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This team is for people who love math, math jokes, math concepts and applications, and cool stuff like that.
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A mathematician went into a bar and ordered a beer . After that, another mathematician walks in and orders one half a beer. Mathematicians continue entering and ordering one half a beer. T
The 30 edges of an icosahedron are distinguished by labeling them 1, 2, . . . , 30. How many different ways are there to paint each edge red, white, or blue such that each of the 20 triangular faces o
Is it possible that the 4th Dimension does exist, but can not be calculated by human brains to become visible? Similarly to how humans cannot visibly see infrared, ultraviolet, microwave, etc.?
i don't know. I'm only 9 years old.
@MoistvonLipwig Or yes, just look at all possible cases in mod 24. :) I like also this variation: Not divisible by 2 and by 3 means p = 6p' +/- 1 and q = 6q' +/- 1. Then p^2 - q^2 = 36p'^2 + 36
Doh, i'm dumb. It can be done in more simple way. a) For any prime p, p ≡ 1 or p ≡ 3 or p ≡ 5 or p ≡ 7 (mod 8), therefore p^2 ≡ 1 (mod 8). b) The same thing for p^2 ≡ 1 (mod 3), as previously. c
This proof isn't exactly pretty but it works so whatever. :D Clearly every prime number greater than 3 is either +1 or -1 mod 6. So the possible mod 24 values are +- (1, 5, 7, 11). Then simply square
The same thing for any numbers p and q that are no divisible by 2 or 3 (not necessarily primes).
Let's p = 2*p' + 1 and q = 2*q' + 1 (as they both are odd) Then p^2 - q^2 = (p+q)(p-q) = (2*p'+2*q'+2)*(2*p'-2*q') = =4*(p'+q'+1)*(p'-q'). Either p'-q' is even, or p'+q'+1 is even - therefore p^2-
Here's a problem: Let p and q be primes greater than 3. Prove that p^2 - q^2 is divisible by 24. good luck!