P. 170 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 16901-17000   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	pcprmpw2 16901*	Self-referential expression for a prime power. (Contributed by Mario Carneiro, 16-Jan-2015.)
⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℕ) → (∃𝑛 ∈ ℕ0 𝐴 ∥ (𝑃↑𝑛) ↔ 𝐴 = (𝑃↑(𝑃 pCnt 𝐴))))
Theorem	pcprmpw 16902*	Self-referential expression for a prime power. (Contributed by Mario Carneiro, 16-Jan-2015.)
⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℕ) → (∃𝑛 ∈ ℕ0 𝐴 = (𝑃↑𝑛) ↔ 𝐴 = (𝑃↑(𝑃 pCnt 𝐴))))
Theorem	dvdsprmpweq 16903*	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 ∥ (𝑃↑𝑁) → ∃𝑛 ∈ ℕ0 𝐴 = (𝑃↑𝑛)))
Theorem	dvdsprmpweqnn 16904*	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 ∥ (𝑃↑𝑁) → ∃𝑛 ∈ ℕ 𝐴 = (𝑃↑𝑛)))
Theorem	dvdsprmpweqle 16905*	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 ∥ (𝑃↑𝑁) → ∃𝑛 ∈ ℕ0 (𝑛 ≤ 𝑁 ∧ 𝐴 = (𝑃↑𝑛))))
Theorem	difsqpwdvds 16906	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 ∈ ℕ0 ∧ 𝐵 ∈ ℕ0 ∧ (𝐵 + 1) < 𝐴) ∧ (𝐶 ∈ ℙ ∧ 𝐷 ∈ ℕ0)) → ((𝐶↑𝐷) = ((𝐴↑2) − (𝐵↑2)) → 𝐶 ∥ (2 · 𝐵)))
Theorem	pcaddlem 16907	Lemma for pcadd 16908. The original numbers 𝐴 and 𝐵 have been decomposed using the prime count function as (𝑃↑𝑀) · (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.)
⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → 𝐴 = ((𝑃↑𝑀) · (𝑅 / 𝑆)))    &   ⊢ (𝜑 → 𝐵 = ((𝑃↑𝑁) · (𝑇 / 𝑈)))    &   ⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘𝑀))    &   ⊢ (𝜑 → (𝑅 ∈ ℤ ∧ ¬ 𝑃 ∥ 𝑅))    &   ⊢ (𝜑 → (𝑆 ∈ ℕ ∧ ¬ 𝑃 ∥ 𝑆))    &   ⊢ (𝜑 → (𝑇 ∈ ℤ ∧ ¬ 𝑃 ∥ 𝑇))    &   ⊢ (𝜑 → (𝑈 ∈ ℕ ∧ ¬ 𝑃 ∥ 𝑈))    ⇒   ⊢ (𝜑 → 𝑀 ≤ (𝑃 pCnt (𝐴 + 𝐵)))
Theorem	pcadd 16908	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.)
⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → 𝐴 ∈ ℚ)    &   ⊢ (𝜑 → 𝐵 ∈ ℚ)    &   ⊢ (𝜑 → (𝑃 pCnt 𝐴) ≤ (𝑃 pCnt 𝐵))    ⇒   ⊢ (𝜑 → (𝑃 pCnt 𝐴) ≤ (𝑃 pCnt (𝐴 + 𝐵)))
Theorem	pcadd2 16909	The inequality of pcadd 16908 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.)
⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → 𝐴 ∈ ℚ)    &   ⊢ (𝜑 → 𝐵 ∈ ℚ)    &   ⊢ (𝜑 → (𝑃 pCnt 𝐴) < (𝑃 pCnt 𝐵))    ⇒   ⊢ (𝜑 → (𝑃 pCnt 𝐴) = (𝑃 pCnt (𝐴 + 𝐵)))
Theorem	pcmptcl 16910	Closure for the prime power map. (Contributed by Mario Carneiro, 12-Mar-2014.)
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (𝑛↑𝐴), 1))    &   ⊢ (𝜑 → ∀𝑛 ∈ ℙ 𝐴 ∈ ℕ0)    ⇒   ⊢ (𝜑 → (𝐹:ℕ⟶ℕ ∧ seq1( · , 𝐹):ℕ⟶ℕ))
Theorem	pcmpt 16911*	Construct a function with given prime count characteristics. (Contributed by Mario Carneiro, 12-Mar-2014.)
