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Theorem List for Metamath Proof Explorer - 15301-15400   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theorempcabs 15301 The prime count of an absolute value. (Contributed by Mario Carneiro, 13-Mar-2014.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℚ) → (𝑃 pCnt (abs‘𝐴)) = (𝑃 pCnt 𝐴))
 
Theorempcdvdstr 15302 The prime count increases under the divisibility relation. (Contributed by Mario Carneiro, 13-Mar-2014.)
((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐴𝐵)) → (𝑃 pCnt 𝐴) ≤ (𝑃 pCnt 𝐵))
 
Theorempcgcd1 15303 The prime count of a GCD is the minimum of the prime counts of the arguments. (Contributed by Mario Carneiro, 3-Oct-2014.)
(((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) ∧ (𝑃 pCnt 𝐴) ≤ (𝑃 pCnt 𝐵)) → (𝑃 pCnt (𝐴 gcd 𝐵)) = (𝑃 pCnt 𝐴))
 
Theorempcgcd 15304 The prime count of a GCD is the minimum of the prime counts of the arguments. (Contributed by Mario Carneiro, 3-Oct-2014.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → (𝑃 pCnt (𝐴 gcd 𝐵)) = if((𝑃 pCnt 𝐴) ≤ (𝑃 pCnt 𝐵), (𝑃 pCnt 𝐴), (𝑃 pCnt 𝐵)))
 
Theorempc2dvds 15305* A characterization of divisibility in terms of prime count. (Contributed by Mario Carneiro, 23-Feb-2014.) (Revised by Mario Carneiro, 3-Oct-2014.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → (𝐴𝐵 ↔ ∀𝑝 ∈ ℙ (𝑝 pCnt 𝐴) ≤ (𝑝 pCnt 𝐵)))
 
Theorempc11 15306* The prime count function, viewed as a function from to (ℕ ↑𝑚 ℙ), is one-to-one. (Contributed by Mario Carneiro, 23-Feb-2014.)
((𝐴 ∈ ℕ0𝐵 ∈ ℕ0) → (𝐴 = 𝐵 ↔ ∀𝑝 ∈ ℙ (𝑝 pCnt 𝐴) = (𝑝 pCnt 𝐵)))
 
Theorempcz 15307* The prime count function can be used as an indicator that a given rational number is an integer. (Contributed by Mario Carneiro, 23-Feb-2014.)
(𝐴 ∈ ℚ → (𝐴 ∈ ℤ ↔ ∀𝑝 ∈ ℙ 0 ≤ (𝑝 pCnt 𝐴)))
 
Theorempcprmpw2 15308* Self-referential expression for a prime power. (Contributed by Mario Carneiro, 16-Jan-2015.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℕ) → (∃𝑛 ∈ ℕ0 𝐴 ∥ (𝑃𝑛) ↔ 𝐴 = (𝑃↑(𝑃 pCnt 𝐴))))
 
Theorempcprmpw 15309* Self-referential expression for a prime power. (Contributed by Mario Carneiro, 16-Jan-2015.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℕ) → (∃𝑛 ∈ ℕ0 𝐴 = (𝑃𝑛) ↔ 𝐴 = (𝑃↑(𝑃 pCnt 𝐴))))
 
Theoremdvdsprmpweq 15310* If a positive integer divides a prime power, it is a prime power. (Contributed by AV, 25-Jul-2021.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℕ ∧ 𝑁 ∈ ℕ0) → (𝐴 ∥ (𝑃𝑁) → ∃𝑛 ∈ ℕ0 𝐴 = (𝑃𝑛)))
 
Theoremdvdsprmpweqnn 15311* If an integer greater than 1 divides a prime power, it is a (proper) prime power. (Contributed by AV, 13-Aug-2021.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ (ℤ‘2) ∧ 𝑁 ∈ ℕ0) → (𝐴 ∥ (𝑃𝑁) → ∃𝑛 ∈ ℕ 𝐴 = (𝑃𝑛)))
 
Theoremdvdsprmpweqle 15312* If a positive integer divides a prime power, it is a prime power with a smaller exponent. (Contributed by AV, 25-Jul-2021.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℕ ∧ 𝑁 ∈ ℕ0) → (𝐴 ∥ (𝑃𝑁) → ∃𝑛 ∈ ℕ0 (𝑛𝑁𝐴 = (𝑃𝑛))))
 
Theoremdifsqpwdvds 15313 If the difference of two squares is a power of a prime, the prime divides twice the second squared number. (Contributed by AV, 13-Aug-2021.)
(((𝐴 ∈ ℕ0𝐵 ∈ ℕ0 ∧ (𝐵 + 1) < 𝐴) ∧ (𝐶 ∈ ℙ ∧ 𝐷 ∈ ℕ0)) → ((𝐶𝐷) = ((𝐴↑2) − (𝐵↑2)) → 𝐶 ∥ (2 · 𝐵)))
 
Theorempcaddlem 15314 Lemma for pcadd 15315. The original numbers 𝐴 and 𝐵 have been decomposed using the prime count function as (𝑃𝑀) · (𝑅 / 𝑆) where 𝑅, 𝑆 are both not divisible by 𝑃 and 𝑀 = (𝑃 pCnt 𝐴), and similarly for 𝐵. (Contributed by Mario Carneiro, 9-Sep-2014.)
(𝜑𝑃 ∈ ℙ)    &   (𝜑𝐴 = ((𝑃𝑀) · (𝑅 / 𝑆)))    &   (𝜑𝐵 = ((𝑃𝑁) · (𝑇 / 𝑈)))    &   (𝜑𝑁 ∈ (ℤ𝑀))    &   (𝜑 → (𝑅 ∈ ℤ ∧ ¬ 𝑃𝑅))    &   (𝜑 → (𝑆 ∈ ℕ ∧ ¬ 𝑃𝑆))    &   (𝜑 → (𝑇 ∈ ℤ ∧ ¬ 𝑃𝑇))    &   (𝜑 → (𝑈 ∈ ℕ ∧ ¬ 𝑃𝑈))       (𝜑𝑀 ≤ (𝑃 pCnt (𝐴 + 𝐵)))
 
Theorempcadd 15315 An inequality for the prime count of a sum. This is the source of the ultrametric inequality for the p-adic metric. (Contributed by Mario Carneiro, 9-Sep-2014.)
(𝜑𝑃 ∈ ℙ)    &   (𝜑𝐴 ∈ ℚ)    &   (𝜑𝐵 ∈ ℚ)    &   (𝜑 → (𝑃 pCnt 𝐴) ≤ (𝑃 pCnt 𝐵))       (𝜑 → (𝑃 pCnt 𝐴) ≤ (𝑃 pCnt (𝐴 + 𝐵)))
 
