let n, x, y, z, t be positive Nat; ( x <= y & y <= z implies ( ((n |^ x) + (n |^ y)) + (n |^ z) = n |^ t iff ( ( n = 2 & y = x & z = x + 1 & t = x + 2 ) or ( n = 3 & y = x & z = x & t = x + 1 ) ) ) )
assume that
A1:
x <= y
and
A2:
y <= z
; ( ((n |^ x) + (n |^ y)) + (n |^ z) = n |^ t iff ( ( n = 2 & y = x & z = x + 1 & t = x + 2 ) or ( n = 3 & y = x & z = x & t = x + 1 ) ) )
thus
( not ((n |^ x) + (n |^ y)) + (n |^ z) = n |^ t or ( n = 2 & y = x & z = x + 1 & t = x + 2 ) or ( n = 3 & y = x & z = x & t = x + 1 ) )
( ( ( n = 2 & y = x & z = x + 1 & t = x + 2 ) or ( n = 3 & y = x & z = x & t = x + 1 ) ) implies ((n |^ x) + (n |^ y)) + (n |^ z) = n |^ t )proof
assume A3:
((n |^ x) + (n |^ y)) + (n |^ z) = n |^ t
;
( ( n = 2 & y = x & z = x + 1 & t = x + 2 ) or ( n = 3 & y = x & z = x & t = x + 1 ) )
reconsider yx =
y - x as
Element of
NAT by A1, INT_1:5;
reconsider zx =
z - x as
Element of
NAT by A1, A2, XXREAL_0:2, INT_1:5;
n <> 1
by A3;
then A4:
n >= 2
by NAT_1:23;
then A5:
n >= 1
by XXREAL_0:2;
then A6:
n |^ x <= n |^ y
by A1, PREPOWER:93;
n |^ y <= n |^ z
by A2, A5, PREPOWER:93;
then A7:
(n |^ x) + (n |^ y) <= (n |^ y) + (n |^ z)
by A6, XREAL_1:7;
per cases
( n = 2 or n > 2 )
by A4, XXREAL_0:1;
suppose A8:
n = 2
;
( ( n = 2 & y = x & z = x + 1 & t = x + 2 ) or ( n = 3 & y = x & z = x & t = x + 1 ) )then reconsider tx =
t - x as non
zero Element of
NAT by INT_1:3;
A11:
( 2
to_power yx = (2 to_power y) / (2 to_power x) & 2
to_power zx = (2 to_power z) / (2 to_power x) & 2
to_power tx = (2 to_power t) / (2 to_power x) )
by POWER:29;
A12:
(((n |^ x) / (n |^ x)) + ((n |^ y) / (n |^ x))) + ((n |^ z) / (n |^ x)) = (n |^ t) / (n |^ x)
by A3;
then A13:
(1 + (2 |^ yx)) + (2 |^ zx) = 2
|^ tx
by A8, A11, XCMPLX_1:60;
A14:
2
to_power ((zx - 1) + 1) = (2 to_power (zx - 1)) * (2 to_power 1)
by POWER:27;
A15:
2
to_power ((tx - 1) + 1) = (2 to_power (tx - 1)) * (2 to_power 1)
by POWER:27;
then A18:
x = y
by A1, XXREAL_0:1;
then A19:
(1 + 1) + (2 |^ zx) = 2
|^ tx
by A13, NEWTON:4;
then
2
+ (2 |^ ((zx - 1) + 1)) = 2
|^ ((tx - 1) + 1)
;
then A20:
1
+ (2 to_power (zx - 1)) = 2
to_power (tx - 1)
by A14, A15;
A21:
now not zx <> 1assume
zx <> 1
;
contradiction end; then
1
+ 1
= 2
to_power (tx - 1)
by A20, POWER:24;
then
2
to_power 1
= 2
to_power (tx - 1)
;
then
tx - 1
= 1
by Th48;
hence
( (
n = 2 &
y = x &
z = x + 1 &
t = x + 2 ) or (
n = 3 &
y = x &
z = x &
t = x + 1 ) )
by A1, A8, A16, A21, XXREAL_0:1;
verum end; suppose A23:
n > 2
;
( ( n = 2 & y = x & z = x + 1 & t = x + 2 ) or ( n = 3 & y = x & z = x & t = x + 1 ) )end; end;
end;
assume
( ( n = 2 & y = x & z = x + 1 & t = x + 2 ) or ( n = 3 & y = x & z = x & t = x + 1 ) )
; ((n |^ x) + (n |^ y)) + (n |^ z) = n |^ t
per cases then
( ( n = 2 & y = x & z = x + 1 & t = x + 2 ) or ( n = 3 & y = x & z = x & t = x + 1 ) )
;
end;