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (𝑛↑𝐴), 1))    &   ⊢ (𝜑 → ∀𝑛 ∈ ℙ 𝐴 ∈ ℕ0)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝑛 = 𝑃 → 𝐴 = 𝐵)    ⇒   ⊢ (𝜑 → (𝑃 pCnt (seq1( · , 𝐹)‘𝑁)) = if(𝑃 ≤ 𝑁, 𝐵, 0))
Theorem	pcmpt2 16912*	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, (𝑛↑𝐴), 1))    &   ⊢ (𝜑 → ∀𝑛 ∈ ℙ 𝐴 ∈ ℕ0)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝑛 = 𝑃 → 𝐴 = 𝐵)    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘𝑁))    ⇒   ⊢ (𝜑 → (𝑃 pCnt ((seq1( · , 𝐹)‘𝑀) / (seq1( · , 𝐹)‘𝑁))) = if((𝑃 ≤ 𝑀 ∧ ¬ 𝑃 ≤ 𝑁), 𝐵, 0))
Theorem	pcmptdvds 16913	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, (𝑛↑𝐴), 1))    &   ⊢ (𝜑 → ∀𝑛 ∈ ℙ 𝐴 ∈ ℕ0)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘𝑁))    ⇒   ⊢ (𝜑 → (seq1( · , 𝐹)‘𝑁) ∥ (seq1( · , 𝐹)‘𝑀))
Theorem	pcprod 16914*	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)
Theorem	sumhash 16915*	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 ∧ 𝐴 ⊆ 𝐵) → Σ𝑘 ∈ 𝐵 if(𝑘 ∈ 𝐴, 1, 0) = (♯‘𝐴))
Theorem	fldivp1 16916	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))
Theorem	pcfaclem 16917	Lemma for pcfac 16918. (Contributed by Mario Carneiro, 20-May-2014.)
⊢ ((𝑁 ∈ ℕ0 ∧ 𝑀 ∈ (ℤ≥‘𝑁) ∧ 𝑃 ∈ ℙ) → (⌊‘(𝑁 / (𝑃↑𝑀))) = 0)
Theorem	pcfac 16918*	Calculate the prime count of a factorial. (Contributed by Mario Carneiro, 11-Mar-2014.) (Revised by Mario Carneiro, 21-May-2014.)
⊢ ((𝑁 ∈ ℕ0 ∧ 𝑀 ∈ (ℤ≥‘𝑁) ∧ 𝑃 ∈ ℙ) → (𝑃 pCnt (!‘𝑁)) = Σ𝑘 ∈ (1...𝑀)(⌊‘(𝑁 / (𝑃↑𝑘))))
Theorem	pcbc 16919*	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 / (𝑃↑𝑘))) − ((⌊‘((𝑁 − 𝐾) / (𝑃↑𝑘))) + (⌊‘(𝐾 / (𝑃↑𝑘))))))
Theorem	qexpz 16920	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 ∧ (𝐴↑𝑁) ∈ ℤ) → 𝐴 ∈ ℤ)
Theorem	expnprm 16921	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)) → ¬ (𝐴↑𝑁) ∈ ℙ)
Theorem	oddprmdvds 16922*	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})𝑝 ∥ 𝐾)
6.2.8  Pocklington's theorem
Theorem	prmpwdvds 16923	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 · (𝑃↑𝑁)) ∧ ¬ 𝐷 ∥ (𝐾 · (𝑃↑(𝑁 − 1))))) → (𝑃↑𝑁) ∥ 𝐷)
Theorem	pockthlem 16924	Lemma for pockthg 16925. (Contributed by Mario Carneiro, 2-Mar-2014.)
⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 < 𝐴)    &   ⊢ (𝜑 → 𝑁 = ((𝐴 · 𝐵) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → 𝑃 ∥ 𝑁)    &   ⊢ (𝜑 → 𝑄 ∈ ℙ)    &   ⊢ (𝜑 → (𝑄 pCnt 𝐴) ∈ ℕ)    &   ⊢ (𝜑 → 𝐶 ∈ 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)))
Theorem	pockthg 16925*	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.)
⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 < 𝐴)    &   ⊢ (𝜑 → 𝑁 = ((𝐴 · 𝐵) + 1))    &   ⊢ (𝜑 → ∀𝑝 ∈ ℙ (𝑝 ∥ 𝐴 → ∃𝑥 ∈ ℤ (((𝑥↑(𝑁 − 1)) mod 𝑁) = 1 ∧ (((𝑥↑((𝑁 − 1) / 𝑝)) − 1) gcd 𝑁) = 1)))    ⇒   ⊢ (𝜑 → 𝑁 ∈ ℙ)
Theorem	pockthi 16926	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 16925 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 · (𝑃↑𝐸))    &   ⊢ 𝐷 < (𝑃↑𝐸)    &   ⊢ ((𝐴↑𝑀) mod 𝑁) = (1 mod 𝑁)    &   ⊢ (((𝐴↑𝐺) − 1) gcd 𝑁) = 1    ⇒   ⊢ 𝑁 ∈ ℙ
6.2.9  Infinite primes theorem
Theorem	unbenlem 16927*	Lemma for unben 16928. (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 ∃𝑛 ∈ 𝐴 𝑚 < 𝑛) → 𝐴 ≈ ω)
Theorem	unben 16928*	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 ∃𝑛 ∈ 𝐴 𝑚 < 𝑛) → 𝐴 ≈ ℕ)
Theorem	infpnlem1 16929*	Lemma for infpn 16931. 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) → 𝑀 ≤ 𝑗)) → (𝑁 < 𝑀 ∧ ∀𝑗 ∈ ℕ ((𝑀 / 𝑗) ∈ ℕ → (𝑗 = 1 ∨ 𝑗 = 𝑀)))))
Theorem	infpnlem2 16930*	Lemma for infpn 16931. 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))))
Theorem	infpn 16931*	There exist infinitely many prime numbers: for any positive integer 𝑁, there exists a prime number 𝑗 greater than 𝑁. (See infpn2 16932 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))))
Theorem	infpn2 16932*	There exist infinitely many prime numbers: the set of all primes 𝑆 is unbounded by infpn 16931, so by unben 16928 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
Theorem	prmunb 16933*	The primes are unbounded. (Contributed by Paul Chapman, 28-Nov-2012.)
⊢ (𝑁 ∈ ℕ → ∃𝑝 ∈ ℙ 𝑁 < 𝑝)
Theorem	prminf 16934	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
Theorem	prmreclem1 16935*	Lemma for prmrec 16941. 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 → ((𝑄‘𝑁) ∈ ℕ ∧ ((𝑄‘𝑁)↑2) ∥ 𝑁 ∧ (𝐾 ∈ (ℤ≥‘2) → ¬ (𝐾↑2) ∥ (𝑁 / ((𝑄‘𝑁)↑2)))))
Theorem	prmreclem2 16936*	Lemma for prmrec 16941. 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))    &   ⊢ (𝜑 → 𝐾 ∈ ℕ)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ 𝑀 = {𝑛 ∈ (1...𝑁) ∣ ∀𝑝 ∈ (ℙ ∖ (1...𝐾)) ¬ 𝑝 ∥ 𝑛}    &   ⊢ 𝑄 = (𝑛 ∈ ℕ ↦ sup({𝑟 ∈ ℕ ∣ (𝑟↑2) ∥ 𝑛}, ℝ, < ))    ⇒   ⊢ (𝜑 → (♯‘{𝑥 ∈ 𝑀 ∣ (𝑄‘𝑥) = 1}) ≤ (2↑𝐾))
Theorem	prmreclem3 16937*	Lemma for prmrec 16941. The main inequality established here is ♯𝑀 ≤ ♯{𝑥 ∈ 𝑀 ∣ (𝑄‘𝑥) = 1} · √𝑁, where {𝑥 ∈ 𝑀 ∣ (𝑄‘𝑥) = 1} is the set of squarefree numbers in 𝑀. This is demonstrated by the map 𝑦 ↦ ⟨𝑦 / (𝑄‘𝑦)↑2, (𝑄‘𝑦)⟩ where 𝑄‘𝑦 is the largest number whose square divides 𝑦. (Contributed by Mario Carneiro, 5-Aug-2014.)