Theorempcadd2 15316 The inequality of pcadd 15315 becomes an equality when one of the factors has prime count strictly less than the other. (Contributed by Mario Carneiro, 16-Jan-2015.) (Revised by Mario Carneiro, 26-Jun-2015.)
(𝜑𝑃 ∈ ℙ)    &   (𝜑𝐴 ∈ ℚ)    &   (𝜑𝐵 ∈ ℚ)    &   (𝜑 → (𝑃 pCnt 𝐴) < (𝑃 pCnt 𝐵))       (𝜑 → (𝑃 pCnt 𝐴) = (𝑃 pCnt (𝐴 + 𝐵)))
 
Theorempcmptcl 15317 Closure for the prime power map. (Contributed by Mario Carneiro, 12-Mar-2014.)
𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (𝑛𝐴), 1))    &   (𝜑 → ∀𝑛 ∈ ℙ 𝐴 ∈ ℕ0)       (𝜑 → (𝐹:ℕ⟶ℕ ∧ seq1( · , 𝐹):ℕ⟶ℕ))
 
Theorempcmpt 15318* Construct a function with given prime count characteristics. (Contributed by Mario Carneiro, 12-Mar-2014.)
𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (𝑛𝐴), 1))    &   (𝜑 → ∀𝑛 ∈ ℙ 𝐴 ∈ ℕ0)    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑃 ∈ ℙ)    &   (𝑛 = 𝑃𝐴 = 𝐵)       (𝜑 → (𝑃 pCnt (seq1( · , 𝐹)‘𝑁)) = if(𝑃𝑁, 𝐵, 0))
 
Theorempcmpt2 15319* Dividing two prime count maps yields a number with all dividing primes confined to an interval. (Contributed by Mario Carneiro, 14-Mar-2014.)
𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (𝑛𝐴), 1))    &   (𝜑 → ∀𝑛 ∈ ℙ 𝐴 ∈ ℕ0)    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑃 ∈ ℙ)    &   (𝑛 = 𝑃𝐴 = 𝐵)    &   (𝜑𝑀 ∈ (ℤ𝑁))       (𝜑 → (𝑃 pCnt ((seq1( · , 𝐹)‘𝑀) / (seq1( · , 𝐹)‘𝑁))) = if((𝑃𝑀 ∧ ¬ 𝑃𝑁), 𝐵, 0))
 
Theorempcmptdvds 15320 The partial products of the prime power map form a divisibility chain. (Contributed by Mario Carneiro, 12-Mar-2014.)
𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (𝑛𝐴), 1))    &   (𝜑 → ∀𝑛 ∈ ℙ 𝐴 ∈ ℕ0)    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑀 ∈ (ℤ𝑁))       (𝜑 → (seq1( · , 𝐹)‘𝑁) ∥ (seq1( · , 𝐹)‘𝑀))
 
Theorempcprod 15321* The product of the primes taken to their respective powers reconstructs the original number. (Contributed by Mario Carneiro, 12-Mar-2014.)
𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (𝑛↑(𝑛 pCnt 𝑁)), 1))       (𝑁 ∈ ℕ → (seq1( · , 𝐹)‘𝑁) = 𝑁)
 
Theoremsumhash 15322* The sum of 1 over a set is the size of the set. (Contributed by Mario Carneiro, 8-Mar-2014.) (Revised by Mario Carneiro, 20-May-2014.)
((𝐵 ∈ Fin ∧ 𝐴𝐵) → Σ𝑘𝐵 if(𝑘𝐴, 1, 0) = (#‘𝐴))
 
Theoremfldivp1 15323 The difference between the floors of adjacent fractions is either 1 or 0. (Contributed by Mario Carneiro, 8-Mar-2014.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℕ) → ((⌊‘((𝑀 + 1) / 𝑁)) − (⌊‘(𝑀 / 𝑁))) = if(𝑁 ∥ (𝑀 + 1), 1, 0))
 
Theorempcfaclem 15324 Lemma for pcfac 15325. (Contributed by Mario Carneiro, 20-May-2014.)
((𝑁 ∈ ℕ0𝑀 ∈ (ℤ𝑁) ∧ 𝑃 ∈ ℙ) → (⌊‘(𝑁 / (𝑃𝑀))) = 0)
 
Theorempcfac 15325* Calculate the prime count of a factorial. (Contributed by Mario Carneiro, 11-Mar-2014.) (Revised by Mario Carneiro, 21-May-2014.)
((𝑁 ∈ ℕ0𝑀 ∈ (ℤ𝑁) ∧ 𝑃 ∈ ℙ) → (𝑃 pCnt (!‘𝑁)) = Σ𝑘 ∈ (1...𝑀)(⌊‘(𝑁 / (𝑃𝑘))))
 
Theorempcbc 15326* Calculate the prime count of a binomial coefficient. (Contributed by Mario Carneiro, 11-Mar-2014.) (Revised by Mario Carneiro, 21-May-2014.)
((𝑁 ∈ ℕ ∧ 𝐾 ∈ (0...𝑁) ∧ 𝑃 ∈ ℙ) → (𝑃 pCnt (𝑁C𝐾)) = Σ𝑘 ∈ (1...𝑁)((⌊‘(𝑁 / (𝑃𝑘))) − ((⌊‘((𝑁𝐾) / (𝑃𝑘))) + (⌊‘(𝐾 / (𝑃𝑘))))))
 
Theoremqexpz 15327 If a power of a rational number is an integer, then the number is an integer. In other words, all n-th roots are irrational unless they are integers (so that the original number is an n-th power). (Contributed by Mario Carneiro, 10-Aug-2015.)
((𝐴 ∈ ℚ ∧ 𝑁 ∈ ℕ ∧ (𝐴𝑁) ∈ ℤ) → 𝐴 ∈ ℤ)
 
Theoremexpnprm 15328 A second or higher power of a rational number is not a prime number. Or by contraposition, the n-th root of a prime number is irrational. Suggested by Norm Megill. (Contributed by Mario Carneiro, 10-Aug-2015.)
((𝐴 ∈ ℚ ∧ 𝑁 ∈ (ℤ‘2)) → ¬ (𝐴𝑁) ∈ ℙ)
 
Theoremoddprmdvds 15329* Every positive integer which is not a power of two is divisible by an odd prime number. (Contributed by AV, 6-Aug-2021.)
((𝐾 ∈ ℕ ∧ ¬ ∃𝑛 ∈ ℕ0 𝐾 = (2↑𝑛)) → ∃𝑝 ∈ (ℙ ∖ {2})𝑝𝐾)
 
