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| Type | Label | Description |
|---|---|---|
| Statement | ||
| Theorem | odzcllem 16701 | - Lemma for odzcl 16702, showing existence of a recurrent point for the exponential. (Contributed by Mario Carneiro, 28-Feb-2014.) (Proof shortened by AV, 26-Sep-2020.) |
| ⊢ ((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1) → (((odℤ‘𝑁)‘𝐴) ∈ ℕ ∧ 𝑁 ∥ ((𝐴↑((odℤ‘𝑁)‘𝐴)) − 1))) | ||
| Theorem | odzcl 16702 | The order of a group element is an integer. (Contributed by Mario Carneiro, 28-Feb-2014.) |
| ⊢ ((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1) → ((odℤ‘𝑁)‘𝐴) ∈ ℕ) | ||
| Theorem | odzid 16703 | Any element raised to the power of its order is 1. (Contributed by Mario Carneiro, 28-Feb-2014.) |
| ⊢ ((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1) → 𝑁 ∥ ((𝐴↑((odℤ‘𝑁)‘𝐴)) − 1)) | ||
| Theorem | odzdvds 16704 | The only powers of 𝐴 that are congruent to 1 are the multiples of the order of 𝐴. (Contributed by Mario Carneiro, 28-Feb-2014.) (Proof shortened by AV, 26-Sep-2020.) |
| ⊢ (((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1) ∧ 𝐾 ∈ ℕ0) → (𝑁 ∥ ((𝐴↑𝐾) − 1) ↔ ((odℤ‘𝑁)‘𝐴) ∥ 𝐾)) | ||
| Theorem | odzphi 16705 | The order of any group element is a divisor of the Euler ϕ function. (Contributed by Mario Carneiro, 28-Feb-2014.) |
| ⊢ ((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1) → ((odℤ‘𝑁)‘𝐴) ∥ (ϕ‘𝑁)) | ||
| Theorem | modprm1div 16706 | A prime number divides an integer minus 1 iff the integer modulo the prime number is 1. (Contributed by Alexander van der Vekens, 17-May-2018.) (Proof shortened by AV, 30-May-2023.) |
| ⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ) → ((𝐴 mod 𝑃) = 1 ↔ 𝑃 ∥ (𝐴 − 1))) | ||
| Theorem | m1dvdsndvds 16707 | If an integer minus 1 is divisible by a prime number, the integer itself is not divisible by this prime number. (Contributed by Alexander van der Vekens, 30-Aug-2018.) |
| ⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ) → (𝑃 ∥ (𝐴 − 1) → ¬ 𝑃 ∥ 𝐴)) | ||
| Theorem | modprminv 16708 | Show an explicit expression for the modular inverse of 𝐴 mod 𝑃. This is an application of prmdiv 16693. (Contributed by Alexander van der Vekens, 15-May-2018.) |
| ⊢ 𝑅 = ((𝐴↑(𝑃 − 2)) mod 𝑃) ⇒ ⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ ¬ 𝑃 ∥ 𝐴) → (𝑅 ∈ (1...(𝑃 − 1)) ∧ ((𝐴 · 𝑅) mod 𝑃) = 1)) | ||
| Theorem | modprminveq 16709 | The modular inverse of 𝐴 mod 𝑃 is unique. (Contributed by Alexander van der Vekens, 17-May-2018.) |
| ⊢ 𝑅 = ((𝐴↑(𝑃 − 2)) mod 𝑃) ⇒ ⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ ¬ 𝑃 ∥ 𝐴) → ((𝑆 ∈ (0...(𝑃 − 1)) ∧ ((𝐴 · 𝑆) mod 𝑃) = 1) ↔ 𝑆 = 𝑅)) | ||
| Theorem | vfermltl 16710 | Variant of Fermat's little theorem if 𝐴 is not a multiple of 𝑃, see theorem 5.18 in [ApostolNT] p. 113. (Contributed by AV, 21-Aug-2020.) (Proof shortened by AV, 5-Sep-2020.) |
| ⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ ¬ 𝑃 ∥ 𝐴) → ((𝐴↑(𝑃 − 1)) mod 𝑃) = 1) | ||
| Theorem | vfermltlALT 16711 | Alternate proof of vfermltl 16710, not using Euler's theorem. (Contributed by AV, 21-Aug-2020.) (New usage is discouraged.) (Proof modification is discouraged.) |
| ⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ ¬ 𝑃 ∥ 𝐴) → ((𝐴↑(𝑃 − 1)) mod 𝑃) = 1) | ||
| Theorem | powm2modprm 16712 | If an integer minus 1 is divisible by a prime number, then the integer to the power of the prime number minus 2 is 1 modulo the prime number. (Contributed by Alexander van der Vekens, 30-Aug-2018.) |
| ⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ) → (𝑃 ∥ (𝐴 − 1) → ((𝐴↑(𝑃 − 2)) mod 𝑃) = 1)) | ||
| Theorem | reumodprminv 16713* | For any prime number and for any positive integer less than this prime number, there is a unique modular inverse of this positive integer. (Contributed by Alexander van der Vekens, 12-May-2018.) |
| ⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ (1..^𝑃)) → ∃!𝑖 ∈ (1...(𝑃 − 1))((𝑁 · 𝑖) mod 𝑃) = 1) | ||
| Theorem | modprm0 16714* | For two positive integers less than a given prime number there is always a nonnegative integer (less than the given prime number) so that the sum of one of the two positive integers and the other of the positive integers multiplied by the nonnegative integer is 0 ( modulo the given prime number). (Contributed by Alexander van der Vekens, 17-May-2018.) |
| ⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ (1..^𝑃) ∧ 𝐼 ∈ (1..^𝑃)) → ∃𝑗 ∈ (0..^𝑃)((𝐼 + (𝑗 · 𝑁)) mod 𝑃) = 0) | ||
| Theorem | nnnn0modprm0 16715* | For a positive integer and a nonnegative integer both less than a given prime number there is always a second nonnegative integer (less than the given prime number) so that the sum of this second nonnegative integer multiplied with the positive integer and the first nonnegative integer is 0 ( modulo the given prime number). (Contributed by Alexander van der Vekens, 8-Nov-2018.) |
| ⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ (1..^𝑃) ∧ 𝐼 ∈ (0..^𝑃)) → ∃𝑗 ∈ (0..^𝑃)((𝐼 + (𝑗 · 𝑁)) mod 𝑃) = 0) | ||
| Theorem | modprmn0modprm0 16716* | For an integer not being 0 modulo a given prime number and a nonnegative integer less than the prime number, there is always a second nonnegative integer (less than the given prime number) so that the sum of this second nonnegative integer multiplied with the integer and the first nonnegative integer is 0 ( modulo the given prime number). (Contributed by Alexander van der Vekens, 10-Nov-2018.) |
| ⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℤ ∧ (𝑁 mod 𝑃) ≠ 0) → (𝐼 ∈ (0..^𝑃) → ∃𝑗 ∈ (0..^𝑃)((𝐼 + (𝑗 · 𝑁)) mod 𝑃) = 0)) | ||
| Theorem | coprimeprodsq 16717 | If three numbers are coprime, and the square of one is the product of the other two, then there is a formula for the other two in terms of gcd and square. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
| ⊢ (((𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℕ0) ∧ ((𝐴 gcd 𝐵) gcd 𝐶) = 1) → ((𝐶↑2) = (𝐴 · 𝐵) → 𝐴 = ((𝐴 gcd 𝐶)↑2))) | ||
| Theorem | coprimeprodsq2 16718 | If three numbers are coprime, and the square of one is the product of the other two, then there is a formula for the other two in terms of gcd and square. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
| ⊢ (((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℕ0 ∧ 𝐶 ∈ ℕ0) ∧ ((𝐴 gcd 𝐵) gcd 𝐶) = 1) → ((𝐶↑2) = (𝐴 · 𝐵) → 𝐵 = ((𝐵 gcd 𝐶)↑2))) | ||
| Theorem | oddprm 16719 | A prime not equal to 2 is odd. (Contributed by Mario Carneiro, 4-Feb-2015.) (Proof shortened by AV, 10-Jul-2022.) |
| ⊢ (𝑁 ∈ (ℙ ∖ {2}) → ((𝑁 − 1) / 2) ∈ ℕ) | ||
| Theorem | nnoddn2prm 16720 | A prime not equal to 2 is an odd positive integer. (Contributed by AV, 28-Jun-2021.) |
| ⊢ (𝑁 ∈ (ℙ ∖ {2}) → (𝑁 ∈ ℕ ∧ ¬ 2 ∥ 𝑁)) | ||
| Theorem | oddn2prm 16721 | A prime not equal to 2 is odd. (Contributed by AV, 28-Jun-2021.) |
| ⊢ (𝑁 ∈ (ℙ ∖ {2}) → ¬ 2 ∥ 𝑁) | ||
