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	pcndvds2 16901	The remainder after dividing out all factors of 𝑃 is not divisible by 𝑃. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → ¬ 𝑃 ∥ (𝑁 / (𝑃↑(𝑃 pCnt 𝑁))))
Theorem	pcdvdsb 16902	𝑃↑𝐴 divides 𝑁 if and only if 𝐴 is at most the count of 𝑃. (Contributed by Mario Carneiro, 3-Oct-2014.)
⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℤ ∧ 𝐴 ∈ ℕ0) → (𝐴 ≤ (𝑃 pCnt 𝑁) ↔ (𝑃↑𝐴) ∥ 𝑁))
Theorem	pcelnn 16903	There are a positive number of powers of a prime 𝑃 in 𝑁 iff 𝑃 divides 𝑁. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → ((𝑃 pCnt 𝑁) ∈ ℕ ↔ 𝑃 ∥ 𝑁))
Theorem	pceq0 16904	There are zero powers of a prime 𝑃 in 𝑁 iff 𝑃 does not divide 𝑁. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → ((𝑃 pCnt 𝑁) = 0 ↔ ¬ 𝑃 ∥ 𝑁))
Theorem	pcidlem 16905	The prime count of a prime power. (Contributed by Mario Carneiro, 12-Mar-2014.)
⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℕ0) → (𝑃 pCnt (𝑃↑𝐴)) = 𝐴)
Theorem	pcid 16906	The prime count of a prime power. (Contributed by Mario Carneiro, 9-Sep-2014.)
⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ) → (𝑃 pCnt (𝑃↑𝐴)) = 𝐴)
Theorem	pcneg 16907	The prime count of a negative number. (Contributed by Mario Carneiro, 13-Mar-2014.)
⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℚ) → (𝑃 pCnt -𝐴) = (𝑃 pCnt 𝐴))
Theorem	pcabs 16908	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))
Theorem	pcdvdstr 16909	The prime count increases under the divisibility relation. (Contributed by Mario Carneiro, 13-Mar-2014.)
⊢ ((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐴 ∥ 𝐵)) → (𝑃 pCnt 𝐴) ≤ (𝑃 pCnt 𝐵))
Theorem	pcgcd1 16910	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))
Theorem	pcgcd 16911	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)))
Theorem	pc2dvds 16912*	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) → (𝐴 ∥ 𝐵 ↔ ∀𝑝 ∈ ℙ (𝑝 pCnt 𝐴) ≤ (𝑝 pCnt 𝐵)))
Theorem	pc11 16913*	The prime count function, viewed as a function from ℕ to (ℕ ↑m ℙ), is one-to-one. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ ((𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ0) → (𝐴 = 𝐵 ↔ ∀𝑝 ∈ ℙ (𝑝 pCnt 𝐴) = (𝑝 pCnt 𝐵)))
Theorem	pcz 16914*	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)))
Theorem	pcprmpw2 16915*	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 16916*	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 16917*	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 16918*	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 16919*	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 16920	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 16921	Lemma for pcadd 16922. 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 16922	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 16923	The inequality of pcadd 16922 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 16924	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 16925*	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 16926*	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 16927	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 16928*	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 16929*	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 16930	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 16931	Lemma for pcfac 16932. (Contributed by Mario Carneiro, 20-May-2014.)
⊢ ((𝑁 ∈ ℕ0 ∧ 𝑀 ∈ (ℤ≥‘𝑁) ∧ 𝑃 ∈ ℙ) → (⌊‘(𝑁 / (𝑃↑𝑀))) = 0)
Theorem	pcfac 16932*	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 16933*	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 16934	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 16935	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 16936*	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 16937	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 16938	Lemma for pockthg 16939. (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 16939*	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 16940	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 16939 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 16941*	Lemma for unben 16942. (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 16942*	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 16943*	Lemma for infpn 16945. 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 16944*	Lemma for infpn 16945. 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 16945*	There exist infinitely many prime numbers: for any positive integer 𝑁, there exists a prime number 𝑗 greater than 𝑁. (See infpn2 16946 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 16946*	There exist infinitely many prime numbers: the set of all primes 𝑆 is unbounded by infpn 16945, so by unben 16942 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 16947*	The primes are unbounded. (Contributed by Paul Chapman, 28-Nov-2012.)
⊢ (𝑁 ∈ ℕ → ∃𝑝 ∈ ℙ 𝑁 < 𝑝)
Theorem	prminf 16948	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 16949*	Lemma for prmrec 16955. 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 16950*	Lemma for prmrec 16955. 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 16951*	Lemma for prmrec 16955. 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 16952*	Lemma for prmrec 16955. 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 16808, 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 16953*	Lemma for prmrec 16955. 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 16952 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 16954*	Lemma for prmrec 16955. 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 16953 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 16955*	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 16956*	Lemma for 1arith 16960. (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 16957*	Lemma for 1arith 16960. (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 16958*	Lemma for 1arith 16960. (Contributed by Mario Carneiro, 30-May-2014.)
⊢ 𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛)))    ⇒   ⊢ (𝑁 ∈ ℕ → (𝑀‘𝑁):ℙ⟶ℕ0)
Theorem	1arithlem4 16959*	Lemma for 1arith 16960. (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 16960*	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 16961*	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 16962	Extend class notation with the set of gaussian integers.
class ℤ[i]
Definition	df-gz 16963	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 16964	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 16965	A gaussian integer is a complex number. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ (𝐴 ∈ ℤ[i] → 𝐴 ∈ ℂ)
Theorem	zgz 16966	An integer is a gaussian integer. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ (𝐴 ∈ ℤ → 𝐴 ∈ ℤ[i])
Theorem	igz 16967	i is a gaussian integer. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ i ∈ ℤ[i]
Theorem	gznegcl 16968	The gaussian integers are closed under negation. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ (𝐴 ∈ ℤ[i] → -𝐴 ∈ ℤ[i])
Theorem	gzcjcl 16969	The gaussian integers are closed under conjugation. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ (𝐴 ∈ ℤ[i] → (∗‘𝐴) ∈ ℤ[i])
Theorem	gzaddcl 16970	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 16971	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 16972	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 16973	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 16974	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 16975	Lemma for 4sq 16997. (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 16976	Lemma for 4sq 16997. (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 16977	Lemma for 4sq 16997. (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 16978	Lemma for 4sq 16997. (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 16979	Lemma for 4sq 16997. (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 16980	Lemma for 4sq 16997. (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 16981*	Lemma for 4sq 16997. 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 16982*	Lemma for 4sq 16997. 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 16983*	Lemma for 4sq 16997. 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 16984*	Lemma for 4sqlem4 16985. (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 16985*	Lemma for 4sq 16997. 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 16986*	Lemma for mul4sq 16987: algebraic manipulations. The extra assumptions involving 𝑀 are for a part of 4sqlem17 16994 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 16987*	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 16986. (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 16988*	Lemma for 4sq 16997. 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 16989*	Lemma for 4sq 16997. 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 16990*	Lemma for 4sq 16997. (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 16991*	Lemma for 4sq 16997. (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 16992*	Lemma for 4sq 16997. (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 16993*	Lemma for 4sq 16997. (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 16994*	Lemma for 4sq 16997. (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 16995*	Lemma for 4sq 16997. 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 16996*	Lemma for 4sq 16997. 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 16995. 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 16987 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 16997*	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 16998	The arithmetic progression function.
class AP
Syntax	cvdwm 16999	The monochromatic arithmetic progression predicate.
class MonoAP
Syntax	cvdwp 17000	The polychromatic arithmetic progression predicate.
class PolyAP