6.2.8  Pocklington's theorem
 
Theoremprmpwdvds 15330 A relation involving divisibility by a prime power. (Contributed by Mario Carneiro, 2-Mar-2014.)
(((𝐾 ∈ ℤ ∧ 𝐷 ∈ ℤ) ∧ (𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) ∧ (𝐷 ∥ (𝐾 · (𝑃𝑁)) ∧ ¬ 𝐷 ∥ (𝐾 · (𝑃↑(𝑁 − 1))))) → (𝑃𝑁) ∥ 𝐷)
 
Theorempockthlem 15331 Lemma for pockthg 15332. (Contributed by Mario Carneiro, 2-Mar-2014.)
(𝜑𝐴 ∈ ℕ)    &   (𝜑𝐵 ∈ ℕ)    &   (𝜑𝐵 < 𝐴)    &   (𝜑𝑁 = ((𝐴 · 𝐵) + 1))    &   (𝜑𝑃 ∈ ℙ)    &   (𝜑𝑃𝑁)    &   (𝜑𝑄 ∈ ℙ)    &   (𝜑 → (𝑄 pCnt 𝐴) ∈ ℕ)    &   (𝜑𝐶 ∈ ℤ)    &   (𝜑 → ((𝐶↑(𝑁 − 1)) mod 𝑁) = 1)    &   (𝜑 → (((𝐶↑((𝑁 − 1) / 𝑄)) − 1) gcd 𝑁) = 1)       (𝜑 → (𝑄 pCnt 𝐴) ≤ (𝑄 pCnt (𝑃 − 1)))
 
Theorempockthg 15332* The generalized Pocklington's theorem. If 𝑁 − 1 = 𝐴 · 𝐵 where 𝐵 < 𝐴, then 𝑁 is prime if and only if for every prime factor 𝑝 of 𝐴, there is an 𝑥 such that 𝑥↑(𝑁 − 1) = 1( mod 𝑁) and gcd (𝑥↑((𝑁 − 1) / 𝑝) − 1, 𝑁) = 1. (Contributed by Mario Carneiro, 2-Mar-2014.)
(𝜑𝐴 ∈ ℕ)    &   (𝜑𝐵 ∈ ℕ)    &   (𝜑𝐵 < 𝐴)    &   (𝜑𝑁 = ((𝐴 · 𝐵) + 1))    &   (𝜑 → ∀𝑝 ∈ ℙ (𝑝𝐴 → ∃𝑥 ∈ ℤ (((𝑥↑(𝑁 − 1)) mod 𝑁) = 1 ∧ (((𝑥↑((𝑁 − 1) / 𝑝)) − 1) gcd 𝑁) = 1)))       (𝜑𝑁 ∈ ℙ)
 
Theorempockthi 15333 Pocklington's theorem, which gives a sufficient criterion for a number 𝑁 to be prime. This is the preferred method for verifying large primes, being much more efficient to compute than trial division. This form has been optimized for application to specific large primes; see pockthg 15332 for a more general closed-form version. (Contributed by Mario Carneiro, 2-Mar-2014.)
𝑃 ∈ ℙ    &   𝐺 ∈ ℕ    &   𝑀 = (𝐺 · 𝑃)    &   𝑁 = (𝑀 + 1)    &   𝐷 ∈ ℕ    &   𝐸 ∈ ℕ    &   𝐴 ∈ ℕ    &   𝑀 = (𝐷 · (𝑃𝐸))    &   𝐷 < (𝑃𝐸)    &   ((𝐴𝑀) mod 𝑁) = (1 mod 𝑁)    &   (((𝐴𝐺) − 1) gcd 𝑁) = 1       𝑁 ∈ ℙ
 
6.2.9  Infinite primes theorem
 
Theoremunbenlem 15334* Lemma for unben 15335. (Contributed by NM, 5-May-2005.) (Revised by Mario Carneiro, 15-Sep-2013.)
𝐺 = (rec((𝑥 ∈ V ↦ (𝑥 + 1)), 1) ↾ ω)       ((𝐴 ⊆ ℕ ∧ ∀𝑚 ∈ ℕ ∃𝑛𝐴 𝑚 < 𝑛) → 𝐴 ≈ ω)
 
Theoremunben 15335* An unbounded set of positive integers is infinite. (Contributed by NM, 5-May-2005.) (Revised by Mario Carneiro, 15-Sep-2013.)
((𝐴 ⊆ ℕ ∧ ∀𝑚 ∈ ℕ ∃𝑛𝐴 𝑚 < 𝑛) → 𝐴 ≈ ℕ)
 
Theoreminfpnlem1 15336* Lemma for infpn 15338. The smallest divisor (greater than 1) 𝑀 of 𝑁! + 1 is a prime greater than 𝑁. (Contributed by NM, 5-May-2005.)
𝐾 = ((!‘𝑁) + 1)       ((𝑁 ∈ ℕ ∧ 𝑀 ∈ ℕ) → (((1 < 𝑀 ∧ (𝐾 / 𝑀) ∈ ℕ) ∧ ∀𝑗 ∈ ℕ ((1 < 𝑗 ∧ (𝐾 / 𝑗) ∈ ℕ) → 𝑀𝑗)) → (𝑁 < 𝑀 ∧ ∀𝑗 ∈ ℕ ((𝑀 / 𝑗) ∈ ℕ → (𝑗 = 1 ∨ 𝑗 = 𝑀)))))
 
Theoreminfpnlem2 15337* Lemma for infpn 15338. For any positive integer 𝑁, there exists a prime number 𝑗 greater than 𝑁. (Contributed by NM, 5-May-2005.)
𝐾 = ((!‘𝑁) + 1)       (𝑁 ∈ ℕ → ∃𝑗 ∈ ℕ (𝑁 < 𝑗 ∧ ∀𝑘 ∈ ℕ ((𝑗 / 𝑘) ∈ ℕ → (𝑘 = 1 ∨ 𝑘 = 𝑗))))
 
Theoreminfpn 15338* There exist infinitely many prime numbers: for any positive integer 𝑁, there exists a prime number 𝑗 greater than 𝑁. (See infpn2 15339 for the equinumerosity version.) (Contributed by NM, 1-Jun-2006.)
(𝑁 ∈ ℕ → ∃𝑗 ∈ ℕ (𝑁 < 𝑗 ∧ ∀𝑘 ∈ ℕ ((𝑗 / 𝑘) ∈ ℕ → (𝑘 = 1 ∨ 𝑘 = 𝑗))))
 