| Theorem | nnoddn2prmb 16722 | A number is a prime number not equal to 2 iff it is an odd prime number. Conversion theorem for two representations of odd primes. (Contributed by AV, 14-Jul-2021.) |
| ⊢ (𝑁 ∈ (ℙ ∖ {2}) ↔ (𝑁 ∈ ℙ ∧ ¬ 2 ∥ 𝑁)) | ||
| Theorem | prm23lt5 16723 | A prime less than 5 is either 2 or 3. (Contributed by AV, 5-Jul-2021.) |
| ⊢ ((𝑃 ∈ ℙ ∧ 𝑃 < 5) → (𝑃 = 2 ∨ 𝑃 = 3)) | ||
| Theorem | prm23ge5 16724 | A prime is either 2 or 3 or greater than or equal to 5. (Contributed by AV, 5-Jul-2021.) |
| ⊢ (𝑃 ∈ ℙ → (𝑃 = 2 ∨ 𝑃 = 3 ∨ 𝑃 ∈ (ℤ≥‘5))) | ||
| Theorem | pythagtriplem1 16725* | Lemma for pythagtrip 16743. Prove a weaker version of one direction of the theorem. (Contributed by Scott Fenton, 28-Mar-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
| ⊢ (∃𝑛 ∈ ℕ ∃𝑚 ∈ ℕ ∃𝑘 ∈ ℕ (𝐴 = (𝑘 · ((𝑚↑2) − (𝑛↑2))) ∧ 𝐵 = (𝑘 · (2 · (𝑚 · 𝑛))) ∧ 𝐶 = (𝑘 · ((𝑚↑2) + (𝑛↑2)))) → ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2)) | ||
| Theorem | pythagtriplem2 16726* | Lemma for pythagtrip 16743. Prove the full version of one direction of the theorem. (Contributed by Scott Fenton, 28-Mar-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
| ⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (∃𝑛 ∈ ℕ ∃𝑚 ∈ ℕ ∃𝑘 ∈ ℕ ({𝐴, 𝐵} = {(𝑘 · ((𝑚↑2) − (𝑛↑2))), (𝑘 · (2 · (𝑚 · 𝑛)))} ∧ 𝐶 = (𝑘 · ((𝑚↑2) + (𝑛↑2)))) → ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2))) | ||
| Theorem | pythagtriplem3 16727 | Lemma for pythagtrip 16743. Show that 𝐶 and 𝐵 are relatively prime under some conditions. (Contributed by Scott Fenton, 8-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
| ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (𝐵 gcd 𝐶) = 1) | ||
| Theorem | pythagtriplem4 16728 | Lemma for pythagtrip 16743. Show that 𝐶 − 𝐵 and 𝐶 + 𝐵 are relatively prime. (Contributed by Scott Fenton, 12-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
| ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → ((𝐶 − 𝐵) gcd (𝐶 + 𝐵)) = 1) | ||
| Theorem | pythagtriplem10 16729 | Lemma for pythagtrip 16743. Show that 𝐶 − 𝐵 is positive. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
| ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2)) → 0 < (𝐶 − 𝐵)) | ||
| Theorem | pythagtriplem6 16730 | Lemma for pythagtrip 16743. Calculate (√‘(𝐶 − 𝐵)). (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
| ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (√‘(𝐶 − 𝐵)) = ((𝐶 − 𝐵) gcd 𝐴)) | ||
| Theorem | pythagtriplem7 16731 | Lemma for pythagtrip 16743. Calculate (√‘(𝐶 + 𝐵)). (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
| ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (√‘(𝐶 + 𝐵)) = ((𝐶 + 𝐵) gcd 𝐴)) | ||
| Theorem | pythagtriplem8 16732 | Lemma for pythagtrip 16743. Show that (√‘(𝐶 − 𝐵)) is a positive integer. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
| ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (√‘(𝐶 − 𝐵)) ∈ ℕ) | ||
| Theorem | pythagtriplem9 16733 | Lemma for pythagtrip 16743. Show that (√‘(𝐶 + 𝐵)) is a positive integer. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
| ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (√‘(𝐶 + 𝐵)) ∈ ℕ) | ||
| Theorem | pythagtriplem11 16734 | Lemma for pythagtrip 16743. Show that 𝑀 (which will eventually be closely related to the 𝑚 in the final statement) is a natural. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
| ⊢ 𝑀 = (((√‘(𝐶 + 𝐵)) + (√‘(𝐶 − 𝐵))) / 2) ⇒ ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝑀 ∈ ℕ) | ||
