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The result must be divisible by 3. Using the example above: 16,499,205,854,376 has four of the digits 1, 4 and 7 and four of the digits 2, 5 and 8; Since 4 − 4 = 0 is a multiple of 3, the number 16,499,205,854,376 is divisible by 3. Subtracting 2 times the last digit from the rest gives a multiple of 3. (Works because 21 is divisible by 3)
The factors 2e, (e – 3f), and (e + 3f) are coprime since 3 cannot divide e: if e were divisible by 3, then 3 would divide u, violating the designation of u and v as coprime. Since the three factors on the right-hand side are coprime, they must individually equal cubes of smaller integers −2e = k 3 e − 3f = l 3 e + 3f = m 3
This proof, [3] [6] discovered by James Ivory [7] and rediscovered by Dirichlet, [8] requires some background in modular arithmetic. Let us assume that a is positive and not divisible by p . The idea is that if we write down the sequence of numbers
If a is not divisible by p, that is, if a is coprime to p, then Fermat's little theorem is equivalent to the statement that a p − 1 − 1 is an integer multiple of p, or in symbols: [1] [2] (). For example, if a = 2 and p = 7, then 2 6 = 64, and 64 − 1 = 63 = 7 × 9 is a multiple of 7.
The case p = 3 was first stated by Abu-Mahmud Khojandi (10th century), but his attempted proof of the theorem was incorrect. [73] [74] In 1770, Leonhard Euler gave a proof of p = 3, [75] but his proof by infinite descent [76] contained a major gap.
Several variations on Euclid's proof exist, including the following: The factorial n! of a positive integer n is divisible by every integer from 2 to n, as it is the product of all of them. Hence, n! + 1 is not divisible by any of the integers from 2 to n, inclusive (it gives a remainder of 1
Exactly one of a, b is divisible by 2 (is even), and the hypotenuse c is always odd. [13] Exactly one of a, b is divisible by 3, but never c. [14] [8]: 23–25 Exactly one of a, b is divisible by 4, [8] but never c (because c is never even). Exactly one of a, b, c is divisible by 5. [8] The largest number that always divides abc is 60. [15]
For example, if p = 19, a = 133, b = 143, then ab = 133 × 143 = 19019, and since this is divisible by 19, the lemma implies that one or both of 133 or 143 must be as well. In fact, 133 = 19 × 7 . The lemma first appeared in Euclid 's Elements , and is a fundamental result in elementary number theory.