Find a pair of distinct positive integers x and y, which satisfy a quartic equation.

The video illustrates step-by-step how to find a pair of distinct positive integers x and y, which satisfy a quartic equation. This equation can be simplified to a quadratic equation.

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[pietergeerkens]

I used the substitution y = k*x for k rational to get
k^2 + k + 1 = - x * (k^3 - 3k^2 + k - 1)
After solving for x and a little algebra this yields:
x = 1 / (1-k) + 3k / (1-k)^3.

For any k = (n-1)/n this gives solution:
x = n + 3*([n-1]/n)*n^3 = n + 3*(n-1)*n^2
y = (n-1) + 3*(n-1)^2*n.

For n = 2:
x = 2 + 3(1)(2^2) = 14; y = (1/2)14 = 7
as above.

For n = 3:
x = 3 + 3(2)(3^2) = 57; y = (2/3)57 = 38
with
19^3 * (3^3 - 2^3) = 19^3 * 19 = 19^4 = (57-38)^4
as required.

etc.

Going one step forward: For any n > 1:
r = ( 1 + 3n(n-1) );
x = n * r
y = (n-1) * r
yields a positive integer solution (x, y).

[brinzanalexandru]

I took y=x+a and after getting a quadratic in x I calculated ∆ to be -12a⁵-3a⁴ from where we get that a€(-Inf;-1/4] so a≤-0, 25 now we need to solve for all a s.t ∆=n² where n is some integer so this is a hard diophantic eqn. But if we rewrite it as 3a⁴(4a+1)=-n² we create a bunch of systems and to solve for all a we also should change the factors like a²*a²(12a+3)=-n² and by creating and solving all possible permutations and combinations we get the only a's to be (a, n)={(-7, 441);(-7, -441);(-19, 5415);(-19, -5415);(-1, 3);(-1, -3)} so basically we have all the info to solve for (x, y) and finally we get (x, y)={(0, -1);(1, 0);(-7, -14);(14, 7);(57, 38);(-38, -57)} and these are all solutions to this eqn.

[sy]

I got infinite solutions for additional problem. And I checked some on them and I found they can be solutions.
(By the way, the method that I used is different from the main problem's one.)

Set √x=X and √y=Y, then, X>Y>10 because x>y>100. The problem can be written as X^2-Y^2=(X-Y)^3.
∴ (X+Y)(X-Y) = (X-Y)*((X-Y)^2)
∴ X+Y = (X-Y)^2 （∵ X-Y is not 0, because X>Y ）
∴ X+Y = X^2-2XY+Y^2
∴ X^2-(2Y+1)X+(Y^2-Y) = 0 ・・・・(i)
Δ = (-(2Y+1))^2 - 4*1*(Y^2-Y) =4Y^2+4^Y+1 - 4Y^2 + 4Y = 8Y+1 ∴ 8Y+1=n^2 (n: natural number)
If n is odd number, n^2-1 can be divided by 8. Set n=2k+1, then Y=k(k+1)/2.　・・・・　(ii)
From (i) and (ii), X=Y+(k+1)=(k+1)(k+2)/2 [Note: The calculation is omitted.]
And k is equal to or more than 5 because Y>10.
Therefore, X=(k+1)(k+2)/2, Y=k(k+1)/2, and k >or= 5.
(From x=X^2 and y=Y^2, ) x=[(k+1)(k+2)/2]^2, y=[k(k+1)/2], k is natural number and at least 5. ・・・・ This is my answer.

[Check:] If k=5, 6, 7, 8, 9・・・ (x, y)=(441, 225), (784, 441), (1296, 784), (2025, 1296), (3025, 2025), ・・・
x-y: 441-225=216, (√x-√y)^3: It is OK.
x-y: 784-441=343, (√x-√y)^3: It is OK.
x-y: 1296-784=512, (√x-√y)^3: It is OK.
x-y: 2025-1296=729, (√x-√y)^3: It is OK.
x-y: 3025-2025=1000, (√x-√y)^3: It is OK.
・・・・・・・
[And if k=4, y=10^2=100. It is not a solution because y should be y>100.]

My method and checks are right, so I guess my theory and calculation is right.
Am I right?

[sy]

I found miss-typing for that I commented a few minutes ago.
I should have written as follows:

x=[(k+1)(k+2)/2]^2, y=[k(k+1)/2]^2, k is natural number and at least 5. ・・・・ This is my answer.

As for "y", I missed "^2" . (Excuse me.)