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 ∈ ℙ → 1 σ P = P + 1

Proof

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