A Nice Olympiad Problem | Diophantine Equations | Integers

A Nice Olympiad Problem | Diophantine Equations | Integers
Welcome to another exciting Math Olympiad problem! 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
#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!

[@michaeldoerr5810]

At the 3:20 mark, is this straregy of adding 28 a common strategy??? I want to know because part of me thinks that this is a version of completing the square. Does that sound right???

[@paulortega5317]

A different way to get to the equations at 4:33
(z+8)^2 = (x+y)^2 = x^2+2xy+y^2= x^2+y^2+2xy = z^2+8 + 2xy
2xy = (z+8)^2 - (z^2+8) = 16z+56
1. xy = 8z+28
2. 8(x+y) = 8x+8y= 8(z+8) = 8z+64
Equation (1) - equation (2): xy-8x-8y = (8z+28)-(8z+64) = -36
-8x+xy-8y+64 = -36+64
(x-8)(y-8) = 28

[@taniacsibi6879]

In al doilea set de soluții a-ați schimbat doar una. ( 36,9,37) iar celelalte nu rezultând astfel doar 7 in loc de 12,in rest fffff ok

[@RealQinnMalloryu4]

{x+x ➖ }=x^2 {y+y ➖}=y^2 {z+z ➖}=z^2 {x^2+y^2+z^2}=xyz^6+{8+8 ➖ }=16{xyz^6+16}=,16xyz^6,4^4xyz^3^2,2^2^2^2xyx^1^2 ,1^1^1^1xyxz^1^2(xyz ➖ 2xyz+1). {x^2+x^2 ➖ }=x^4 {y^2+y^2 ➖}=y^4 {z^2+z^2 ➖ }=z^4 {x^4+y^4+z^4}=xyz^12 +{8+8 ➖ }=16 {xyz^12+16}=16xyz^12 4^4xyz^3^4 2^2^2^2xyz^3^2^2.1^1^1^1xyz^3^1^2 xyz^3^2 (xyz ➖ 3xyz+2).