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.

[@johnpride4209]

x=1 and y=0 also works

[@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..

[@nasrullahhusnan2289]

For the 2nd method, to avoid invalid solution value w/o checking, note that

[@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)