Beyond Linear: Tackling Nonlinear Diophantine Equations

Step beyond the realm of linear equations and embrace the challenge of nonlinear Diophantine equations. In this thought-provoking video, we explore the fascinating world of nonlinear equations and their unique properties. Journey with us as we tackle these intriguing puzzles, using advanced techniques and methods tailored specifically for nonlinear Diophantine equations. Prepare to push the boundaries of mathematical exploration and conquer the realm of nonlinear Diophantine equations. Watch now!

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[kodychoi]

Using AM-GM on x^3, y^3 and 1, we have x^3 + y^3 + 1 >= 3xy, and equality holds if and only if x = y = 1

[schukark]

I am a little bit late to the video, but i want to propose another solution:

Consider this equation mod 3.
if x -> 0, then x^3 -> 0
if x ->1 then x^3 -> 1
if x->2 then x^3 ->1

So cubes can only be 0 or 1 mod 3.
Then looking at our equation, right side is always 0 mod 3, so x^3 + y^3 needs to be 2 mod 3.
The only possibility is x mod 3 = 1, y mod 3 = 1.

x = 3k+1, y = 3p+1
After substittuion and cancellation we get
k^3+k^2+p^3+p^2=kp

Wlog k>=p

Case 1: p>= 1 then k^3>=k^2, p^3>=p^2
k^3+k^2+p^3+p^2>=2(k^2+p^2)>k^2>=kp, so there are no solutions

Case 2: k>= 1, but p is <1.
As y = 3p+1 is positive integer, then p <1 can only be 0 => y = 1 =>
x^3+1+1=3 => x^3 = 1 => x = 1

Case 3: 1>= k>= p
As mentioned above x and y > 0 so k = p = 0 => x = y = 1 (already seen)

That leaves us with the answer x = y = 1

[myththelegendtyson]

Hi man I really like this solution although I did find it quite difficult to follow with those confusing variable names, a lot of the time I couldn’t tell what was what and they looked just like squiggles

[moeberry]

Good video bro. Keep it up. I want a degree 10 polynomial involving lots of algebra and substitution. And I want it to have 8 non real roots and 2 distinct real roots.