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Theorem List for Metamath Proof Explorer - 15401-15500   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theorempcrec 15401 Prime power of a reciprocal. (Contributed by Mario Carneiro, 10-Aug-2015.)
((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℚ ∧ 𝐴 ≠ 0)) → (𝑃 pCnt (1 / 𝐴)) = -(𝑃 pCnt 𝐴))
 
Theorempcexp 15402 Prime power of an exponential. (Contributed by Mario Carneiro, 10-Aug-2015.)
((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℚ ∧ 𝐴 ≠ 0) ∧ 𝑁 ∈ ℤ) → (𝑃 pCnt (𝐴𝑁)) = (𝑁 · (𝑃 pCnt 𝐴)))
 
Theorempcxcl 15403 Extended real closure of the general prime count function. (Contributed by Mario Carneiro, 3-Oct-2014.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℚ) → (𝑃 pCnt 𝑁) ∈ ℝ*)
 
Theorempcge0 15404 The prime count of an integer is greater or equal to zero. (Contributed by Mario Carneiro, 3-Oct-2014.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℤ) → 0 ≤ (𝑃 pCnt 𝑁))
 
Theorempczdvds 15405 Defining property of the prime count function. (Contributed by Mario Carneiro, 9-Sep-2014.)
((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝑃↑(𝑃 pCnt 𝑁)) ∥ 𝑁)
 
Theorempcdvds 15406 Defining property of the prime count function. (Contributed by Mario Carneiro, 23-Feb-2014.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → (𝑃↑(𝑃 pCnt 𝑁)) ∥ 𝑁)
 
Theorempczndvds 15407 Defining property of the prime count function. (Contributed by Mario Carneiro, 3-Oct-2014.)
((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → ¬ (𝑃↑((𝑃 pCnt 𝑁) + 1)) ∥ 𝑁)
 
Theorempcndvds 15408 Defining property of the prime count function. (Contributed by Mario Carneiro, 23-Feb-2014.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → ¬ (𝑃↑((𝑃 pCnt 𝑁) + 1)) ∥ 𝑁)
 
Theorempczndvds2 15409 The remainder after dividing out all factors of 𝑃 is not divisible by 𝑃. (Contributed by Mario Carneiro, 9-Sep-2014.)
((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → ¬ 𝑃 ∥ (𝑁 / (𝑃↑(𝑃 pCnt 𝑁))))
 
Theorempcndvds2 15410 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))))
 
Theorempcdvdsb 15411 𝑃𝐴 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) ↔ (𝑃𝐴) ∥ 𝑁))
 
Theorempcelnn 15412 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 ↔ 𝑃𝑁))
 
Theorempceq0 15413 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 ↔ ¬ 𝑃𝑁))
 
Theorempcidlem 15414 The prime count of a prime power. (Contributed by Mario Carneiro, 12-Mar-2014.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℕ0) → (𝑃 pCnt (𝑃𝐴)) = 𝐴)
 
Theorempcid 15415 The prime count of a prime power. (Contributed by Mario Carneiro, 9-Sep-2014.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ) → (𝑃 pCnt (𝑃𝐴)) = 𝐴)
 
Theorempcneg 15416 The prime count of a negative number. (Contributed by Mario Carneiro, 13-Mar-2014.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℚ) → (𝑃 pCnt -𝐴) = (𝑃 pCnt 𝐴))
 
Theorempcabs 15417 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 15418 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 15419 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 15420 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 15421* 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 15422* 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 15423* 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 15424* 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 15425* 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 15426* 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 15427* 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 15428* 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 15429 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 15430 Lemma for pcadd 15431. 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 15431 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 15432 The inequality of pcadd 15431 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 15433 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 15434* 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 15435* 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 15436 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 15437* 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 15438* 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 15439 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 15440 Lemma for pcfac 15441. (Contributed by Mario Carneiro, 20-May-2014.)
((𝑁 ∈ ℕ0𝑀 ∈ (ℤ𝑁) ∧ 𝑃 ∈ ℙ) → (⌊‘(𝑁 / (𝑃𝑀))) = 0)
 