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (1 / 𝑛), 0))    &   ⊢ (𝜑 → 𝐾 ∈ ℕ)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ 𝑀 = {𝑛 ∈ (1...𝑁) ∣ ∀𝑝 ∈ (ℙ ∖ (1...𝐾)) ¬ 𝑝 ∥ 𝑛}    &   ⊢ 𝑄 = (𝑛 ∈ ℕ ↦ sup({𝑟 ∈ ℕ ∣ (𝑟↑2) ∥ 𝑛}, ℝ, < ))    ⇒   ⊢ (𝜑 → (♯‘𝑀) ≤ ((2↑𝐾) · (√‘𝑁)))
Theorem	prmreclem4 16938*	Lemma for prmrec 16941. Show by induction that the indexed (nondisjoint) union 𝑊‘𝑘 is at most the size of the prime reciprocal series. The key counting lemma is hashdvds 16793, 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))    &   ⊢ (𝜑 → 𝐾 ∈ ℕ)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ 𝑀 = {𝑛 ∈ (1...𝑁) ∣ ∀𝑝 ∈ (ℙ ∖ (1...𝐾)) ¬ 𝑝 ∥ 𝑛}    &   ⊢ (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 ∧ 𝑝 ∥ 𝑛)})    ⇒   ⊢ (𝜑 → (𝑁 ∈ (ℤ≥‘𝐾) → (♯‘∪ 𝑘 ∈ ((𝐾 + 1)...𝑁)(𝑊‘𝑘)) ≤ (𝑁 · Σ𝑘 ∈ ((𝐾 + 1)...𝑁)if(𝑘 ∈ ℙ, (1 / 𝑘), 0))))
Theorem	prmreclem5 16939*	Lemma for prmrec 16941. 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 𝑊‘𝑘 that divide the prime 𝑘. By prmreclem4 16938 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))    &   ⊢ (𝜑 → 𝐾 ∈ ℕ)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ 𝑀 = {𝑛 ∈ (1...𝑁) ∣ ∀𝑝 ∈ (ℙ ∖ (1...𝐾)) ¬ 𝑝 ∥ 𝑛}    &   ⊢ (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 ∧ 𝑝 ∥ 𝑛)})    ⇒   ⊢ (𝜑 → (𝑁 / 2) < ((2↑𝐾) · (√‘𝑁)))
Theorem	prmreclem6 16940*	Lemma for prmrec 16941. 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 16939 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 ⇝
Theorem	prmrec 16941*	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
Theorem	1arithlem1 16942*	Lemma for 1arith 16946. (Contributed by Mario Carneiro, 30-May-2014.)
⊢ 𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛)))    ⇒   ⊢ (𝑁 ∈ ℕ → (𝑀‘𝑁) = (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑁)))
Theorem	1arithlem2 16943*	Lemma for 1arith 16946. (Contributed by Mario Carneiro, 30-May-2014.)
⊢ 𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛)))    ⇒   ⊢ ((𝑁 ∈ ℕ ∧ 𝑃 ∈ ℙ) → ((𝑀‘𝑁)‘𝑃) = (𝑃 pCnt 𝑁))
Theorem	1arithlem3 16944*	Lemma for 1arith 16946. (Contributed by Mario Carneiro, 30-May-2014.)
⊢ 𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛)))    ⇒   ⊢ (𝑁 ∈ ℕ → (𝑀‘𝑁):ℙ⟶ℕ0)
Theorem	1arithlem4 16945*	Lemma for 1arith 16946. (Contributed by Mario Carneiro, 30-May-2014.)
⊢ 𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛)))    &   ⊢ 𝐺 = (𝑦 ∈ ℕ ↦ if(𝑦 ∈ ℙ, (𝑦↑(𝐹‘𝑦)), 1))    &   ⊢ (𝜑 → 𝐹:ℙ⟶ℕ0)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ ((𝜑 ∧ (𝑞 ∈ ℙ ∧ 𝑁 ≤ 𝑞)) → (𝐹‘𝑞) = 0)    ⇒   ⊢ (𝜑 → ∃𝑥 ∈ ℕ 𝐹 = (𝑀‘𝑥))
Theorem	1arith 16946*	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 ∈ (ℕ0 ↑m ℙ) ∣ (◡𝑒 “ ℕ) ∈ Fin}    ⇒   ⊢ 𝑀:ℕ–1-1-onto→𝑅
Theorem	1arith2 16947*	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 ∈ (ℕ0 ↑m ℙ) ∣ (◡𝑒 “ ℕ) ∈ Fin}    ⇒   ⊢ ∀𝑧 ∈ ℕ ∃!𝑔 ∈ 𝑅 (𝑀‘𝑧) = 𝑔
6.2.12  Lagrange's four-square theorem
Syntax	cgz 16948	Extend class notation with the set of gaussian integers.
