let m be natural number ; :: thesis: for n0 being non zero natural number st sigma n0 = n0 + m & m divides n0 & n0 <> m holds
( m = 1 & n0 is prime )
let n0 be non zero natural number ; :: thesis: ( sigma n0 = n0 + m & m divides n0 & n0 <> m implies ( m = 1 & n0 is prime ) )
assume A1:
sigma n0 = n0 + m
; :: thesis: ( not m divides n0 or not n0 <> m or ( m = 1 & n0 is prime ) )
assume A2:
m divides n0
; :: thesis: ( not n0 <> m or ( m = 1 & n0 is prime ) )
assume A3:
n0 <> m
; :: thesis: ( m = 1 & n0 is prime )
per cases
( n0 = 1 or n0 <> 1 )
;
suppose A4:
n0 <> 1
;
:: thesis: ( m = 1 & n0 is prime )consider k being
natural number such that A5:
n0 = m * k
by A2, NAT_D:def 3;
A6:
k <> m
k = n0
proof
assume A7:
k <> n0
;
:: thesis: contradiction
A8:
k divides n0
by A5, NAT_D:def 3;
A9:
k <> 1
by A5, A3;
m <> 1
by A5, A7;
then
((1 + m) + k) + n0 <= sigma n0
by A2, A3, A7, A8, A9, A6, Th32;
then
(((1 + m) + k) + n0) - (m + n0) <= (n0 + m) - (m + n0)
by A1, XREAL_1:11;
then
1
+ k <= 0
;
hence
contradiction
;
:: thesis: verum
end; then
n0 / n0 = m
by A5, XCMPLX_1:90;
hence A10:
m = 1
by XCMPLX_1:60;
:: thesis: n0 is prime A11:
n0 > 1
by A4, NAT_1:26;
for
k being
natural number holds
( not
k divides n0 or
k = 1 or
k = n0 )
hence
n0 is
prime
by A11, INT_2:def 5;
:: thesis: verum end; end;