Theoreminfpn2 15339* There exist infinitely many prime numbers: the set of all primes 𝑆 is unbounded by infpn 15338, so by unben 15335 it is infinite. This is Metamath 100 proof #11. (Contributed by NM, 5-May-2005.)
𝑆 = {𝑛 ∈ ℕ ∣ (1 < 𝑛 ∧ ∀𝑚 ∈ ℕ ((𝑛 / 𝑚) ∈ ℕ → (𝑚 = 1 ∨ 𝑚 = 𝑛)))}       𝑆 ≈ ℕ
 
Theoremprmunb 15340* The primes are unbounded. (Contributed by Paul Chapman, 28-Nov-2012.)
(𝑁 ∈ ℕ → ∃𝑝 ∈ ℙ 𝑁 < 𝑝)
 
Theoremprminf 15341 There are an infinite number of primes. Theorem 1.7 in [ApostolNT] p. 16. (Contributed by Paul Chapman, 28-Nov-2012.)
ℙ ≈ ℕ
 
6.2.10  Sum of prime reciprocals
 
Theoremprmreclem1 15342* Lemma for prmrec 15348. Properties of the "square part" function, which extracts the 𝑚 of the decomposition 𝑁 = 𝑟𝑚↑2, with 𝑚 maximal and 𝑟 squarefree. (Contributed by Mario Carneiro, 5-Aug-2014.)
𝑄 = (𝑛 ∈ ℕ ↦ sup({𝑟 ∈ ℕ ∣ (𝑟↑2) ∥ 𝑛}, ℝ, < ))       (𝑁 ∈ ℕ → ((𝑄𝑁) ∈ ℕ ∧ ((𝑄𝑁)↑2) ∥ 𝑁 ∧ (𝐾 ∈ (ℤ‘2) → ¬ (𝐾↑2) ∥ (𝑁 / ((𝑄𝑁)↑2)))))
 
Theoremprmreclem2 15343* Lemma for prmrec 15348. There are at most 2↑𝐾 squarefree numbers which divide no primes larger than 𝐾. (We could strengthen this to 2↑#(ℙ ∩ (1...𝐾)) but there's no reason to.) We establish the inequality by showing that the prime counts of the number up to 𝐾 completely determine it because all higher prime counts are zero, and they are all at most 1 because no square divides the number, so there are at most 2↑𝐾 possibilities. (Contributed by Mario Carneiro, 5-Aug-2014.)
𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (1 / 𝑛), 0))    &   (𝜑𝐾 ∈ ℕ)    &   (𝜑𝑁 ∈ ℕ)    &   𝑀 = {𝑛 ∈ (1...𝑁) ∣ ∀𝑝 ∈ (ℙ ∖ (1...𝐾)) ¬ 𝑝𝑛}    &   𝑄 = (𝑛 ∈ ℕ ↦ sup({𝑟 ∈ ℕ ∣ (𝑟↑2) ∥ 𝑛}, ℝ, < ))       (𝜑 → (#‘{𝑥𝑀 ∣ (𝑄𝑥) = 1}) ≤ (2↑𝐾))
 
Theoremprmreclem3 15344* Lemma for prmrec 15348. The main inequality established here is #𝑀 ≤ #{𝑥𝑀 ∣ (𝑄𝑥) = 1} · √𝑁, where {𝑥𝑀 ∣ (𝑄𝑥) = 1} is the set of squarefree numbers in 𝑀. This is demonstrated by the map 𝑦 ↦ ⟨𝑦 / (𝑄𝑦)↑2, (𝑄𝑦)⟩ where 𝑄𝑦 is the largest number whose square divides 𝑦. (Contributed by Mario Carneiro, 5-Aug-2014.)
𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (1 / 𝑛), 0))    &   (𝜑𝐾 ∈ ℕ)    &   (𝜑𝑁 ∈ ℕ)    &   𝑀 = {𝑛 ∈ (1...𝑁) ∣ ∀𝑝 ∈ (ℙ ∖ (1...𝐾)) ¬ 𝑝𝑛}    &   𝑄 = (𝑛 ∈ ℕ ↦ sup({𝑟 ∈ ℕ ∣ (𝑟↑2) ∥ 𝑛}, ℝ, < ))       (𝜑 → (#‘𝑀) ≤ ((2↑𝐾) · (√‘𝑁)))
 
Theoremprmreclem4 15345* Lemma for prmrec 15348. Show by induction that the indexed (nondisjoint) union 𝑊𝑘 is at most the size of the prime reciprocal series. The key counting lemma is hashdvds 15196, to show that the number of numbers in 1...𝑁 that divide 𝑘 is at most 𝑁 / 𝑘. (Contributed by Mario Carneiro, 6-Aug-2014.)
𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (1 / 𝑛), 0))    &   (𝜑𝐾 ∈ ℕ)    &   (𝜑𝑁 ∈ ℕ)    &   𝑀 = {𝑛 ∈ (1...𝑁) ∣ ∀𝑝 ∈ (ℙ ∖ (1...𝐾)) ¬ 𝑝𝑛}    &   (𝜑 → seq1( + , 𝐹) ∈ dom ⇝ )    &   (𝜑 → Σ𝑘 ∈ (ℤ‘(𝐾 + 1))if(𝑘 ∈ ℙ, (1 / 𝑘), 0) < (1 / 2))    &   𝑊 = (𝑝 ∈ ℕ ↦ {𝑛 ∈ (1...𝑁) ∣ (𝑝 ∈ ℙ ∧ 𝑝𝑛)})       (𝜑 → (𝑁 ∈ (ℤ𝐾) → (#‘ 𝑘 ∈ ((𝐾 + 1)...𝑁)(𝑊𝑘)) ≤ (𝑁 · Σ𝑘 ∈ ((𝐾 + 1)...𝑁)if(𝑘 ∈ ℙ, (1 / 𝑘), 0))))
 
Theoremprmreclem5 15346* Lemma for prmrec 15348. Here we show the inequality 𝑁 / 2 < #𝑀 by decomposing the set (1...𝑁) into the disjoint union of the set 𝑀 of those numbers that are not divisible by any "large" primes (above 𝐾) and the indexed union over 𝐾 < 𝑘 of the numbers 𝑊𝑘 that divide the prime 𝑘. By prmreclem4 15345 the second of these has size less than 𝑁 times the prime reciprocal series, which is less than 1 / 2 by assumption, we find that the complementary part 𝑀 must be at least 𝑁 / 2 large. (Contributed by Mario Carneiro, 6-Aug-2014.)
𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (1 / 𝑛), 0))    &   (𝜑𝐾 ∈ ℕ)    &   (𝜑𝑁 ∈ ℕ)    &   𝑀 = {𝑛 ∈ (1...𝑁) ∣ ∀𝑝 ∈ (ℙ ∖ (1...𝐾)) ¬ 𝑝𝑛}    &   (𝜑 → seq1( + , 𝐹) ∈ dom ⇝ )    &   (𝜑 → Σ𝑘 ∈ (ℤ‘(𝐾 + 1))if(𝑘 ∈ ℙ, (1 / 𝑘), 0) < (1 / 2))    &   𝑊 = (𝑝 ∈ ℕ ↦ {𝑛 ∈ (1...𝑁) ∣ (𝑝 ∈ ℙ ∧ 𝑝𝑛)})       (𝜑 → (𝑁 / 2) < ((2↑𝐾) · (√‘𝑁)))
 