class ℤ[i]
Definition	df-gz 16949	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)}
Theorem	elgz 16950	Elementhood in the gaussian integers. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ (𝐴 ∈ ℤ[i] ↔ (𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ ℤ ∧ (ℑ‘𝐴) ∈ ℤ))
Theorem	gzcn 16951	A gaussian integer is a complex number. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ (𝐴 ∈ ℤ[i] → 𝐴 ∈ ℂ)
Theorem	zgz 16952	An integer is a gaussian integer. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ (𝐴 ∈ ℤ → 𝐴 ∈ ℤ[i])
Theorem	igz 16953	i is a gaussian integer. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ i ∈ ℤ[i]
Theorem	gznegcl 16954	The gaussian integers are closed under negation. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ (𝐴 ∈ ℤ[i] → -𝐴 ∈ ℤ[i])
Theorem	gzcjcl 16955	The gaussian integers are closed under conjugation. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ (𝐴 ∈ ℤ[i] → (∗‘𝐴) ∈ ℤ[i])
Theorem	gzaddcl 16956	The gaussian integers are closed under addition. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ ((𝐴 ∈ ℤ[i] ∧ 𝐵 ∈ ℤ[i]) → (𝐴 + 𝐵) ∈ ℤ[i])
Theorem	gzmulcl 16957	The gaussian integers are closed under multiplication. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ ((𝐴 ∈ ℤ[i] ∧ 𝐵 ∈ ℤ[i]) → (𝐴 · 𝐵) ∈ ℤ[i])
Theorem	gzreim 16958	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])
Theorem	gzsubcl 16959	The gaussian integers are closed under subtraction. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ ((𝐴 ∈ ℤ[i] ∧ 𝐵 ∈ ℤ[i]) → (𝐴 − 𝐵) ∈ ℤ[i])
Theorem	gzabssqcl 16960	The squared norm of a gaussian integer is an integer. (Contributed by Mario Carneiro, 16-Jul-2014.)
⊢ (𝐴 ∈ ℤ[i] → ((abs‘𝐴)↑2) ∈ ℕ0)
Theorem	4sqlem5 16961	Lemma for 4sq 16983. (Contributed by Mario Carneiro, 15-Jul-2014.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ 𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    ⇒   ⊢ (𝜑 → (𝐵 ∈ ℤ ∧ ((𝐴 − 𝐵) / 𝑀) ∈ ℤ))
Theorem	4sqlem6 16962	Lemma for 4sq 16983. (Contributed by Mario Carneiro, 15-Jul-2014.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ 𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    ⇒   ⊢ (𝜑 → (-(𝑀 / 2) ≤ 𝐵 ∧ 𝐵 < (𝑀 / 2)))
Theorem	4sqlem7 16963	Lemma for 4sq 16983. (Contributed by Mario Carneiro, 15-Jul-2014.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ 𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    ⇒   ⊢ (𝜑 → (𝐵↑2) ≤ (((𝑀↑2) / 2) / 2))
Theorem	4sqlem8 16964	Lemma for 4sq 16983. (Contributed by Mario Carneiro, 15-Jul-2014.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ 𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    ⇒   ⊢ (𝜑 → 𝑀 ∥ ((𝐴↑2) − (𝐵↑2)))
Theorem	4sqlem9 16965	Lemma for 4sq 16983. (Contributed by Mario Carneiro, 15-Jul-2014.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ 𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ ((𝜑 ∧ 𝜓) → (𝐵↑2) = 0)    ⇒   ⊢ ((𝜑 ∧ 𝜓) → (𝑀↑2) ∥ (𝐴↑2))
Theorem	4sqlem10 16966	Lemma for 4sq 16983. (Contributed by Mario Carneiro, 16-Jul-2014.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ 𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ ((𝜑 ∧ 𝜓) → ((((𝑀↑2) / 2) / 2) − (𝐵↑2)) = 0)    ⇒   ⊢ ((𝜑 ∧ 𝜓) → (𝑀↑2) ∥ ((𝐴↑2) − (((𝑀↑2) / 2) / 2)))
Theorem	4sqlem1 16967*	Lemma for 4sq 16983. 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
