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 ∈ ℕ0
3 sgmppw ( ( 1 ∈ ℂ ∧ 𝑃 ∈ ℙ ∧ 1 ∈ ℕ0 ) → ( 1 σ ( 𝑃 ↑ 1 ) ) = Σ 𝑘 ∈ ( 0 ... 1 ) ( ( 𝑃𝑐 1 ) ↑ 𝑘 ) )
4 1 2 3 mp3an13 ( 𝑃 ∈ ℙ → ( 1 σ ( 𝑃 ↑ 1 ) ) = Σ 𝑘 ∈ ( 0 ... 1 ) ( ( 𝑃𝑐 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 ) ) → ( 𝑃𝑐 1 ) = 𝑃 )
11 10 oveq1d ( ( 𝑃 ∈ ℙ ∧ 𝑘 ∈ ( 0 ... 1 ) ) → ( ( 𝑃𝑐 1 ) ↑ 𝑘 ) = ( 𝑃𝑘 ) )
12 11 sumeq2dv ( 𝑃 ∈ ℙ → Σ 𝑘 ∈ ( 0 ... 1 ) ( ( 𝑃𝑐 1 ) ↑ 𝑘 ) = Σ 𝑘 ∈ ( 0 ... 1 ) ( 𝑃𝑘 ) )
13 1m1e0 ( 1 − 1 ) = 0
14 13 oveq2i ( 0 ... ( 1 − 1 ) ) = ( 0 ... 0 )
15 14 sumeq1i Σ 𝑘 ∈ ( 0 ... ( 1 − 1 ) ) ( 𝑃𝑘 ) = Σ 𝑘 ∈ ( 0 ... 0 ) ( 𝑃𝑘 )
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 → ( 𝑃𝑘 ) = ( 𝑃 ↑ 0 ) )
20 19 fsum1 ( ( 0 ∈ ℤ ∧ ( 𝑃 ↑ 0 ) ∈ ℂ ) → Σ 𝑘 ∈ ( 0 ... 0 ) ( 𝑃𝑘 ) = ( 𝑃 ↑ 0 ) )
21 16 18 20 sylancr ( 𝑃 ∈ ℙ → Σ 𝑘 ∈ ( 0 ... 0 ) ( 𝑃𝑘 ) = ( 𝑃 ↑ 0 ) )
22 21 17 eqtrd ( 𝑃 ∈ ℙ → Σ 𝑘 ∈ ( 0 ... 0 ) ( 𝑃𝑘 ) = 1 )
23 15 22 eqtrid ( 𝑃 ∈ ℙ → Σ 𝑘 ∈ ( 0 ... ( 1 − 1 ) ) ( 𝑃𝑘 ) = 1 )
24 23 7 oveq12d ( 𝑃 ∈ ℙ → ( Σ 𝑘 ∈ ( 0 ... ( 1 − 1 ) ) ( 𝑃𝑘 ) + ( 𝑃 ↑ 1 ) ) = ( 1 + 𝑃 ) )
25 2 a1i ( 𝑃 ∈ ℙ → 1 ∈ ℕ0 )
26 nn0uz 0 = ( ℤ ‘ 0 )
27 25 26 eleqtrdi ( 𝑃 ∈ ℙ → 1 ∈ ( ℤ ‘ 0 ) )
28 elfznn0 ( 𝑘 ∈ ( 0 ... 1 ) → 𝑘 ∈ ℕ0 )
29 expcl ( ( 𝑃 ∈ ℂ ∧ 𝑘 ∈ ℕ0 ) → ( 𝑃𝑘 ) ∈ ℂ )
30 6 28 29 syl2an ( ( 𝑃 ∈ ℙ ∧ 𝑘 ∈ ( 0 ... 1 ) ) → ( 𝑃𝑘 ) ∈ ℂ )
31 oveq2 ( 𝑘 = 1 → ( 𝑃𝑘 ) = ( 𝑃 ↑ 1 ) )
32 27 30 31 fsumm1 ( 𝑃 ∈ ℙ → Σ 𝑘 ∈ ( 0 ... 1 ) ( 𝑃𝑘 ) = ( Σ 𝑘 ∈ ( 0 ... ( 1 − 1 ) ) ( 𝑃𝑘 ) + ( 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 → Σ 𝑘 ∈ ( 0 ... 1 ) ( 𝑃𝑘 ) = ( 𝑃 + 1 ) )
36 12 35 eqtrd ( 𝑃 ∈ ℙ → Σ 𝑘 ∈ ( 0 ... 1 ) ( ( 𝑃𝑐 1 ) ↑ 𝑘 ) = ( 𝑃 + 1 ) )
37 4 8 36 3eqtr3d ( 𝑃 ∈ ℙ → ( 1 σ 𝑃 ) = ( 𝑃 + 1 ) )