The British Mathematical Olympiad Diophantine Equation

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  • Опубліковано 18 чер 2024
  • The British Mathematical Olympiad Diophantine Equation
    Join us as we delve into the intricate world of Diophantine equations with a challenge straight from the British Mathematical Olympiad! In this video, we explore the elegance and complexity of solving Diophantine equations, a cornerstone of algebraic problem-solving. Get ready to sharpen your mathematical skills and embark on a journey of discovery with this captivating BMO challenge!
    Topics Covered:
    1. Understanding the basics of radical Diophantine equation in integers
    2. Analyzing the unique properties and substitution of the given equation.
    3. Step-by-step approach to solving the BMO Challenge for positive integers.
    4. Tips and tricks for handling three variable Diophantine equation like a pro.
    5. Algebraic identities and manipulations while solving equations.
    Timestamps:
    0:00 Introduction
    1:14 Algebraic identities
    3:41 Algebraic manipulations
    4:45 Prime factorization
    8:45 Positive integers
    11:25 All positive triples
    #matholympiad #BMO #diophantineequations #numbertheory #mathchallenge #algebra #problemsolving #mathematics #mathematicscompetition #mathematicseducation #maths
    🎯 This video is perfect for students, math enthusiasts, or anyone seeking to sharpen their problem-solving skills and gain confidence in dealing with radical equations. 🎓📈
    🔔 Challenge yourself and see if you can solve the equation before we do! Hit the like button if you're up for the challenge and remember to subscribe for more exhilarating math content! 🛎️🔔
    Don't forget to like, comment, and subscribe to join our math-loving community. Let's get started on this exciting journey together! 🤝🌟
    Thanks for Watching!
    @infyGyan

КОМЕНТАРІ • 6

  • @mohammedsaysrashid3587
    @mohammedsaysrashid3587 Місяць тому

    A wonderful explanation...smartness think 🤔..thanks

  • @user-ji5su2uq9m
    @user-ji5su2uq9m Місяць тому +1

    from equation 1, z = x + y - 12, substituting this to equation 2
    x^2 + y^2 - (x + y - 12)^2 = -2xy + 24x + 24y - 144 = 12
    which gives xy -12x -12y + 78 =0 and (x - 12) * (y - 12) = 66

  • @woobjun2582
    @woobjun2582 Місяць тому

    Wow! who would think of +66 to both sides in the middle of process. Good Job! ^.^

  • @SrisailamNavuluri
    @SrisailamNavuluri Місяць тому +1

    I tried sir.I did not get any idea.Thank you.