Solving A Cool Diophantine Equation With Integers

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[@lagomoof]

The unsaid fact: n(n+1)/2 is the nth triangular number. Also, any time 8n+1 (where n is any variable or expression representing an integer) turns up alone under a square root, you can be sure that triangular numbers are lurking somewhere. A bit like if n^2-n-1 shows up then there's usually a golden ratio afoot.

[@FisicTrapella]

If x + y = k^2 then x - y can be k or (-k). Hence the 2 solutions you got with the 1st method.

[@barberickarc3460]

took me a while of bashing my skull in to the wall with congruence, parity, perfect squares and divisibility, but then the solution hit me and i felt kind of dumb..
Set x-y=n, therefore x=y+n
Replace in our original equation to get n+y+y=n^2
Cleaning up you get y = ½(n^2 - n)
Do the same for y=x-n, x= ½(n^2 + n)
The solution set to our original equation is (x, y) = [½(n^2 + n), ½(n^2 - n)] for every integer n

[@ShortsOfSyber]

That type of struggle makes us better. Nice! 😍

[@johnpride4209]

x=1 and y=0 also works

[@nasrullahhusnan2289]

For the 2nd method, to avoid invalid solution value w/o checking, note that
x+y=(x-y)² --> x+y is positive
It means that
• both x and y can't be negative
• both x and y are positive or
• one of x and y is positive and the other one is negative but the absolute value of the negative one is less than the positive. To be clearer, let say y

[@mr.d8747]

*x and y are both equal to 0*

[@broytingaravsol]

only for (0, 0)?

[@dardoburgos3179]

X= 0, Y=0.

[@mcwulf25]

In #2 k and -k give (x,y) and (y,x)