Theoremprmreclem6 15347* Lemma for prmrec 15348. If the series 𝐹 was convergent, there would be some 𝑘 such that the sum starting from 𝑘 + 1 sums to less than 1 / 2; this is a sufficient hypothesis for prmreclem5 15346 to produce the contradictory bound 𝑁 / 2 < (2↑𝑘)√𝑁, which is false for 𝑁 = 2↑(2𝑘 + 2). (Contributed by Mario Carneiro, 6-Aug-2014.)
𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (1 / 𝑛), 0))        ¬ seq1( + , 𝐹) ∈ dom ⇝
 
Theoremprmrec 15348* The sum of the reciprocals of the primes diverges. Theorem 1.13 in [ApostolNT] p. 18. This is the "second" proof at http://en.wikipedia.org/wiki/Prime_harmonic_series, attributed to Paul Erdős. This is Metamath 100 proof #81. (Contributed by Mario Carneiro, 6-Aug-2014.)
𝐹 = (𝑛 ∈ ℕ ↦ Σ𝑘 ∈ (ℙ ∩ (1...𝑛))(1 / 𝑘))        ¬ 𝐹 ∈ dom ⇝
 
6.2.11  Fundamental theorem of arithmetic
 
Theorem1arithlem1 15349* Lemma for 1arith 15353. (Contributed by Mario Carneiro, 30-May-2014.)
𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛)))       (𝑁 ∈ ℕ → (𝑀𝑁) = (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑁)))
 
Theorem1arithlem2 15350* Lemma for 1arith 15353. (Contributed by Mario Carneiro, 30-May-2014.)
𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛)))       ((𝑁 ∈ ℕ ∧ 𝑃 ∈ ℙ) → ((𝑀𝑁)‘𝑃) = (𝑃 pCnt 𝑁))
 
Theorem1arithlem3 15351* Lemma for 1arith 15353. (Contributed by Mario Carneiro, 30-May-2014.)
𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛)))       (𝑁 ∈ ℕ → (𝑀𝑁):ℙ⟶ℕ0)
 
Theorem1arithlem4 15352* Lemma for 1arith 15353. (Contributed by Mario Carneiro, 30-May-2014.)
𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛)))    &   𝐺 = (𝑦 ∈ ℕ ↦ if(𝑦 ∈ ℙ, (𝑦↑(𝐹𝑦)), 1))    &   (𝜑𝐹:ℙ⟶ℕ0)    &   (𝜑𝑁 ∈ ℕ)    &   ((𝜑 ∧ (𝑞 ∈ ℙ ∧ 𝑁𝑞)) → (𝐹𝑞) = 0)       (𝜑 → ∃𝑥 ∈ ℕ 𝐹 = (𝑀𝑥))
 
Theorem1arith 15353* Fundamental theorem of arithmetic, where a prime factorization is represented as a sequence of prime exponents, for which only finitely many primes have nonzero exponent. The function 𝑀 maps the set of positive integers one-to-one onto the set of prime factorizations 𝑅. (Contributed by Paul Chapman, 17-Nov-2012.) (Proof shortened by Mario Carneiro, 30-May-2014.)
𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛)))    &   𝑅 = {𝑒 ∈ (ℕ0𝑚 ℙ) ∣ (𝑒 “ ℕ) ∈ Fin}       𝑀:ℕ–1-1-onto𝑅
 
Theorem1arith2 15354* Fundamental theorem of arithmetic, where a prime factorization is represented as a finite monotonic 1-based sequence of primes. Every positive integer has a unique prime factorization. Theorem 1.10 in [ApostolNT] p. 17. This is Metamath 100 proof #80. (Contributed by Paul Chapman, 17-Nov-2012.) (Revised by Mario Carneiro, 30-May-2014.)
𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛)))    &   𝑅 = {𝑒 ∈ (ℕ0𝑚 ℙ) ∣ (𝑒 “ ℕ) ∈ Fin}       𝑧 ∈ ℕ ∃!𝑔𝑅 (𝑀𝑧) = 𝑔
 
6.2.12  Lagrange's four-square theorem
 
Syntaxcgz 15355 Extend class notation with the set of gaussian integers.
class ℤ[i]
 
Definitiondf-gz 15356 Define the set of gaussian integers, which are complex numbers whose real and imaginary parts are integers. (Note that the [i] is actually part of the symbol token and has no independent meaning.) (Contributed by Mario Carneiro, 14-Jul-2014.)
ℤ[i] = {𝑥 ∈ ℂ ∣ ((ℜ‘𝑥) ∈ ℤ ∧ (ℑ‘𝑥) ∈ ℤ)}
 
Theoremelgz 15357 Elementhood in the gaussian integers. (Contributed by Mario Carneiro, 14-Jul-2014.)
(𝐴 ∈ ℤ[i] ↔ (𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ ℤ ∧ (ℑ‘𝐴) ∈ ℤ))
 
Theoremgzcn 15358 A gaussian integer is a complex number. (Contributed by Mario Carneiro, 14-Jul-2014.)
(𝐴 ∈ ℤ[i] → 𝐴 ∈ ℂ)
 
Theoremzgz 15359 An integer is a gaussian integer. (Contributed by Mario Carneiro, 14-Jul-2014.)
(𝐴 ∈ ℤ → 𝐴 ∈ ℤ[i])
 
Theoremigz 15360 i is a gaussian integer. (Contributed by Mario Carneiro, 14-Jul-2014.)
i ∈ ℤ[i]
 
Theoremgznegcl 15361 The gaussian integers are closed under negation. (Contributed by Mario Carneiro, 14-Jul-2014.)
(𝐴 ∈ ℤ[i] → -𝐴 ∈ ℤ[i])
 
Theoremgzcjcl 15362 The gaussian integers are closed under conjugation. (Contributed by Mario Carneiro, 14-Jul-2014.)
(𝐴 ∈ ℤ[i] → (∗‘𝐴) ∈ ℤ[i])
 