| Theorem | pythagtriplem12 16735 | Lemma for pythagtrip 16743. Calculate the square of 𝑀. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
| ⊢ 𝑀 = (((√‘(𝐶 + 𝐵)) + (√‘(𝐶 − 𝐵))) / 2) ⇒ ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (𝑀↑2) = ((𝐶 + 𝐴) / 2)) | ||
| Theorem | pythagtriplem13 16736 | Lemma for pythagtrip 16743. Show that 𝑁 (which will eventually be closely related to the 𝑛 in the final statement) is a natural. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
| ⊢ 𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶 − 𝐵))) / 2) ⇒ ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝑁 ∈ ℕ) | ||
| Theorem | pythagtriplem14 16737 | Lemma for pythagtrip 16743. Calculate the square of 𝑁. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
| ⊢ 𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶 − 𝐵))) / 2) ⇒ ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (𝑁↑2) = ((𝐶 − 𝐴) / 2)) | ||
| Theorem | pythagtriplem15 16738 | Lemma for pythagtrip 16743. Show the relationship between 𝑀, 𝑁, and 𝐴. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
| ⊢ 𝑀 = (((√‘(𝐶 + 𝐵)) + (√‘(𝐶 − 𝐵))) / 2) & ⊢ 𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶 − 𝐵))) / 2) ⇒ ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝐴 = ((𝑀↑2) − (𝑁↑2))) | ||
| Theorem | pythagtriplem16 16739 | Lemma for pythagtrip 16743. Show the relationship between 𝑀, 𝑁, and 𝐵. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
| ⊢ 𝑀 = (((√‘(𝐶 + 𝐵)) + (√‘(𝐶 − 𝐵))) / 2) & ⊢ 𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶 − 𝐵))) / 2) ⇒ ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝐵 = (2 · (𝑀 · 𝑁))) | ||
| Theorem | pythagtriplem17 16740 | Lemma for pythagtrip 16743. Show the relationship between 𝑀, 𝑁, and 𝐶. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
| ⊢ 𝑀 = (((√‘(𝐶 + 𝐵)) + (√‘(𝐶 − 𝐵))) / 2) & ⊢ 𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶 − 𝐵))) / 2) ⇒ ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝐶 = ((𝑀↑2) + (𝑁↑2))) | ||
| Theorem | pythagtriplem18 16741* | Lemma for pythagtrip 16743. Wrap the previous 𝑀 and 𝑁 up in quantifiers. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
| ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → ∃𝑛 ∈ ℕ ∃𝑚 ∈ ℕ (𝐴 = ((𝑚↑2) − (𝑛↑2)) ∧ 𝐵 = (2 · (𝑚 · 𝑛)) ∧ 𝐶 = ((𝑚↑2) + (𝑛↑2)))) | ||
| Theorem | pythagtriplem19 16742* | Lemma for pythagtrip 16743. Introduce 𝑘 and remove the relative primality requirement. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
| ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ¬ 2 ∥ (𝐴 / (𝐴 gcd 𝐵))) → ∃𝑛 ∈ ℕ ∃𝑚 ∈ ℕ ∃𝑘 ∈ ℕ (𝐴 = (𝑘 · ((𝑚↑2) − (𝑛↑2))) ∧ 𝐵 = (𝑘 · (2 · (𝑚 · 𝑛))) ∧ 𝐶 = (𝑘 · ((𝑚↑2) + (𝑛↑2))))) | ||
| Theorem | pythagtrip 16743* | Parameterize the Pythagorean triples. If 𝐴, 𝐵, and 𝐶 are naturals, then they obey the Pythagorean triple formula iff they are parameterized by three naturals. This proof follows the Isabelle proof at http://afp.sourceforge.net/entries/Fermat3_4.shtml. This is Metamath 100 proof #23. (Contributed by Scott Fenton, 19-Apr-2014.) |
| ⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) → (((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ↔ ∃𝑛 ∈ ℕ ∃𝑚 ∈ ℕ ∃𝑘 ∈ ℕ ({𝐴, 𝐵} = {(𝑘 · ((𝑚↑2) − (𝑛↑2))), (𝑘 · (2 · (𝑚 · 𝑛)))} ∧ 𝐶 = (𝑘 · ((𝑚↑2) + (𝑛↑2)))))) | ||
| Theorem | iserodd 16744* | Collect the odd terms in a sequence. (Contributed by Mario Carneiro, 7-Apr-2015.) (Proof shortened by AV, 10-Jul-2022.) |
