Cracking a Diophantine Challenge from Olympiads!

Cracking a Diophantine Challenge from Olympiads!
Dive into the world of high-level mathematics with this exciting Diophantine challenge from Olympiads! 🧠✨ In this video, we'll tackle a complex Diophantine equation step-by-step, exploring the strategies and techniques used to solve it. Perfect for math enthusiasts, students preparing for competitions, or anyone looking to challenge their problem-solving skills. Join us and see if you can crack this Olympiad-level puzzle! Don't forget to like, comment, and subscribe for more math challenges and tutorials. 📏🏆
Topics Covered:
1. Understanding the basics of Diophantine equation in integers
2. Analyzing the unique properties of the given equation.
3. Step-by-step approach to solving the radical equation for integers.
4. Tips and tricks for handling tricky equation like a pro.
5. Algebraic identities and manipulations while solving equations.
#numbertheory #matholympiad #mathchallenge #diophantineequation #mathtutorial #mathpuzzles #education #stem #mathhelp #algebra
Timestamps:
0:00 Introduction
1:22 Algebraic manipulations
2:20 Algebraic identities
3:57 Prime factors
6:10 Solving system of equations
9:38 Integer ordered pairs
9:50 Verification
🎯 This video is perfect for students, math enthusiasts, or anyone seeking to sharpen their problem-solving skills and gain confidence in dealing with Diophantine 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

[@woobjun2582]

Multiplying 4 and adding 5 in the process was tricky but wonderful, sir ^.^

[@mohammedsaysrashid3587]

Wonderful introduction.....thanks for sharing...( x,y )= (0,+1) ( 0,-1) ...

[@user-ji5su2uq9m]

case a and d can be immediatedly rejected because 2x^3+2y^2+3 >= 2x^3-2y^2+3

[@user-ji5su2uq9m]

(x, y) = (0, ±1)

[@user-ji5su2uq9m]

another approach
let a = x^3 +1, y^4 + 1 = a(a+1)
=> a^2 + a - (y^4 +1) = 0
discriminant D = 4y^4 + 5 = k^2 for some positive integer k
4y^4 - k^2 = (2y^2 + k) * (2y^2 - k) = -5
1st term is positive, only (5, -1) and (1,-5) are possible cases.
case1 (2y^2 + k) = 5, (2y^2 - k) = -1
=> y = ±1, k = 3 => a = (-1 ± k)/2 = 1 or -2
a = x^3 + 1 = 1 or -2 => x = 0 is only integer solution.
(x, y) = (0, ±1)
case2 (2y^2 + k) = 1, (2y^2 - k) = -5
=> y^2 = -1 => no integer(real) solution.

[@Maiyut27tgb]

Thanks bro