When is p^2-p+1 a Cube? | Balkan MO 2005

Поділитися
Вставка
  • Опубліковано 24 гру 2024

КОМЕНТАРІ • 23

  • @cassidoodledookevorkian274
    @cassidoodledookevorkian274 2 роки тому +6

    Always the best math vids on UA-cam 👍

  • @richardfredlund8846
    @richardfredlund8846 Рік тому +1

    Amazing .. I love the way the logic flows.

  • @victorchoripapa2232
    @victorchoripapa2232 2 роки тому +2

    Wowow... Nice exercise

  • @Manisphesto
    @Manisphesto 2 роки тому +5

    Why do questions like the shape cube instead of x³?

  • @mathcanbeeasy
    @mathcanbeeasy 2 роки тому +3

    This is art. 🥰

  • @drynshock1
    @drynshock1 6 місяців тому

    How would you choose the squares if you got something like m^4 + 3m^3 + ... with the cubic term

  • @bowlteajuicesandlemon
    @bowlteajuicesandlemon Рік тому

    How do we know that if p divides (x²+x+1), then a factor of p-1 doesn't also divide (x²+x+1) which would mean (x-1) doesn't have to be divisible by p-1?

    • @saadhaider9576
      @saadhaider9576 11 місяців тому

      Because p-1 = (x^2+x+1) * ((x-1)/p)
      Since (x-1)/p is an integer, x^2+x+1 divides p-1

  • @JohnWick-hj4ep
    @JohnWick-hj4ep 2 роки тому +4

    Amazing! However i have a question if anyone could answer. I was able to follow most of it, but didn't understand how he chose the 2 close perfect squares, around 4:50 - 6:27.
    P.s tips about how to improve would also help :D

    • @Leardakhan3
      @Leardakhan3 2 роки тому +2

      I think the strategy is to look at the higher order terms, since there is a^4 - 6a^2, he looked for (a^2 - 3)^2, then you can look for which values of a the linear term and the constant are higher or lower than each other, that is my take at least.

    • @bait6652
      @bait6652 2 роки тому

      Radic has to be a square dictated by the 2 Pn terms of highest power.
      Its a bounding trick he often uses when Pn w/o sq repr. has to be square. Bound it to 2-3 values or soln doesnt exist. Look for his n^4 problems ...theres like 3 of them

    • @bosorot
      @bosorot 2 роки тому +1

      To make you feel better this is standard olympic math technique. No normal kids can think how to do this on the spot for the first time. They learnt and practice and practice ... . If you can do this technique invented on your own , then you are a genius .

  • @jpolowin0
    @jpolowin0 2 роки тому +2

    I suppose that 1 isn't counted as a prime. :-)

    • @alessiodaniotti264
      @alessiodaniotti264 2 роки тому +4

      the definition of prime is based on non trivial divisibility/factorization, so unitary elements are never primes.

  • @johns.8246
    @johns.8246 2 роки тому +1

    So how about when is p^2-p-1 a cube? There's at least one obvious solution.

  • @cassidoodledookevorkian274
    @cassidoodledookevorkian274 2 роки тому +1

    🔥🔥

  • @harris5140
    @harris5140 2 роки тому +1

    p=1 doesn't work? why?

    • @quanquin3822
      @quanquin3822 2 роки тому +2

      1 is not prime.

    • @10names55
      @10names55 2 роки тому +1

      You ment why not x=1
      Well p^2-p is always >0

  • @lo-fi_community
    @lo-fi_community 2 роки тому +1

  • @piman9280
    @piman9280 2 роки тому +1

    It's a short exercise to see that p = 19 produces 343, which is 7^3. Once again we have a problem whose only solution can be found by very simple mathematics. I still enjoyed the robust solution.

  • @와우-m1y
    @와우-m1y 2 роки тому +1

    asnwer=1 os isit 🤣🤣🤣🤣

  • @DontStudy
    @DontStudy 2 роки тому

    N9