| ⊢ ((𝜑 ∧ 𝑘 ∈ ℕ0) → 𝐶 ∈ ℂ) & ⊢ (𝑛 = ((2 · 𝑘) + 1) → 𝐵 = 𝐶) ⇒ ⊢ (𝜑 → (seq0( + , (𝑘 ∈ ℕ0 ↦ 𝐶)) ⇝ 𝐴 ↔ seq1( + , (𝑛 ∈ ℕ ↦ if(2 ∥ 𝑛, 0, 𝐵))) ⇝ 𝐴)) | ||
| Syntax | cpc 16745 | Extend class notation with the prime count function. |
| class pCnt | ||
| Definition | df-pc 16746* | Define the prime count function, which returns the largest exponent of a given prime (or other positive integer) that divides the number. For rational numbers, it returns negative values according to the power of a prime in the denominator. (Contributed by Mario Carneiro, 23-Feb-2014.) |
| ⊢ pCnt = (𝑝 ∈ ℙ, 𝑟 ∈ ℚ ↦ if(𝑟 = 0, +∞, (℩𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑟 = (𝑥 / 𝑦) ∧ 𝑧 = (sup({𝑛 ∈ ℕ0 ∣ (𝑝↑𝑛) ∥ 𝑥}, ℝ, < ) − sup({𝑛 ∈ ℕ0 ∣ (𝑝↑𝑛) ∥ 𝑦}, ℝ, < )))))) | ||
| Theorem | pclem 16747* | - Lemma for the prime power pre-function's properties. (Contributed by Mario Carneiro, 23-Feb-2014.) |
| ⊢ 𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑁} ⇒ ⊢ ((𝑃 ∈ (ℤ≥‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝐴 ⊆ ℤ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℤ ∀𝑦 ∈ 𝐴 𝑦 ≤ 𝑥)) | ||
| Theorem | pcprecl 16748* | Closure of the prime power pre-function. (Contributed by Mario Carneiro, 23-Feb-2014.) |
| ⊢ 𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑁} & ⊢ 𝑆 = sup(𝐴, ℝ, < ) ⇒ ⊢ ((𝑃 ∈ (ℤ≥‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝑆 ∈ ℕ0 ∧ (𝑃↑𝑆) ∥ 𝑁)) | ||
| Theorem | pcprendvds 16749* | Non-divisibility property of the prime power pre-function. (Contributed by Mario Carneiro, 23-Feb-2014.) |
| ⊢ 𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑁} & ⊢ 𝑆 = sup(𝐴, ℝ, < ) ⇒ ⊢ ((𝑃 ∈ (ℤ≥‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → ¬ (𝑃↑(𝑆 + 1)) ∥ 𝑁) | ||
| Theorem | pcprendvds2 16750* | Non-divisibility property of the prime power pre-function. (Contributed by Mario Carneiro, 23-Feb-2014.) |
| ⊢ 𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑁} & ⊢ 𝑆 = sup(𝐴, ℝ, < ) ⇒ ⊢ ((𝑃 ∈ (ℤ≥‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → ¬ 𝑃 ∥ (𝑁 / (𝑃↑𝑆))) | ||
| Theorem | pcpre1 16751* | Value of the prime power pre-function at 1. (Contributed by Mario Carneiro, 23-Feb-2014.) (Revised by Mario Carneiro, 26-Apr-2016.) |
| ⊢ 𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑁} & ⊢ 𝑆 = sup(𝐴, ℝ, < ) ⇒ ⊢ ((𝑃 ∈ (ℤ≥‘2) ∧ 𝑁 = 1) → 𝑆 = 0) | ||
| Theorem | pcpremul 16752* | Multiplicative property of the prime count pre-function. Note that the primality of 𝑃 is essential for this property; (4 pCnt 2) = 0 but (4 pCnt (2 · 2)) = 1 ≠ 2 · (4 pCnt 2) = 0. Since this is needed to show uniqueness for the real prime count function (over ℚ), we don't bother to define it off the primes. (Contributed by Mario Carneiro, 23-Feb-2014.) |
| ⊢ 𝑆 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑀}, ℝ, < ) & ⊢ 𝑇 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑁}, ℝ, < ) & ⊢ 𝑈 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ (𝑀 · 𝑁)}, ℝ, < ) ⇒ ⊢ ((𝑃 ∈ ℙ ∧ (𝑀 ∈ ℤ ∧ 𝑀 ≠ 0) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝑆 + 𝑇) = 𝑈) | ||
| Theorem | pcval 16753* | The value of the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.) (Revised by Mario Carneiro, 3-Oct-2014.) |
| ⊢ 𝑆 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑥}, ℝ, < ) & ⊢ 𝑇 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑦}, ℝ, < ) ⇒ ⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℚ ∧ 𝑁 ≠ 0)) → (𝑃 pCnt 𝑁) = (℩𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆 − 𝑇)))) | ||
