A Hard Diophantine Challenge | 99% Failed to Solve! | Algebra

A Hard Diophantine Challenge | 99% Failed to Solve! | Algebra
Welcome to another exciting algebra challenge! In this video, we tackle a nice & hard Diophantine equation that's sure to test our problem-solving skills. Diophantine equations are a classic type of problem that require finding positive 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 positive integers
2. Analyzing the unique properties of the equation
3. Step-by-step approach to solving the Diophantine equation
4. Tips and tricks for handling tough 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 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! 🛎️🔔
Additional Resources:
• Solving an Intriguing ...
• Diophantine Dilemma: S...
• Diophantine Delights: ...
• Cracking the Diophanti...
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Thanks for Watching!

[@roberttelarket4934]

Very nice.

[@johnlv12]

Why does (sqrt(x)-4) (sqrt(y)-4)=6 and x, y positive integers imply that the factors on the left hand side both need to be integers? I can accept it is true in this case, but it is not always true for all functions f and g that f(x) g(y) equals an integer and x and y are integers implies f(x) and f(y) are integers, right? Is there some theorem that applies for this specific scenario where f(x)=g(x)=sqrt(x)-4?

[@gregevgeni1864]

Obviously x, y, z non negative integers .
Let √x = t, √y = s, √z = r (t, s ,r ≥ 0)
The given system of equations is written { t + s - r = 4 and t^2 + s^2 - r^2 = 4 } { r = t + s - 4 and
t^2 + s^2 - ( t + s - 4)^2 = 4 }
{ r = t + s - 4 and ts - 4t - 4s + 10 =0 } { r = t + s - 4 and
(s - 4)(t - 4) = 6 } .
6 = 1•6 = 2•3 = (-1)(-6) =(-2)(-3) . .
So s - 4 =1 and t - 4 = 6 =>
s = 5 and t = 10 and r = 11
Hence x = 25, y = 100 and z = 121 ..
Similarly for the other cases.
And x = 36, y = 49, z = 81 is a solution ..

[@moeberry8226]

You forgot another pair of numbers -2 and -3 are another set of factors of 6. Which will give x=4, y=1 and z=-1. But since they have to be positive Zaigen numbers then this isn’t an ordered triple as a solution. But we still had to check.

[@9허공]

also -1 x -6

[@moeberry8226]

@@9허공 you cannot have those factors seeing that straight away since sqrt(x)-4=-6 that would imply sqrt(x)=-2 and that cannot happen.

[@9허공]

@@moeberry8226 i know. but he had to comment this.

[@moeberry8226]

@@9허공 your 100 percent right that’s true.

[@RealQinnMalloryu4]

(4)+(4)=8 ➖ (4)=4 2^2 (xyz ➖ 4xyz+4) (4)+(4)=8 ➖ (4)=4 (xyz ➖ 4xyz+4).

[@RyanLewis-Johnson-wq6xs]

Sqrt[x]+Sqrt[y]-Sqrt[z]=4 x+y-z=4 (x,y,z)=(25,100,121),(100,25,121),(36,49,81),(49,36,81) It’s in my head.