Reducing a quadratic Diophantine to find solutions

Will readily admit do some posts to talk things where it also helps when I just get curious as well. As found myself thinking should do something new showing my method for reducing binary quadratic Diophantine equations, and eventually decided to go with:

x² + y² = xy + x+y + 102

There is a LOT of deliberate easy there and found out could still get more difficult with that last.

Where it took me a bit to pick that where I was too ambitious at first and tried 1000 which had NO integer solutions! Played around there for a bit and decided to use smaller numbers and 100 and 101 had no solutions. Luckily 102 did, as get tired of such things quickly.

(If you want to test your math knowledge, find solutions before reading further.

Can you prove that with 1000, 100 or 101 instead of 102, no integer solutions?)

Oh, and first thing is to standardize to the form I used with my method to reduce! As forgot to do that at first and was confusing myself:

x² - xy + y² = 102 + x+y

What follows relies on my post showing how to reduce and you can reference it to see how I get A, B and C as well as the rest.

Ok, so then A = 9 + 4(-3) = -3, and B = -3(0) + 2(-3) = -6, C =-4*102(-3) = 1224

Which means:

(-3(x+y) + 6)² + 3s² = 36 +3672 = 3708

And dividing off 9, gives:

(-(x+y) + 2)² + s²/3 = 412

And -(x+y) + 2 = 20  works, as gives s = 6, as a solution. I like to use the positive solutions as just want an answer. This thing is picky though! Was maybe a bit surprised had to search a bit, but glad 102 wasn't too far from where I started.

x+ y = -18, so y = -x-18, and substituting with original:

x² - x(-x-18) + (x+18)² = 102 +x - x - 18

So:

x² + x² + 18x + x² + 36x +324  = 84

3x² + 54x +240 = 0, which is: x² + 18x +80 = 0

And (x+8)(x+10) = x² + 18x + 80

So have two possible solutions and will use x = -10, so y = -8. Oh there's a symmetry thing going there.

So original was: x² + y² = xy + x+y + 102

And: 100 + 64 = 80 -18 + 102 = 164

And that's just one set. Noticed also that s = 33 works. And can get:

x = 3, y = -8

9 + 64 = -24 + 3 - 8 + 102 = 73

Is interesting to me that coming back to play with the math often for me is an adventure, especially as get some distance from the discovery.

Here it also kind of intrigues me that there are NO integer solutions for:

x² + y² = xy + x + y + 1000

Had a vague feeling that some combination would work, and that vague feeling was wrong. And then there weren't any integer solutions with 100 or 101 either.

Those integers can be SO picky.

Numbers just stay interesting to me.


James Harris