Theorem	4sqlem2 16968*	Lemma for 4sq 16983. 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)))}    ⇒   ⊢ (𝐴 ∈ 𝑆 ↔ ∃𝑎 ∈ ℤ ∃𝑏 ∈ ℤ ∃𝑐 ∈ ℤ ∃𝑑 ∈ ℤ 𝐴 = (((𝑎↑2) + (𝑏↑2)) + ((𝑐↑2) + (𝑑↑2))))
Theorem	4sqlem3 16969*	Lemma for 4sq 16983. 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)
Theorem	4sqlem4a 16970*	Lemma for 4sqlem4 16971. (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)
Theorem	4sqlem4 16971*	Lemma for 4sq 16983. 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)))}    ⇒   ⊢ (𝐴 ∈ 𝑆 ↔ ∃𝑢 ∈ ℤ[i] ∃𝑣 ∈ ℤ[i] 𝐴 = (((abs‘𝑢)↑2) + ((abs‘𝑣)↑2)))
Theorem	mul4sqlem 16972*	Lemma for mul4sq 16973: algebraic manipulations. The extra assumptions involving 𝑀 are for a part of 4sqlem17 16980 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)))}    &   ⊢ (𝜑 → 𝐴 ∈ ℤ[i])    &   ⊢ (𝜑 → 𝐵 ∈ ℤ[i])    &   ⊢ (𝜑 → 𝐶 ∈ ℤ[i])    &   ⊢ (𝜑 → 𝐷 ∈ ℤ[i])    &   ⊢ 𝑋 = (((abs‘𝐴)↑2) + ((abs‘𝐵)↑2))    &   ⊢ 𝑌 = (((abs‘𝐶)↑2) + ((abs‘𝐷)↑2))    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ (𝜑 → ((𝐴 − 𝐶) / 𝑀) ∈ ℤ[i])    &   ⊢ (𝜑 → ((𝐵 − 𝐷) / 𝑀) ∈ ℤ[i])    &   ⊢ (𝜑 → (𝑋 / 𝑀) ∈ ℕ0)    ⇒   ⊢ (𝜑 → ((𝑋 / 𝑀) · (𝑌 / 𝑀)) ∈ 𝑆)
Theorem	mul4sq 16973*	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 16972. (For the curious, the explicit formula that is used is ( ∣ 𝑎 ∣ ↑2 + ∣ 𝑏 ∣ ↑2)( ∣ 𝑐 ∣ ↑2 + ∣ 𝑑 ∣ ↑2) = ∣ 𝑎∗ · 𝑐 + 𝑏 · 𝑑∗ ∣ ↑2 + ∣ 𝑎∗ · 𝑑 − 𝑏 · 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)))}    ⇒   ⊢ ((𝐴 ∈ 𝑆 ∧ 𝐵 ∈ 𝑆) → (𝐴 · 𝐵) ∈ 𝑆)
Theorem	4sqlem11 16974*	Lemma for 4sq 16983. 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)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ 𝐴 = {𝑢 ∣ ∃𝑚 ∈ (0...𝑁)𝑢 = ((𝑚↑2) mod 𝑃)}    &   ⊢ 𝐹 = (𝑣 ∈ 𝐴 ↦ ((𝑃 − 1) − 𝑣))    ⇒   ⊢ (𝜑 → (𝐴 ∩ ran 𝐹) ≠ ∅)
Theorem	4sqlem12 16975*	Lemma for 4sq 16983. 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)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ 𝐴 = {𝑢 ∣ ∃𝑚 ∈ (0...𝑁)𝑢 = ((𝑚↑2) mod 𝑃)}    &   ⊢ 𝐹 = (𝑣 ∈ 𝐴 ↦ ((𝑃 − 1) − 𝑣))    ⇒   ⊢ (𝜑 → ∃𝑘 ∈ (1...(𝑃 − 1))∃𝑢 ∈ ℤ[i] (((abs‘𝑢)↑2) + 1) = (𝑘 · 𝑃))
Theorem	4sqlem13 16976*	Lemma for 4sq 16983. (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)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   ⊢ 𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   ⊢ 𝑀 = inf(𝑇, ℝ, < )    ⇒   ⊢ (𝜑 → (𝑇 ≠ ∅ ∧ 𝑀 < 𝑃))
Theorem	4sqlem14 16977*	Lemma for 4sq 16983. (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)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   ⊢ 𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   ⊢ 𝑀 = inf(𝑇, ℝ, < )    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘2))    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐶 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ 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 → 𝑅 ∈ ℕ0)
Theorem	4sqlem15 16978*	Lemma for 4sq 16983. (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)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   ⊢ 𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   ⊢ 𝑀 = inf(𝑇, ℝ, < )    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘2))    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐶 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ 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 ∧ 𝑅 = 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)))