| Theorem | pceulem 16754* | Lemma for pceu 16755. (Contributed by Mario Carneiro, 23-Feb-2014.) |
| ⊢ 𝑆 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑥}, ℝ, < ) & ⊢ 𝑇 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑦}, ℝ, < ) & ⊢ 𝑈 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑠}, ℝ, < ) & ⊢ 𝑉 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑡}, ℝ, < ) & ⊢ (𝜑 → 𝑃 ∈ ℙ) & ⊢ (𝜑 → 𝑁 ≠ 0) & ⊢ (𝜑 → (𝑥 ∈ ℤ ∧ 𝑦 ∈ ℕ)) & ⊢ (𝜑 → 𝑁 = (𝑥 / 𝑦)) & ⊢ (𝜑 → (𝑠 ∈ ℤ ∧ 𝑡 ∈ ℕ)) & ⊢ (𝜑 → 𝑁 = (𝑠 / 𝑡)) ⇒ ⊢ (𝜑 → (𝑆 − 𝑇) = (𝑈 − 𝑉)) | ||
| Theorem | pceu 16755* | Uniqueness for the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.) |
| ⊢ 𝑆 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑥}, ℝ, < ) & ⊢ 𝑇 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑦}, ℝ, < ) ⇒ ⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℚ ∧ 𝑁 ≠ 0)) → ∃!𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆 − 𝑇))) | ||
| Theorem | pczpre 16756* | Connect the prime count pre-function to the actual prime count function, when restricted to the integers. (Contributed by Mario Carneiro, 23-Feb-2014.) (Proof shortened by Mario Carneiro, 24-Dec-2016.) |
| ⊢ 𝑆 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑁}, ℝ, < ) ⇒ ⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝑃 pCnt 𝑁) = 𝑆) | ||
| Theorem | pczcl 16757 | Closure of the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.) |
| ⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝑃 pCnt 𝑁) ∈ ℕ0) | ||
| Theorem | pccl 16758 | Closure of the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.) |
| ⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → (𝑃 pCnt 𝑁) ∈ ℕ0) | ||
| Theorem | pccld 16759 | Closure of the prime power function. (Contributed by Mario Carneiro, 29-May-2016.) |
| ⊢ (𝜑 → 𝑃 ∈ ℙ) & ⊢ (𝜑 → 𝑁 ∈ ℕ) ⇒ ⊢ (𝜑 → (𝑃 pCnt 𝑁) ∈ ℕ0) | ||
| Theorem | pcmul 16760 | Multiplication property of the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.) |
| ⊢ ((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℤ ∧ 𝐴 ≠ 0) ∧ (𝐵 ∈ ℤ ∧ 𝐵 ≠ 0)) → (𝑃 pCnt (𝐴 · 𝐵)) = ((𝑃 pCnt 𝐴) + (𝑃 pCnt 𝐵))) | ||
| Theorem | pcdiv 16761 | Division property of the prime power function. (Contributed by Mario Carneiro, 1-Mar-2014.) |
| ⊢ ((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℤ ∧ 𝐴 ≠ 0) ∧ 𝐵 ∈ ℕ) → (𝑃 pCnt (𝐴 / 𝐵)) = ((𝑃 pCnt 𝐴) − (𝑃 pCnt 𝐵))) | ||
| Theorem | pcqmul 16762 | Multiplication property of the prime power function. (Contributed by Mario Carneiro, 9-Sep-2014.) |
| ⊢ ((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℚ ∧ 𝐴 ≠ 0) ∧ (𝐵 ∈ ℚ ∧ 𝐵 ≠ 0)) → (𝑃 pCnt (𝐴 · 𝐵)) = ((𝑃 pCnt 𝐴) + (𝑃 pCnt 𝐵))) | ||
| Theorem | pc0 16763 | The value of the prime power function at zero. (Contributed by Mario Carneiro, 3-Oct-2014.) |
| ⊢ (𝑃 ∈ ℙ → (𝑃 pCnt 0) = +∞) | ||
| Theorem | pc1 16764 | Value of the prime count function at 1. (Contributed by Mario Carneiro, 23-Feb-2014.) |
| ⊢ (𝑃 ∈ ℙ → (𝑃 pCnt 1) = 0) | ||
| Theorem | pcqcl 16765 | Closure of the general prime count function. (Contributed by Mario Carneiro, 23-Feb-2014.) |
| ⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℚ ∧ 𝑁 ≠ 0)) → (𝑃 pCnt 𝑁) ∈ ℤ) | ||
| Theorem | pcqdiv 16766 | Division property of the prime power function. (Contributed by Mario Carneiro, 10-Aug-2015.) |
