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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!
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I was doing a small bit of research on a number sequence that popped-out at me in the course of my work. Specifically, a modified (x, y) graph (grid). That research is a bit complex to explain, but, i
Mathematician Presh Talwalkar tackles some tricky math problems from around the world, particularly in Asia. Math problems include this: https://m.youtube.com/watch?v=xnE_sO7PbBs