Solving x+y=(x-y)^2, a Diophantine Equation

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#NumberTheoryProblems #DiophantineEquations

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[davidfriday]

For anybody interested in an oversimplified form of the final answer for this, x and y are consecutive triangle numbers. This leads to a pretty cool geometric interpretation if you feel like drawing pretty pictures.

[MichaelRothwell]

Method 1 gives two sets of parametric solutions:
1. ((k²+k)/2, (k²-k)/2), or (k(k+1)/2, k(k-1)/2)
2. ((k²-3k+2)/2, (k²-k)/2), or ((k-1)(k-2)/2, k(k-1)/2)
However, Method 2 just gives one set (the first).
In fact, the second set of solutions in method 1 is the same as the first. To see this, just replace k by -k+1 in the first set and you get the 2nd set. This is easy to see using the factorised form.

[jarikosonen]

it is checked that the square root gives integer out, but not checked the quadratic result is integer (that it is divided by two is still integer).

[Blaqjaqshellaq]

Notice that the positive solutions for either variable are triangular numbers!

[satrajitghosh]

Writing u for x-y one gets
u - 2y = u^2
or (u -1/2)^2 = 1 -2y
or u = 1/2 + √(1-2y), 1/2 - √(1-2y)

[yoav]

Nice, the second method blowed my mind!💥💥

[Jack_Callcott_AU]

It's interesting to note at 7:40 that all those number are triangular numbers. Triangular numbers occur when 2 consecutive integers are multiplied together and divided by 2, and they equal the sum of the first n, n-1, n-2 integers respectively.

[ffggddss]

x + y = (x – y)²After fiddling around with that equation from the thumbnail, I came up with 3 solutions: (x, y) = (0, 0), (1, 0), (6, 3).
[Also note the symmetry in x vs y; swapping them preserves equality/inequality. So (0, 1) and (3, 6) are also solutions.
And we need only seek solutions with x > y; (0, 0) is the only solution with x = y.]

To get a bigger picture, you could plot z = LHS and z = RHS, in (x, y, z)-space, and see where they intersect (which will be a curve in 3-space).
Then try to find where intersections are integers.
z = x + y will be a plane that intersects the z = 0 plane in the line y = –x, and slopes upward with gradient vector ∇z = (1, 1).
z = (x – y)² is a parabolic cylinder opening upward, with "trough line" y=x, z=0; i.e., (x, x, 0) ∀x.

That's as far as I've gone for now. Let's see what else there is . . .

Just now seeing the comment from michael purtell, and I think he's got it:
• Choose any integer n.
• (x, y) = (½n[n+1],  ½n[n–1]) is a solution . . Note that these are both integers, b/c "n" or the other factor must be even. They are, in fact, just consecutive triangular numbers.
Then LHS = x + y = n² ; x – y = n, so that RHS = (x – y)² = n²
And as stated earlier, (y, x) will also be a solution. Those would appear to be all possible solutions.

Hat-tip and kudos to michael purtell.

Fred

[tomctutor]

A few comments below refer to triangular numbers.✓
if we let x -y = n, then x+y = n+2y
n +2y = n^2 
y = (n^2-n)/2 = n(n-1)/2 for some integer n = ∑_(m=0, m=n-1) m = 0+1+2+...n-1
x = y+n = ∑_(m=0, m=n) m = 0+ 1 +2+ ...n
or vica versa by symmetry of solution set (x, y)
both x, y are pairs of successive triangular numbers;
e.g. 
(0, 0), (0, 1), (1, 3), (3, 6) ... and so on

[michaelpurtell]

Must have missed something. To me it seems that there are an infinite number of solutions pick any N sum the consecutive integers including N that is X subtract N that is Y ( or the same thing multiple any two consecutive number divide by 2 that is X subtract the lesser one that is Y)

[mthtactics]

I challenge you to find x in A^x^x + Bx^x + C = R in terms of A, B, C

[echandler]

Nice problem. Using method 2, replacing k with -k interchanges x and y. Done.

[vladimirkaplun]

you should have noticed that the sum of 2 neighbor sums of the arithmetic progression is a full square. it is what n(n+1)/2 and n(n-1)/2 are

[MgtowRubicon]

The solutions are pairs of the sum of consecutive integers. k(k-1)/2 and k(k+1)/2. The other "solution" is just a variant.

[SuperYoonHo]

Hey? I saw this problem somewhere but you'r solution is REALLY cool! AWESOME video!😎👌

[tanishkraj]

with the help of your videos i have developed interest in mathematics thanks

[rakenzarnsworld]

Level 1:
x = 1, y = 0
Level 2:
x = 3, y = 1
Level 3:
x = 6, y = 3
Level 4:
x = 10, y = 6
Level 5:
x = 15, y = 10
Level n:
x = {n(n+1)}/2, y = {n(n-1)}/2

[MrLidless]

Without loss of generality, let x>= y, ie x = y + n, n>=0.
2y + n = n²
2y = n² - n = n(n - 1)
y = n(n - 1)/2
x = n(n + 1)/2

[maxm]

Assume y>=x and y=x+a, a>=0. Then LHS=2x+a.
If x<0 and y>0, then because x, y are integers x<=-1, y<=-1, then LHS=a-2|x|<a<a^2=RHS which contradicts the given.

If x<0 and y<0 then LHS=x+y<0 but RHS>=0, so this cannot be the case.

So x>=0 and y>=0.

Then LHS=2x+a, RHS=a^2. So effectively 2x=a^2-a=(a-1)a. So there is an answer for every a>=0, where x=(a-1)a/2, y=x+a=(a+1)a/2

[angelmendez-rivera]

Let m = x – y, hence m + 2·y = m^2, so 2·y = m^2 – m, and y = m·(m – 1)/2. Therefore, m = x – m·(m – 1)/2, which implies x = m·(m + 1)/2. From here, it is easy to prove that these satisfy the equation for all integers m. In fact, if y(m) = m·(m – 1)/2, then x(m) = y(m + 1), and every pair (x(m), y(m)) satisfies the equation.