| ⊢ ((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℚ ∧ 𝐴 ≠ 0) ∧ (𝐵 ∈ ℚ ∧ 𝐵 ≠ 0)) → (𝑃 pCnt (𝐴 / 𝐵)) = ((𝑃 pCnt 𝐴) − (𝑃 pCnt 𝐵))) | ||
| Theorem | pcrec 16767 | 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)) | ||
| Theorem | pcexp 16768 | Prime power of an exponential. (Contributed by Mario Carneiro, 10-Aug-2015.) |
| ⊢ ((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℚ ∧ 𝐴 ≠ 0) ∧ 𝑁 ∈ ℤ) → (𝑃 pCnt (𝐴↑𝑁)) = (𝑁 · (𝑃 pCnt 𝐴))) | ||
| Theorem | pcxnn0cl 16769 | Extended nonnegative integer closure of the general prime count function. (Contributed by Jim Kingdon, 13-Oct-2024.) |
| ⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℤ) → (𝑃 pCnt 𝑁) ∈ ℕ0*) | ||
| Theorem | pcxcl 16770 | Extended real closure of the general prime count function. (Contributed by Mario Carneiro, 3-Oct-2014.) |
| ⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℚ) → (𝑃 pCnt 𝑁) ∈ ℝ*) | ||
| Theorem | pcge0 16771 | The prime count of an integer is greater than or equal to zero. (Contributed by Mario Carneiro, 3-Oct-2014.) |
| ⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℤ) → 0 ≤ (𝑃 pCnt 𝑁)) | ||
| Theorem | pczdvds 16772 | 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) | ||
| Theorem | pcdvds 16773 | Defining property of the prime count function. (Contributed by Mario Carneiro, 23-Feb-2014.) |
| ⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → (𝑃↑(𝑃 pCnt 𝑁)) ∥ 𝑁) | ||
| Theorem | pczndvds 16774 | 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) | ||
| Theorem | pcndvds 16775 | Defining property of the prime count function. (Contributed by Mario Carneiro, 23-Feb-2014.) |
| ⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → ¬ (𝑃↑((𝑃 pCnt 𝑁) + 1)) ∥ 𝑁) | ||
| Theorem | pczndvds2 16776 | 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)))) | ||
| Theorem | pcndvds2 16777 | 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 16778 | 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 16779 | 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 16780 | 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 16781 | 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 16782 | 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 16783 | 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 16784 | 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 16785 | 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 16786 | 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 16787 | 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 16788* | 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 16789* | 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 16790* | 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 16791* | 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 16792* | 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 16793* | 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 16794* | 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 16795* | 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 16796 | 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 16797 | Lemma for pcadd 16798. 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 16798 | 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 16799 | The inequality of pcadd 16798 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 16800 | 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)) | ||
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