Theorempcfac 15441* 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 15442* 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 15443 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 15444 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 15445* 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 15446 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 15447 Lemma for pockthg 15448. (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 15448* 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 15449 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 15448 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 15450* Lemma for unben 15451. (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 15451* 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 15452* Lemma for infpn 15454. 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 15453* Lemma for infpn 15454. 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 15454* There exist infinitely many prime numbers: for any positive integer 𝑁, there exists a prime number 𝑗 greater than 𝑁. (See infpn2 15455 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 15455* There exist infinitely many prime numbers: the set of all primes 𝑆 is unbounded by infpn 15454, so by unben 15451 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 15456* The primes are unbounded. (Contributed by Paul Chapman, 28-Nov-2012.)
(𝑁 ∈ ℕ → ∃𝑝 ∈ ℙ 𝑁 < 𝑝)
 
Theoremprminf 15457 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 15458* Lemma for prmrec 15464. 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 15459* Lemma for prmrec 15464. 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 15460* Lemma for prmrec 15464. 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 15461* Lemma for prmrec 15464. 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 15318, 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 15462* Lemma for prmrec 15464. 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 15461 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 15463* Lemma for prmrec 15464. 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 15462 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 15464* 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 15465* Lemma for 1arith 15469. (Contributed by Mario Carneiro, 30-May-2014.)
𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛)))       (𝑁 ∈ ℕ → (𝑀𝑁) = (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑁)))
 
Theorem1arithlem2 15466* Lemma for 1arith 15469. (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 15467* Lemma for 1arith 15469. (Contributed by Mario Carneiro, 30-May-2014.)
𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛)))       (𝑁 ∈ ℕ → (𝑀𝑁):ℙ⟶ℕ0)
 
Theorem1arithlem4 15468* Lemma for 1arith 15469. (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 15469* 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 15470* 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 15471 Extend class notation with the set of gaussian integers.
class ℤ[i]
 
Definitiondf-gz 15472 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 15473 Elementhood in the gaussian integers. (Contributed by Mario Carneiro, 14-Jul-2014.)
(𝐴 ∈ ℤ[i] ↔ (𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ ℤ ∧ (ℑ‘𝐴) ∈ ℤ))
 
Theoremgzcn 15474 A gaussian integer is a complex number. (Contributed by Mario Carneiro, 14-Jul-2014.)
(𝐴 ∈ ℤ[i] → 𝐴 ∈ ℂ)
 
Theoremzgz 15475 An integer is a gaussian integer. (Contributed by Mario Carneiro, 14-Jul-2014.)
(𝐴 ∈ ℤ → 𝐴 ∈ ℤ[i])
 
Theoremigz 15476 i is a gaussian integer. (Contributed by Mario Carneiro, 14-Jul-2014.)
i ∈ ℤ[i]
 
Theoremgznegcl 15477 The gaussian integers are closed under negation. (Contributed by Mario Carneiro, 14-Jul-2014.)
(𝐴 ∈ ℤ[i] → -𝐴 ∈ ℤ[i])
 
Theoremgzcjcl 15478 The gaussian integers are closed under conjugation. (Contributed by Mario Carneiro, 14-Jul-2014.)
(𝐴 ∈ ℤ[i] → (∗‘𝐴) ∈ ℤ[i])
 
Theoremgzaddcl 15479 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 15480 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 15481 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 15482 The gaussian integers are closed under subtraction. (Contributed by Mario Carneiro, 14-Jul-2014.)
((𝐴 ∈ ℤ[i] ∧ 𝐵 ∈ ℤ[i]) → (𝐴𝐵) ∈ ℤ[i])
 
Theoremgzabssqcl 15483 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 15484 Lemma for 4sq 15506. (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 15485 Lemma for 4sq 15506. (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 15486 Lemma for 4sq 15506. (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 15487 Lemma for 4sq 15506. (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 15488 Lemma for 4sq 15506. (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 15489 Lemma for 4sq 15506. (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 15490* Lemma for 4sq 15506. 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 15491* Lemma for 4sq 15506. 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 15492* Lemma for 4sq 15506. 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 15493* Lemma for 4sqlem4 15494. (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 15494* Lemma for 4sq 15506. 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 15495* Lemma for mul4sq 15496: algebraic manipulations. The extra assumptions involving 𝑀 are for a part of 4sqlem17 15503 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 15496* 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 15495. (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 15497* Lemma for 4sq 15506. 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 15498* Lemma for 4sq 15506. 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 15499* Lemma for 4sq 15506. (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 15500* Lemma for 4sq 15506. (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)
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