Theorem	4sqlem16 16979*	Lemma for 4sq 16983. (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)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   ⊢ 𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   ⊢ 𝑀 = inf(𝑇, ℝ, < )    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘2))    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐶 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ 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 → (𝑅 ≤ 𝑀 ∧ ((𝑅 = 0 ∨ 𝑅 = 𝑀) → (𝑀↑2) ∥ (𝑀 · 𝑃))))
Theorem	4sqlem17 16980*	Lemma for 4sq 16983. (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)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   ⊢ 𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   ⊢ 𝑀 = inf(𝑇, ℝ, < )    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘2))    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐶 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ 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
Theorem	4sqlem18 16981*	Lemma for 4sq 16983. 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)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   ⊢ 𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   ⊢ 𝑀 = inf(𝑇, ℝ, < )    ⇒   ⊢ (𝜑 → 𝑃 ∈ 𝑆)
Theorem	4sqlem19 16982*	Lemma for 4sq 16983. 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 16981. If 𝑘 is 0, 1, 2, we show 𝑘 ∈ 𝑆 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 16973 𝑘 ∈ 𝑆. (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)))}    ⇒   ⊢ ℕ0 = 𝑆
Theorem	4sq 16983*	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
Syntax	cvdwa 16984	The arithmetic progression function.
class AP
Syntax	cvdwm 16985	The monochromatic arithmetic progression predicate.
class MonoAP
Syntax	cvdwp 16986	The polychromatic arithmetic progression predicate.
class PolyAP
Definition	df-vdwap 16987*	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)))))
Definition	df-vdwmc 16988*	Define the "contains a monochromatic AP" predicate. (Contributed by Mario Carneiro, 18-Aug-2014.)
⊢ MonoAP = {⟨𝑘, 𝑓⟩ ∣ ∃𝑐(ran (AP‘𝑘) ∩ 𝒫 (◡𝑓 “ {𝑐})) ≠ ∅}
Definition	df-vdwpc 16989*	Define the "contains a polychromatic collection of APs" predicate. See vdwpc 16999 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 + (𝑑‘𝑖))(AP‘𝑘)(𝑑‘𝑖)) ⊆ (◡𝑓 “ {(𝑓‘(𝑎 + (𝑑‘𝑖)))}) ∧ (♯‘ran (𝑖 ∈ (1...𝑚) ↦ (𝑓‘(𝑎 + (𝑑‘𝑖))))) = 𝑚)}
Theorem	vdwapfval 16990*	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)))))
Theorem	vdwapf 16991	The arithmetic progression function is a function. (Contributed by Mario Carneiro, 18-Aug-2014.)
⊢ (𝐾 ∈ ℕ0 → (AP‘𝐾):(ℕ × ℕ)⟶𝒫 ℕ)
Theorem	vdwapval 16992*	Value of the arithmetic progression function. (Contributed by Mario Carneiro, 18-Aug-2014.)
⊢ ((𝐾 ∈ ℕ0 ∧ 𝐴 ∈ ℕ ∧ 𝐷 ∈ ℕ) → (𝑋 ∈ (𝐴(AP‘𝐾)𝐷) ↔ ∃𝑚 ∈ (0...(𝐾 − 1))𝑋 = (𝐴 + (𝑚 · 𝐷))))
Theorem	vdwapun 16993	Remove the first element of an arithmetic progression. (Contributed by Mario Carneiro, 11-Sep-2014.)
⊢ ((𝐾 ∈ ℕ0 ∧ 𝐴 ∈ ℕ ∧ 𝐷 ∈ ℕ) → (𝐴(AP‘(𝐾 + 1))𝐷) = ({𝐴} ∪ ((𝐴 + 𝐷)(AP‘𝐾)𝐷)))
Theorem	vdwapid1 16994	The first element of an arithmetic progression. (Contributed by Mario Carneiro, 12-Sep-2014.)
⊢ ((𝐾 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐷 ∈ ℕ) → 𝐴 ∈ (𝐴(AP‘𝐾)𝐷))
Theorem	vdwap0 16995	Value of a length-1 arithmetic progression. (Contributed by Mario Carneiro, 18-Aug-2014.)
⊢ ((𝐴 ∈ ℕ ∧ 𝐷 ∈ ℕ) → (𝐴(AP‘0)𝐷) = ∅)
Theorem	vdwap1 16996	Value of a length-1 arithmetic progression. (Contributed by Mario Carneiro, 18-Aug-2014.)