Theoremgzaddcl 15363 The gaussian integers are closed under addition. (Contributed by Mario Carneiro, 14-Jul-2014.)
((𝐴 ∈ ℤ[i] ∧ 𝐵 ∈ ℤ[i]) → (𝐴 + 𝐵) ∈ ℤ[i])
 
Theoremgzmulcl 15364 The gaussian integers are closed under multiplication. (Contributed by Mario Carneiro, 14-Jul-2014.)
((𝐴 ∈ ℤ[i] ∧ 𝐵 ∈ ℤ[i]) → (𝐴 · 𝐵) ∈ ℤ[i])
 
Theoremgzreim 15365 Construct a gaussian integer from real and imaginary parts. (Contributed by Mario Carneiro, 16-Jul-2014.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → (𝐴 + (i · 𝐵)) ∈ ℤ[i])
 
Theoremgzsubcl 15366 The gaussian integers are closed under subtraction. (Contributed by Mario Carneiro, 14-Jul-2014.)
((𝐴 ∈ ℤ[i] ∧ 𝐵 ∈ ℤ[i]) → (𝐴𝐵) ∈ ℤ[i])
 
Theoremgzabssqcl 15367 The squared norm of a gaussian integer is an integer. (Contributed by Mario Carneiro, 16-Jul-2014.)
(𝐴 ∈ ℤ[i] → ((abs‘𝐴)↑2) ∈ ℕ0)
 
Theorem4sqlem5 15368 Lemma for 4sq 15390. (Contributed by Mario Carneiro, 15-Jul-2014.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝑀 ∈ ℕ)    &   𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))       (𝜑 → (𝐵 ∈ ℤ ∧ ((𝐴𝐵) / 𝑀) ∈ ℤ))
 
Theorem4sqlem6 15369 Lemma for 4sq 15390. (Contributed by Mario Carneiro, 15-Jul-2014.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝑀 ∈ ℕ)    &   𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))       (𝜑 → (-(𝑀 / 2) ≤ 𝐵𝐵 < (𝑀 / 2)))
 
Theorem4sqlem7 15370 Lemma for 4sq 15390. (Contributed by Mario Carneiro, 15-Jul-2014.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝑀 ∈ ℕ)    &   𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))       (𝜑 → (𝐵↑2) ≤ (((𝑀↑2) / 2) / 2))
 
Theorem4sqlem8 15371 Lemma for 4sq 15390. (Contributed by Mario Carneiro, 15-Jul-2014.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝑀 ∈ ℕ)    &   𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))       (𝜑𝑀 ∥ ((𝐴↑2) − (𝐵↑2)))
 
Theorem4sqlem9 15372 Lemma for 4sq 15390. (Contributed by Mario Carneiro, 15-Jul-2014.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝑀 ∈ ℕ)    &   𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ((𝜑𝜓) → (𝐵↑2) = 0)       ((𝜑𝜓) → (𝑀↑2) ∥ (𝐴↑2))
 
Theorem4sqlem10 15373 Lemma for 4sq 15390. (Contributed by Mario Carneiro, 16-Jul-2014.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝑀 ∈ ℕ)    &   𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ((𝜑𝜓) → ((((𝑀↑2) / 2) / 2) − (𝐵↑2)) = 0)       ((𝜑𝜓) → (𝑀↑2) ∥ ((𝐴↑2) − (((𝑀↑2) / 2) / 2)))
 
Theorem4sqlem1 15374* Lemma for 4sq 15390. The set 𝑆 is the set of all numbers that are expressible as a sum of four squares. Our goal is to show that 𝑆 = ℕ0; here we show one subset direction. (Contributed by Mario Carneiro, 14-Jul-2014.)
𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}       𝑆 ⊆ ℕ0
 
Theorem4sqlem2 15375* Lemma for 4sq 15390. Change bound variables in 𝑆. (Contributed by Mario Carneiro, 14-Jul-2014.)
𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}       (𝐴𝑆 ↔ ∃𝑎 ∈ ℤ ∃𝑏 ∈ ℤ ∃𝑐 ∈ ℤ ∃𝑑 ∈ ℤ 𝐴 = (((𝑎↑2) + (𝑏↑2)) + ((𝑐↑2) + (𝑑↑2))))
 
Theorem4sqlem3 15376* Lemma for 4sq 15390. Sufficient condition to be in 𝑆. (Contributed by Mario Carneiro, 14-Jul-2014.)
𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}       (((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) ∧ (𝐶 ∈ ℤ ∧ 𝐷 ∈ ℤ)) → (((𝐴↑2) + (𝐵↑2)) + ((𝐶↑2) + (𝐷↑2))) ∈ 𝑆)
 
Theorem4sqlem4a 15377* Lemma for 4sqlem4 15378. (Contributed by Mario Carneiro, 14-Jul-2014.)
𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}       ((𝐴 ∈ ℤ[i] ∧ 𝐵 ∈ ℤ[i]) → (((abs‘𝐴)↑2) + ((abs‘𝐵)↑2)) ∈ 𝑆)
 
Theorem4sqlem4 15378* Lemma for 4sq 15390. We can express the four-square property more compactly in terms of gaussian integers, because the norms of gaussian integers are exactly sums of two squares. (Contributed by Mario Carneiro, 14-Jul-2014.)
𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}       (𝐴𝑆 ↔ ∃𝑢 ∈ ℤ[i] ∃𝑣 ∈ ℤ[i] 𝐴 = (((abs‘𝑢)↑2) + ((abs‘𝑣)↑2)))
 
Theoremmul4sqlem 15379* Lemma for mul4sq 15380: algebraic manipulations. The extra assumptions involving 𝑀 are for a part of 4sqlem17 15387 which needs to know not just that the product is a sum of squares, but also that it preserves divisibility by 𝑀. (Contributed by Mario Carneiro, 14-Jul-2014.)
𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   (𝜑𝐴 ∈ ℤ[i])    &   (𝜑𝐵 ∈ ℤ[i])    &   (𝜑𝐶 ∈ ℤ[i])    &   (𝜑𝐷 ∈ ℤ[i])    &   𝑋 = (((abs‘𝐴)↑2) + ((abs‘𝐵)↑2))    &   𝑌 = (((abs‘𝐶)↑2) + ((abs‘𝐷)↑2))    &   (𝜑𝑀 ∈ ℕ)    &   (𝜑 → ((𝐴𝐶) / 𝑀) ∈ ℤ[i])    &   (𝜑 → ((𝐵𝐷) / 𝑀) ∈ ℤ[i])    &   (𝜑 → (𝑋 / 𝑀) ∈ ℕ0)       (𝜑 → ((𝑋 / 𝑀) · (𝑌 / 𝑀)) ∈ 𝑆)
 
