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That's Mathematics!!!! TEAM

This team is for people who love math, math jokes, math concepts and applications, and cool stuff like that.
Location: within the R of any Nth dimentional space

Forum (456)

  1. Hard Math henryliu2009

    i don't know. I'm only 9 years old.

  2. Hard Math BOT centaur01

    @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

  3. Hard Math BOT centaur01

    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

  4. Hard Math MoistvonLipwig

    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

  5. Hard Math BOT centaur01

    The same thing for any numbers p and q that are no divisible by 2 or 3 (not necessarily primes).

  6. Hard Math BOT centaur01

    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-

  7. Hard Math mathpluschess

    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!

  8. I have youtube channel called antichess nikhil2609

    Please everyone subscribe it

  9. The Antelope Graph (Modified Knight) MrCharles

    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

  10. MindYourDecisions YT channel Classicler

    Mathematician Presh Talwalkar tackles some tricky math problems from around the world, particularly in Asia. Math problems include this:

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