Squaring Primes - Numberphile
Why Mathematics Stack Exchange recommends it
Mathematics Stack Exchange users link it as the standard explanation of why p^2 is congruent to 1 mod 24 for primes above 3, and cite it when answering questions about primes and modular arithmetic.
Recommended by 1 people across 2 different months, 2019 to 2024.
What it is
A Numberphile video in which Matt Parker shows that the square of any prime greater than 3, minus one, is a multiple of 24. He walks through the pattern and explains why it holds, with a follow-up look at related ideas.
What people said
Recommends m-stgt · 2024-02 · Mathematics Stack Exchange · on “Mirror symmetry in distances of remaining numbers of Eratosthenes-sieve”
@berthold -- Maybe not exactly the links you're asking for, as starting point about 'wheels' [LINK] , and a special property of primes > 3 [this video]
Recommends m-stgt · 2024-09 · Mathematics Stack Exchange · on “"Non-random" non-trivial structure in the prime numbers”
@user25406 -- No need for primes, also odd number which are not multiple of $3$ show this characteristic. E.g. $35^2=1225=51*24+1$. There's a clip with two explanations [this video] , I link to the end of the 1st to save time and -- one explanation is sufficient me think.
Cites H. Halv · 2019-01 · Mathematics Stack Exchange · on “Why are all primes squared less one a multiple of 24, but not also a multiple of 12?”
So I can't wrap my head around this problem from numberphile ( [this video] ). Why does this work for multiples of 24 but not 12?
Cites user645636 · 2019-06 · Mathematics Stack Exchange · on “$p^2 - 2 q^2 = 5039$ for primes $p, q$”
$$p^2-2q^2=24k+23$$ Things to note are: mod 9 we have squares 0,1,4,7 the first of which can't happen with primes. $24k+23\equiv 6k+5\bmod 9$ $p^2\equiv q^2\bmod 9\implies -(p^2)\equiv 6k+5\bmod 9$ $p^2\equiv q^2+3\bmod 9 \implies -(q^2)+3\equiv 6k+5\bmod 9\implies -(q^2)\equiv 6k+2\bmod 9$ $p^2\equiv q^2-3\bmod 9 \implies -(q^2)-3\equiv 6k+5\bmod 9\implies -(q^2)\equiv 6(k+1)+2\bmod 9$ $p^2\equiv q^2+6\bmod 9 \implies -(q^2)+6\equiv 6k+5\bmod 9\implies -(q^2)\equiv 6(k-1)+5\bmod 9$ $p^2\equiv q^2-6\bmod 9 \implies -(q^2)-6\equiv 6k+5\bmod 9\implies -(q^2)\equiv 6(k+1)+5\bmod 9$ using the fact that primes>3 squared are 1 mod 24 [this video] we get that the negatives are 5 mod 6. which then determ
Critical emacs drives me nuts · 2022-03 · Mathematics Stack Exchange · on “Does this alternative way of calculating Twin Primes help to prove that there are an infinite number of Twin Primes?”
That result is trivial: any pair of primes other than $(3,5)$ has the form $(3n-1,3n+1)$ because none of the primes is a multiple of 3. Hence the product of the primes is $$(3n-1)(3n+1)=9n^2-1\equiv-1\equiv8\mod9.$$ As an aside, $n$ must also be even because neither of the primes is even, thus the prime pair can be represented as $(6m-1,6m+1)$ . So we have the stronger result $$(6m-1)(6m+1) = 36m^2-1 \equiv -1 \mod 36.$$ The fact that a natural number has the same remainder mod 9 like the sum of its digits follows from $10\equiv 1 \mod 9$ : $$n =\sum a_k10^k \equiv \sum a_k1^k \equiv \sum a_k = \operatorname{sum\_of\_digits}(n) \mod 9$$ where the $a_k$ are the decimal figures of $n$ . As far