOOLE_0:def 5;
then ((Sgm (rng (f mod p))) | n) . d <= p / 2 by A83;
then A84: ((Sgm (rng (f mod p))) | n) . d <= (p -' 1) div 2 by A17, INT_1:54;
not ((Sgm (rng (f mod p))) | n) . d in {0} by A49, A83, XBOOLE_0:3;
then ((Sgm (rng (f mod p))) | n) . d <> 0 by TARSKI:def 1;
hence ( f4 . d > 0 & f4 . d <= (p -' 1) div 2 ) by A81, A84, FINSEQ_1:def 7; :: thesis:

end;

suppose ex l being Nat st
( l in dom f3 & d = (len ((Sgm (rng (f mod p))) | n)) + l ) ; :: thesis:

then consider l being Element of NAT such that
A85: l in dom f3 and
A86: d = (len ((Sgm (rng (f mod p))) | n)) + l ;
f4 . d = f3 . l by A85, A86, FINSEQ_1:def 7;
hence ( f4 . d > 0 & f4 . d <= (p -' 1) div 2 ) by A68, A85; :: thesis:

end;

A87: Sgm (rng (f mod p)) = ((Sgm (rng (f mod p))) | n) ^ ((Sgm (rng (f mod p))) /^ n) by RFINSEQ:8;
then A88: (Sgm (rng (f mod p))) /^ n is one-to-one by A50, FINSEQ_3:91;
for d, e being Nat st 1 <= d & d < e & e <= len f3 holds
f3 . d <> f3 . e

proof

let d, e be Nat; :: thesis:
assume that
A89: 1 <= d and
A90: d < e and
A91: e <= len f3 ; :: thesis:
1 <= e by A89, A90, XXREAL_0:2;
then A92: e in dom f3 by A91, FINSEQ_3:25;
then A93: f3 . e = p - (((Sgm (rng (f mod p))) /^ n) . e) by A64;
d < len f3 by A90, A91, XXREAL_0:2;
then A94: d in dom f3 by A89, FINSEQ_3:25;
then f3 . d = p - (((Sgm (rng (f mod p))) /^ n) . d) by A64;
hence f3 . d <> f3 . e by A88, A57, A90, A94, A92, A93; :: thesis:

end;

then len f3 = card (rng f3) by GRAPH_5:7;
then A95: f3 is one-to-one by FINSEQ_4:62;
A96: len (Sgm (rng (f mod p))) = (p -' 1) div 2 by ppp, A13, A5, A42, FINSEQ_3:39;
then A97: n <= len (Sgm (rng (f mod p))) by XREAL_1:43;
A98: rng ((Sgm (rng (f mod p))) | n) misses rng f3

proof

assume rng ((Sgm (rng (f mod p))) | n) meets rng f3 ; :: thesis:
then consider x being object such that
A99: x in rng ((Sgm (rng (f mod p))) | n) and
A100: x in rng f3 by XBOOLE_0:3;
consider e being Nat such that
A101: e in dom f3 and
A102: f3 . e = x by A100, FINSEQ_2:10;
x = p - (((Sgm (rng (f mod p))) /^ n) . e) by A64, A101, A102;
then A103: x = p - ((Sgm (rng (f mod p))) . (e + n)) by A97, A57, A101, RFINSEQ:def 1;
e + n in dom (Sgm (rng (f mod p))) by A57, A101, FINSEQ_5:26;
then consider e1 being Nat such that
A104: e1 in dom (f mod p) and
A105: (f mod p) . e1 = (Sgm (rng (f mod p))) . (e + n) by A54, FINSEQ_2:10, FUNCT_1:3;
A106: e1 in dom f by A5, A104, FINSEQ_3:29;
A107: e1 <= (p -' 1) div 2 by A13, A5, A104, FINSEQ_3:25;
rng ((Sgm (rng (f mod p))) | n) c= rng (Sgm (rng (f mod p))) by FINSEQ_5:19;
then consider d1 being Nat such that
A108: d1 in dom (f mod p) and
A109: (f mod p) . d1 = x by A54, A99, FINSEQ_2:10;
d1 <= (p -' 1) div 2 by A13, A5, A108, FINSEQ_3:25;
then d1 + e1 <= ((p -' 1) div 2) + ((p -' 1) div 2) by A107, XREAL_1:7;
then A110: d1 + e1 < p by A11, A18, XREAL_1:146, XXREAL_0:2;
x = (f . d1) mod p by A44, A108, A109, EULER_2:def 2;
then ((f . d1) mod p) + ((Sgm (rng (f mod p))) . (e + n)) = p by A103;
then ((f . d1) mod p) + ((f . e1) mod p) = p by A105, A106, EULER_2:def 2;
then (((f . d1) mod p) + ((f . e1) mod p)) mod p = 0 by NAT_D:25;
then ((f . d1) + (f . e1)) mod p = 0 by NAT_D:66;
then p divides (f . d1) + (f . e1) by PEPIN:6;
then p divides (d1 * a) + (f . e1) by A25, A44, A108;
then p divides (d1 * a) + (e1 * a) by A25, A106;
then A111: p divides (d1 + e1) * a ;
d1 >= 1 by A108, FINSEQ_3:25;
hence contradiction by A2, A111, A110, NAT_D:7, WSIERP_1:30; :: thesis:

end;

(Sgm (rng (f mod p))) | n is one-to-one by A50, A87, FINSEQ_3:91;
then A112: f4 is one-to-one by A95, A98, FINSEQ_3:91;
A113: for d being Nat st d in dom f3 holds
f3 . d, - (((Sgm (rng (f mod p))) /^ n) . d) are_congruent_mod p

proof

let d be Nat; :: thesis:
assume d in dom f3 ; :: thesis:
then (f3 . d) mod p = (p - (((Sgm (rng (f mod p))) /^ n) . d)) mod p by A64
.= ((1 * p) + (- (((Sgm (rng (f mod p))) /^ n) . d))) mod p
.= (- (((Sgm (rng (f mod p))) /^ n) . d)) mod p by NAT_D:61 ;
hence f3 . d, - (((Sgm (rng (f mod p))) /^ n) . d) are_congruent_mod p by NAT_D:64; :: thesis:

end;

per cases ) |^ m = 1 or (- 1) |^ m = - 1 ) by A77;

suppose A119: (- 1) |^ m = 1 ; :: thesis:

then A120: Lege (a,p) <> - 1 by A118, A1, NAT_D:7;

now :: thesis:

assume Lege (a,p) = 0 ; :: thesis:
then p divides 1 - 0 by A118, A119;
then p = 1 by WSIERP_1:15;
hence contradiction by A1; :: thesis:

end;

hence Lege (a,p) = (- 1) |^ m by A119, Th25, A120; :: thesis:

end;

suppose A121: (- 1) |^ m = - 1 ; :: thesis:

A122: now :: thesis:

assume Lege (a,p) = 1 ; :: thesis:
then p divides - 2 by A118, A121;
then p divides 2 by INT_2:10;
hence contradiction by A1, NAT_D:7; :: thesis:

end;

now :: thesis:

assume Lege (a,p) = 0 ; :: thesis:
then p divides (- 1) - 0 by A118, A121;
then p = 1 by WSIERP_1:15, INT_2:10;
hence contradiction by A1; :: thesis:

end;

hence Lege (a,p) = (- 1) |^ m by A121, Th25, A122; :: thesis:

end;