Theoremmul4sq 15380* Euler's four-square identity: The product of two sums of four squares is also a sum of four squares. This is usually quoted as an explicit formula involving eight real variables; we save some time by working with complex numbers (gaussian integers) instead, so that we only have to work with four variables, and also hiding the actual formula for the product in the proof of mul4sqlem 15379. (For the curious, the explicit formula that is used is ( ∣ 𝑎 ∣ ↑2 + ∣ 𝑏 ∣ ↑2)( ∣ 𝑐 ∣ ↑2 + ∣ 𝑑 ∣ ↑2) = 𝑎∗ · 𝑐 + 𝑏 · 𝑑∗ ∣ ↑2 + ∣ 𝑎∗ · 𝑑𝑏 · 𝑐∗ ∣ ↑2.) (Contributed by Mario Carneiro, 14-Jul-2014.)
𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}       ((𝐴𝑆𝐵𝑆) → (𝐴 · 𝐵) ∈ 𝑆)
 
Theorem4sqlem11 15381* Lemma for 4sq 15390. Use the pigeonhole principle to show that the sets {𝑚↑2 ∣ 𝑚 ∈ (0...𝑁)} and {-1 − 𝑛↑2 ∣ 𝑛 ∈ (0...𝑁)} have a common element, mod 𝑃. (Contributed by Mario Carneiro, 15-Jul-2014.)
𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑃 = ((2 · 𝑁) + 1))    &   (𝜑𝑃 ∈ ℙ)    &   𝐴 = {𝑢 ∣ ∃𝑚 ∈ (0...𝑁)𝑢 = ((𝑚↑2) mod 𝑃)}    &   𝐹 = (𝑣𝐴 ↦ ((𝑃 − 1) − 𝑣))       (𝜑 → (𝐴 ∩ ran 𝐹) ≠ ∅)
 
Theorem4sqlem12 15382* Lemma for 4sq 15390. For any odd prime 𝑃, there is a 𝑘 < 𝑃 such that 𝑘𝑃 − 1 is a sum of two squares. (Contributed by Mario Carneiro, 15-Jul-2014.)
𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑃 = ((2 · 𝑁) + 1))    &   (𝜑𝑃 ∈ ℙ)    &   𝐴 = {𝑢 ∣ ∃𝑚 ∈ (0...𝑁)𝑢 = ((𝑚↑2) mod 𝑃)}    &   𝐹 = (𝑣𝐴 ↦ ((𝑃 − 1) − 𝑣))       (𝜑 → ∃𝑘 ∈ (1...(𝑃 − 1))∃𝑢 ∈ ℤ[i] (((abs‘𝑢)↑2) + 1) = (𝑘 · 𝑃))
 
Theorem4sqlem13 15383* Lemma for 4sq 15390. (Contributed by Mario Carneiro, 16-Jul-2014.) (Revised by AV, 14-Sep-2020.)
𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑃 = ((2 · 𝑁) + 1))    &   (𝜑𝑃 ∈ ℙ)    &   (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   𝑀 = inf(𝑇, ℝ, < )       (𝜑 → (𝑇 ≠ ∅ ∧ 𝑀 < 𝑃))
 
Theorem4sqlem14 15384* Lemma for 4sq 15390. (Contributed by Mario Carneiro, 16-Jul-2014.) (Revised by AV, 14-Sep-2020.)
𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑃 = ((2 · 𝑁) + 1))    &   (𝜑𝑃 ∈ ℙ)    &   (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   𝑀 = inf(𝑇, ℝ, < )    &   (𝜑𝑀 ∈ (ℤ‘2))    &   (𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   (𝜑𝐶 ∈ ℤ)    &   (𝜑𝐷 ∈ ℤ)    &   𝐸 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   𝐹 = (((𝐵 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   𝐺 = (((𝐶 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   𝐻 = (((𝐷 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   𝑅 = ((((𝐸↑2) + (𝐹↑2)) + ((𝐺↑2) + (𝐻↑2))) / 𝑀)    &   (𝜑 → (𝑀 · 𝑃) = (((𝐴↑2) + (𝐵↑2)) + ((𝐶↑2) + (𝐷↑2))))       (𝜑𝑅 ∈ ℕ0)
 
Theorem4sqlem15 15385* Lemma for 4sq 15390. (Contributed by Mario Carneiro, 16-Jul-2014.) (Revised by AV, 14-Sep-2020.)
𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑃 = ((2 · 𝑁) + 1))    &   (𝜑𝑃 ∈ ℙ)    &   (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   𝑀 = inf(𝑇, ℝ, < )    &   (𝜑𝑀 ∈ (ℤ‘2))    &   (𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   (𝜑𝐶 ∈ ℤ)    &   (𝜑𝐷 ∈ ℤ)    &   𝐸 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   𝐹 = (((𝐵 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   𝐺 = (((𝐶 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   𝐻 = (((𝐷 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   𝑅 = ((((𝐸↑2) + (𝐹↑2)) + ((𝐺↑2) + (𝐻↑2))) / 𝑀)    &   (𝜑 → (𝑀 · 𝑃) = (((𝐴↑2) + (𝐵↑2)) + ((𝐶↑2) + (𝐷↑2))))       ((𝜑𝑅 = 𝑀) → ((((((𝑀↑2) / 2) / 2) − (𝐸↑2)) = 0 ∧ ((((𝑀↑2) / 2) / 2) − (𝐹↑2)) = 0) ∧ (((((𝑀↑2) / 2) / 2) − (𝐺↑2)) = 0 ∧ ((((𝑀↑2) / 2) / 2) − (𝐻↑2)) = 0)))
 
Theorem4sqlem16 15386* Lemma for 4sq 15390. (Contributed by Mario Carneiro, 16-Jul-2014.) (Revised by AV, 14-Sep-2020.)
𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑃 = ((2 · 𝑁) + 1))    &   (𝜑𝑃 ∈ ℙ)    &   (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   𝑀 = inf(𝑇, ℝ, < )    &   (𝜑𝑀 ∈ (ℤ‘2))    &   (𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   (𝜑𝐶 ∈ ℤ)    &   (𝜑𝐷 ∈ ℤ)    &   𝐸 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   𝐹 = (((𝐵 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   𝐺 = (((𝐶 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   𝐻 = (((𝐷 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   𝑅 = ((((𝐸↑2) + (𝐹↑2)) + ((𝐺↑2) + (𝐻↑2))) / 𝑀)    &   (𝜑 → (𝑀 · 𝑃) = (((𝐴↑2) + (𝐵↑2)) + ((𝐶↑2) + (𝐷↑2))))       (𝜑 → (𝑅𝑀 ∧ ((𝑅 = 0 ∨ 𝑅 = 𝑀) → (𝑀↑2) ∥ (𝑀 · 𝑃))))
 
