A Nice Math Olympiad Problem | Diophantine Equation

A Nice Math Olympiad Problem | Diophantine Equation
Welcome to another exciting Math Olympiad challenge! In this video, we tackle a nice Diophantine equation that's sure to test problem-solving skills. Diophantine equations are a classic type of problem that require finding integer solutions, and they're a staple in math competitions.
Join us as we break down the problem, explore different strategies, and find the solution step-by-step. This problem is perfect for anyone preparing for a Math Olympiad or just looking to improve their mathematical reasoning.
Difficulty Level: Intermediate to Advanced
Topics Covered:
1. Understanding the basics of Diophantine equation in integers
2. Analyzing the unique properties and substitution of the given equation
3. Step-by-step approach to solving the Diophantine equation for integer
4. Tips and tricks for handling tricky equation like a pro
5. Algebraic identities and manipulations while solving equations
Time-stamps:
0:00 Introduction
1:20 Substitution
3:52 Algebraic manipulations
4:25 Factors
7:12 Solving system of equations
13:30 Integer ordered pairs
#mathematics #diophantineequations #integers #problemsolving #algebra #education #numbertheory #matholympiad #matholympics
🎯 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! 🛎️🔔
Additional Resources:
• Solving an Intriguing ...
• Diophantine Dilemma: S...
• Diophantine Delights: ...
• Cracking the Diophanti...
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Thanks for Watching!

[@ronbannon]

Thanks for sharing. Great problem.

[@mohammedsaysrashid3587]

It was a wonderful introduction thank you Sir 🙏 for sharing

[@gnanadesikansenthilnathan6750]

Understood this problem

[@gregevgeni1864]

Let (1) the initial equation.
(1) (x+y)^2 -2xy -xy(x+y)+1 =0 (2).
x+y=s , xy=p with s, p integers as sum and product of integers x, y .
(2) : s^2 -2p -ps+1 =0 (3) .
The discriminant if (3) it is necessary a perfect square because the (3) have integers roots.
Thus (-p)^2 - 4(-2p+1) = k^2
(p+4)^2 - k^2 =20
(p+4+k)(p+4-k) = 20
Factors of 20 : 1, 2, 4, 5, 10, 20 and
-1, -2, -4, -5, -10, -20 ..
We solve some systems,
p+4+k=1 and p+4-k=20 etc .. 😂
Accepted only integer solutions. .
....
Another most sophisticated solution.
(1) x^2 •(y-1) +y^2 •(x-1) =1 (2) .
Let x=r+1 and y=t+1 and
(2) : (r+1)^2 •t + (t+1)^2 •r =1 after some algebra . .
(rt+1)(r+t+4) =5 and same method as higher up.
Solutions:
(1, 2), (-5, 2), (2, 1), (2, -5) (symmetry).

[@kassuskassus6263]

(x,y)=(1,2); (2,1)

[@RajeshKumar-wu7ox]

X=y, x^3 (x-1)=1

[@user-kp2rd5qv8g]

(x,y) = (2,1), (1,2).

[@Chacal0152]

Dos casos más -1×-5 =- 5 × -1= 5

[@RealQinnMalloryu4]

xyxy4/xyxyxy^6=xyxyxyxyxy1.2 (xyxyxyxyxy ➖ 2xyxyxyxyxy+1)