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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 (461)

  1. Stop me if yo already heard this one. HalfSicilian

    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

  2. Math Problem Quantum_Physics

    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

  3. 4th Dimension Existence MathematicChess

    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.?

  4. Hard Math henryliu2009

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

  5. 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

  6. 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

  7. 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

  8. Hard Math BOT centaur01

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

  9. 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-

  10. 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!

That's Mathematics!!!! Forum »