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RYD date created : 2025-02-06T18:13:59.638697Z
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Top Comments of this video!! :3
These comes from 2 well-known facts from number theory.
1. Every prime >3 is 1 apart from a multiple of 6.
2. In base 10 the sum of the digits has the same remainder modulo 9 as the number itself.
Applying (1) we get the the two primes are necessarily (6k-1) and (6k+1), so their product is (6k-1)•(6k+1) = 36•k^2-1.
Now lets apply (2) and reduce it modulo 9.
36•k^2 - 1 = 9(4k^2) - 1 = -1 = 8 (mod 9)
Here you go.
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When you have a pair (a, b) that is 2 apart (in other words, b = a+2) and both are greater than 3 , then either
- one of (a,b) is a positive multiple of 3 (but not 3 itself) and hence is not prime; or
- the pair (a,b) is congruent to one of the three following cases:
(a,b) = (2, 4) mod 9 OR
(a,b) = (5, 7) mod 9 = (-4, -2) mod 9 OR
(a,b) = (8, 1) mod 9 ,
and hence their product a*b = 8 mod 9 .
In other words, the property also applies to "twins" that aren't both primes and of which neither is a multiple of 3 ; for example:
8 * 10 = 80 ==> sum of digits: 8+0 = 8
23 * 25 = 575 ==> sum of digits: 5+7+5 = 17 , 1+7 = 8
119 *121 = 14399 ==> sum of digits: 1+4+3+9+9 = 26 , 2+6 = 8
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I have also noticed that the product of two twin primes is always one less than a perfect square, which is proved as follows:
the product of 2 twin primes
p(p+2)
where p+2 is also prime is p² + 2p, which is one less than p² + 2p + 1, or (p+1)², so p(p+2) = (p+1)² -1,
or, the product of 2 twin primes is always 1 less than a perfect square.
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@isaiahgreenlee9273
6 months ago
For the curious:
This happens because in base 10, adding the digits together repeatedly in this way reduces a number modulo 9 (gives the remainder when divided by 9)
Now, call the primes n+1 and n-1, where n is the number between them. Multiplying them together yields:
(n+1)(n-1) = n^2 - 1
For any three numbers n-1, n, and n+1, one of the three must be divisible by 3, and since n+1 and n-1 are prime, n must be the multiple of 3 (unless one of the twin primes is 3, in which case this property doesn't hold), which means n^2 is a multiple of 9. So subtracting 1 from n^2 always gives a result that is 1 less than a multiple of 9, or 8 more than a multiple of nine. This is where the 8 comes from when reducing mod 9 by repeatedly adding digits.
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