Theorem4sqlem17 15387* Lemma for 4sq 15390. (Contributed by Mario Carneiro, 16-Jul-2014.) (Revised by AV, 14-Sep-2020.)
𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑃 = ((2 · 𝑁) + 1))    &   (𝜑𝑃 ∈ ℙ)    &   (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   𝑀 = inf(𝑇, ℝ, < )    &   (𝜑𝑀 ∈ (ℤ‘2))    &   (𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   (𝜑𝐶 ∈ ℤ)    &   (𝜑𝐷 ∈ ℤ)    &   𝐸 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   𝐹 = (((𝐵 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   𝐺 = (((𝐶 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   𝐻 = (((𝐷 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   𝑅 = ((((𝐸↑2) + (𝐹↑2)) + ((𝐺↑2) + (𝐻↑2))) / 𝑀)    &   (𝜑 → (𝑀 · 𝑃) = (((𝐴↑2) + (𝐵↑2)) + ((𝐶↑2) + (𝐷↑2))))        ¬ 𝜑
 
Theorem4sqlem18 15388* Lemma for 4sq 15390. Inductive step, odd prime case. (Contributed by Mario Carneiro, 16-Jul-2014.) (Revised by AV, 14-Sep-2020.)
𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑃 = ((2 · 𝑁) + 1))    &   (𝜑𝑃 ∈ ℙ)    &   (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   𝑀 = inf(𝑇, ℝ, < )       (𝜑𝑃𝑆)
 
Theorem4sqlem19 15389* Lemma for 4sq 15390. The proof is by strong induction - we show that if all the integers less than 𝑘 are in 𝑆, then 𝑘 is as well. In this part of the proof we do the induction argument and dispense with all the cases except the odd prime case, which is sent to 4sqlem18 15388. If 𝑘 is 0, 1, 2, we show 𝑘𝑆 directly; otherwise if 𝑘 is composite, 𝑘 is the product of two numbers less than it (and hence in 𝑆 by assumption), so by mul4sq 15380 𝑘𝑆. (Contributed by Mario Carneiro, 14-Jul-2014.) (Revised by Mario Carneiro, 20-Jun-2015.)
𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}       0 = 𝑆
 
Theorem4sq 15390* Lagrange's four-square theorem, or Bachet's conjecture: every nonnegative integer is expressible as a sum of four squares. This is Metamath 100 proof #19. (Contributed by Mario Carneiro, 16-Jul-2014.)
(𝐴 ∈ ℕ0 ↔ ∃𝑎 ∈ ℤ ∃𝑏 ∈ ℤ ∃𝑐 ∈ ℤ ∃𝑑 ∈ ℤ 𝐴 = (((𝑎↑2) + (𝑏↑2)) + ((𝑐↑2) + (𝑑↑2))))
 
6.2.13  Van der Waerden's theorem
 
Syntaxcvdwa 15391 The arithmetic progression function.
class AP
 
Syntaxcvdwm 15392 The monochromatic arithmetic progression predicate.
class MonoAP
 
Syntaxcvdwp 15393 The polychromatic arithmetic progression predicate.
class PolyAP
 
Definitiondf-vdwap 15394* Define the arithmetic progression function, which takes as input a length 𝑘, a start point 𝑎, and a step 𝑑 and outputs the set of points in this progression. (Contributed by Mario Carneiro, 18-Aug-2014.)
AP = (𝑘 ∈ ℕ0 ↦ (𝑎 ∈ ℕ, 𝑑 ∈ ℕ ↦ ran (𝑚 ∈ (0...(𝑘 − 1)) ↦ (𝑎 + (𝑚 · 𝑑)))))
 
Definitiondf-vdwmc 15395* Define the "contains a monochromatic AP" predicate. (Contributed by Mario Carneiro, 18-Aug-2014.)
MonoAP = {⟨𝑘, 𝑓⟩ ∣ ∃𝑐(ran (AP‘𝑘) ∩ 𝒫 (𝑓 “ {𝑐})) ≠ ∅}
 
Definitiondf-vdwpc 15396* Define the "contains a polychromatic collection of APs" predicate. See vdwpc 15406 for more information. (Contributed by Mario Carneiro, 18-Aug-2014.)
PolyAP = {⟨⟨𝑚, 𝑘⟩, 𝑓⟩ ∣ ∃𝑎 ∈ ℕ ∃𝑑 ∈ (ℕ ↑𝑚 (1...𝑚))(∀𝑖 ∈ (1...𝑚)((𝑎 + (𝑑𝑖))(AP‘𝑘)(𝑑𝑖)) ⊆ (𝑓 “ {(𝑓‘(𝑎 + (𝑑𝑖)))}) ∧ (#‘ran (𝑖 ∈ (1...𝑚) ↦ (𝑓‘(𝑎 + (𝑑𝑖))))) = 𝑚)}
 
Theoremvdwapfval 15397* Define the arithmetic progression function, which takes as input a length 𝑘, a start point 𝑎, and a step 𝑑 and outputs the set of points in this progression. (Contributed by Mario Carneiro, 18-Aug-2014.)
(𝐾 ∈ ℕ0 → (AP‘𝐾) = (𝑎 ∈ ℕ, 𝑑 ∈ ℕ ↦ ran (𝑚 ∈ (0...(𝐾 − 1)) ↦ (𝑎 + (𝑚 · 𝑑)))))
 
Theoremvdwapf 15398 The arithmetic progression function is a function. (Contributed by Mario Carneiro, 18-Aug-2014.)
(𝐾 ∈ ℕ0 → (AP‘𝐾):(ℕ × ℕ)⟶𝒫 ℕ)
 
Theoremvdwapval 15399* Value of the arithmetic progression function. (Contributed by Mario Carneiro, 18-Aug-2014.)
((𝐾 ∈ ℕ0𝐴 ∈ ℕ ∧ 𝐷 ∈ ℕ) → (𝑋 ∈ (𝐴(AP‘𝐾)𝐷) ↔ ∃𝑚 ∈ (0...(𝐾 − 1))𝑋 = (𝐴 + (𝑚 · 𝐷))))
 
Theoremvdwapun 15400 Remove the first element of an arithmetic progression. (Contributed by Mario Carneiro, 11-Sep-2014.)
((𝐾 ∈ ℕ0𝐴 ∈ ℕ ∧ 𝐷 ∈ ℕ) → (𝐴(AP‘(𝐾 + 1))𝐷) = ({𝐴} ∪ ((𝐴 + 𝐷)(AP‘𝐾)𝐷)))
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