Solution 62: Cleverly Manipulating a Diophantine Equation with x^2018

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  • Published on Oct 18, 2018
  • Let's prove that there is a unique solution to an initially complicated-looking Diophantine equation.
    Congratulations to Gabriel N., Allaizn, Theodore Leebrant, Bruno Alejandro Andrades, Nicholas Patel, Beshoy Nabil, Jaleb, adandap, Luis Alba Sarria, and Alex Edwards for successfully solving this math challenge question! Gabriel N. was the first person to solve the question.
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Comments • 28

  • Rakesh Kumar
    Rakesh Kumar 2 months ago

    Just wowwww....!!!!

  • 라테우스
    라테우스 4 months ago

    너무 대단....

  • Amit Kumar
    Amit Kumar 6 months ago

    Respected sir,plz solve this question-find the value of sine^-1(sine20),where 20 is in radians.Answer is pi-20.

  • GetReadyNow
    GetReadyNow 6 months ago

    We can notice that as x-1 divides x^2018 +1 and x-1 also divides x^2018-1 (sum of geometric sequence) x-1 is 1 or 2. But since x is odd then x-1 =2 , x=3

  • Ian Brown
    Ian Brown 6 months ago

    In what class is modular arithmetic covered?

  • SUPRATIM SANTRA
    SUPRATIM SANTRA 8 months ago

    Sir good problem with excellent approach... Please continue ur uploading

  • Felipe Lorenzzon
    Felipe Lorenzzon 10 months ago +1

    I imagine how incredible would be a number theory question with integer imaginary numbers. I think it is yet a big field to study

  • E Rock
    E Rock 11 months ago

    Modular arithmetic is a usual suspect for many Diophantine equations but I wouldn’t have thought to try polynomial division. Very slick!

  • Shivam Malluri
    Shivam Malluri 11 months ago

    I have an integral question for you
    Integrate the function :
    (1-x²)/(x⁴+3x²+1)
    It is an indefinite integral

  • Luis Ernesto Morales Cordova

    u make me feel stupid

  • Jezus Jr
    Jezus Jr 11 months ago +1

    I don't know what accent this is. Where are you from?

    • Yikes 888
      Yikes 888 7 months ago +1

      pure asian english accent

    • Yikes 888
      Yikes 888 7 months ago

      @IceSCream4u stfu lol

    • IceSCream4u
      IceSCream4u 11 months ago

      Pretty sure it is an italian accent

    • Victor P.
      Victor P. 11 months ago

      Sounds a bit east asian because he slurs his R, but idk this is the first time I've watch his videos.

  • Just in Y.
    Just in Y. 11 months ago +5

    I studied modular Arithmetic last year out of curiosity. Thanks for revealing its importance.

  • Shenghui Yang
    Shenghui Yang 11 months ago +4

    Both solutions are elegant

  • Ben Burdick
    Ben Burdick 11 months ago +10

    I would have never thought to do polynomial division to that. A great problem though!

  • δτ
    δτ 11 months ago +10

    Nice video and interesting solutions.
    I have never heard of synthetic long division of polynomials before, and I appreciate learning something new 👍

    • RGP Maths
      RGP Maths 11 months ago

      Technically the synthetic division method is more precisely equivalent to using the Remainder Theorem to show the remainder is 2. But that would mean the remainder is both 2 and 0. It's easier to explain intelligibly using the Factor Theorem instead.

    • LetsSolveMathProblems
      LetsSolveMathProblems  11 months ago +3

      That is an excellent insight. It does make the solution almost instantaneous once we reach the part you mentioned. =)

    • RGP Maths
      RGP Maths 11 months ago +4

      It seemed to me that the synthetic division method was equivalent to, but more elaborate than, appeal to the Factor Theorem. Having established that (x - 1) divides (x^2018 + 1), and since (x - 1) also obviously divides (x^2018 - 1) by the Factor Theorem, then (x - 1) divides their difference, which is 2.
      I wish I could have found the solution myself but, when you got to that stage the next step seemed obvious as above, whereas the division method made it seem more awkward.
      Great problem though!

  • Amitayush Thakur
    Amitayush Thakur 11 months ago +6

    I wanted to know if this solution was correct. I originally posted this solution on the challenge video. The solution goes like:
    Let x = 2n + 1 and y = 2m + 1, by rearranging the original equation we can get (y^2 + 1)*(x-1) = 2(x^2018 + 1), Substituting for x and y we get the following equation ((2m + 1)^2 + 1)(2n + 1 -1) = 2 ((2n + 1)^2018 + 1), now simplifying this we get (4m^2 + 4m + 2)(n) = (2n+1)^2018 + 1. Now on RHS using binomial expansion we get RHS = (1 + C(2018,1)*(2n) + C(2018,2)*(2n)^2 + ... + (2n)^2018) + 1, this RHS can be re-written as RHS = 2 + 2nk for some nutural number k.Clearly if n > 1 then RHS gives remainder 2 when divided by 2n and if n = 1 then RHS is divisible by 2n. But we have LHS = (4m^2 + 4m + 2)(n) = (2m^2 + 2m + 1)(2n), clearly LHS is divisible by 2n. So from our conclusion about RHS we must have n = 1. If n = 1 this means x = 2n + 1 = 3, substituting the value back in the original equation we get 3y^2 + 3 = 2*3^2018 + y^2 + 3, which simplifies to y^2 = 3^2018 which means y = 3^1009 so log_x y = 1009.

    • Amitayush Thakur
      Amitayush Thakur 11 months ago

      @ripan sharma Still less than the hard work done in uploading the videos with solution. Thanks to LetsSolveMathProblems, want more diophantine equation question :D

    • ripan sharma
      ripan sharma 11 months ago

      That's a lot of hard work done in typing the entire solution

    • ripan sharma
      ripan sharma 11 months ago

      👍👍👍..I also solved it in the same way