Metamath Proof Explorer


Theorem 1sgmprm

Description: The sum of divisors for a prime is P + 1 because the only divisors are 1 and P . (Contributed by Mario Carneiro, 17-May-2016)

Ref Expression
Assertion 1sgmprm P 1 σ P = P + 1

Proof

Step Hyp Ref Expression
1 ax-1cn 1
2 1nn0 1 0
3 sgmppw 1 P 1 0 1 σ P 1 = k = 0 1 P 1 k
4 1 2 3 mp3an13 P 1 σ P 1 = k = 0 1 P 1 k
5 prmnn P P
6 5 nncnd P P
7 6 exp1d P P 1 = P
8 7 oveq2d P 1 σ P 1 = 1 σ P
9 6 adantr P k 0 1 P
10 9 cxp1d P k 0 1 P 1 = P
11 10 oveq1d P k 0 1 P 1 k = P k
12 11 sumeq2dv P k = 0 1 P 1 k = k = 0 1 P k
13 1m1e0 1 1 = 0
14 13 oveq2i 0 1 1 = 0 0
15 14 sumeq1i k = 0 1 1 P k = k = 0 0 P k
16 0z 0
17 6 exp0d P P 0 = 1
18 17 1 eqeltrdi P P 0
19 oveq2 k = 0 P k = P 0
20 19 fsum1 0 P 0 k = 0 0 P k = P 0
21 16 18 20 sylancr P k = 0 0 P k = P 0
22 21 17 eqtrd P k = 0 0 P k = 1
23 15 22 eqtrid P k = 0 1 1 P k = 1
24 23 7 oveq12d P k = 0 1 1 P k + P 1 = 1 + P
25 2 a1i P 1 0
26 nn0uz 0 = 0
27 25 26 eleqtrdi P 1 0
28 elfznn0 k 0 1 k 0
29 expcl P k 0 P k
30 6 28 29 syl2an P k 0 1 P k
31 oveq2 k = 1 P k = P 1
32 27 30 31 fsumm1 P k = 0 1 P k = k = 0 1 1 P k + P 1
33 addcom P 1 P + 1 = 1 + P
34 6 1 33 sylancl P P + 1 = 1 + P
35 24 32 34 3eqtr4d P k = 0 1 P k = P + 1
36 12 35 eqtrd P k = 0 1 P 1 k = P + 1
37 4 8 36 3eqtr3d P 1 σ P = P + 1