⊢ ((𝐴 ∈ ℕ ∧ 𝐷 ∈ ℕ) → (𝐴(AP‘1)𝐷) = {𝐴})
Theorem	vdwmc 16997*	The predicate " The ⟨𝑅, 𝑁⟩-coloring 𝐹 contains a monochromatic AP of length 𝐾". (Contributed by Mario Carneiro, 18-Aug-2014.)
⊢ 𝑋 ∈ V    &   ⊢ (𝜑 → 𝐾 ∈ ℕ0)    &   ⊢ (𝜑 → 𝐹:𝑋⟶𝑅)    ⇒   ⊢ (𝜑 → (𝐾 MonoAP 𝐹 ↔ ∃𝑐∃𝑎 ∈ ℕ ∃𝑑 ∈ ℕ (𝑎(AP‘𝐾)𝑑) ⊆ (◡𝐹 “ {𝑐})))
Theorem	vdwmc2 16998*	Expand out the definition of an arithmetic progression. (Contributed by Mario Carneiro, 18-Aug-2014.)
⊢ 𝑋 ∈ V    &   ⊢ (𝜑 → 𝐾 ∈ ℕ0)    &   ⊢ (𝜑 → 𝐹:𝑋⟶𝑅)    &   ⊢ (𝜑 → 𝐴 ∈ 𝑋)    ⇒   ⊢ (𝜑 → (𝐾 MonoAP 𝐹 ↔ ∃𝑐 ∈ 𝑅 ∃𝑎 ∈ ℕ ∃𝑑 ∈ ℕ ∀𝑚 ∈ (0...(𝐾 − 1))(𝑎 + (𝑚 · 𝑑)) ∈ (◡𝐹 “ {𝑐})))
Theorem	vdwpc 16999*	The predicate " The coloring 𝐹 contains a polychromatic 𝑀-tuple of AP's of length 𝐾". A polychromatic 𝑀-tuple of AP's is a set of AP's with the same base point but different step lengths, such that each individual AP is monochromatic, but the AP's all have mutually distinct colors. (The common basepoint is not required to have the same color as any of the AP's.) (Contributed by Mario Carneiro, 18-Aug-2014.)
⊢ 𝑋 ∈ V    &   ⊢ (𝜑 → 𝐾 ∈ ℕ0)    &   ⊢ (𝜑 → 𝐹:𝑋⟶𝑅)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ 𝐽 = (1...𝑀)    ⇒   ⊢ (𝜑 → (⟨𝑀, 𝐾⟩ PolyAP 𝐹 ↔ ∃𝑎 ∈ ℕ ∃𝑑 ∈ (ℕ ↑m 𝐽)(∀𝑖 ∈ 𝐽 ((𝑎 + (𝑑‘𝑖))(AP‘𝐾)(𝑑‘𝑖)) ⊆ (◡𝐹 “ {(𝐹‘(𝑎 + (𝑑‘𝑖)))}) ∧ (♯‘ran (𝑖 ∈ 𝐽 ↦ (𝐹‘(𝑎 + (𝑑‘𝑖))))) = 𝑀)))
Theorem	vdwlem1 17000*	Lemma for vdw 17013. (Contributed by Mario Carneiro, 12-Sep-2014.)
⊢ (𝜑 → 𝑅 ∈ Fin)    &   ⊢ (𝜑 → 𝐾 ∈ ℕ)    &   ⊢ (𝜑 → 𝑊 ∈ ℕ)    &   ⊢ (𝜑 → 𝐹:(1...𝑊)⟶𝑅)    &   ⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ (𝜑 → 𝐷:(1...𝑀)⟶ℕ)    &   ⊢ (𝜑 → ∀𝑖 ∈ (1...𝑀)((𝐴 + (𝐷‘𝑖))(AP‘𝐾)(𝐷‘𝑖)) ⊆ (◡𝐹 “ {(𝐹‘(𝐴 + (𝐷‘𝑖)))}))    &   ⊢ (𝜑 → 𝐼 ∈ (1...𝑀))    &   ⊢ (𝜑 → (𝐹‘𝐴) = (𝐹‘(𝐴 + (𝐷‘𝐼))))    ⇒   ⊢ (𝜑 → (𝐾 + 1) MonoAP 𝐹)