from (2):$x-y=a^2>0---(*)$
from (3):$x+y=\dfrac{b^2-1}{3}=\dfrac{(b+1)(b-1)}{3}$
$\therefore (b+1)\,\, mod \,\ 3=0\,\, or \,\, (b-1)\,\, mod \,\, 3=0$
if $x+y \,\, even \,\, then \,\, b \,\, must \,\, be\,\ odd$
take $b=13$,we have $x+y=56$
if we let $x-y=a^2=4,\,\, and \,\ x+y=56$
the solution $x=30,y=26$ found
the solution of $a,b$ will meet the following equation:
$3a^2+6ab-b^2+1=0\,\, (a,b\in N)$
I wrote a computer program if b<1000000
the corresponding solution of :
[TABLE="width: 183"]
[TR]
[TD="width: 133, bgcolor: transparent"]x=$a^2+ab$
[/TD]
[TD="width: 110, bgcolor: transparent"]y=$ab$
[/TD]
[/TR]
[TR]
[TD="bgcolor: transparent"]30
[/TD]
[TD="bgcolor: transparent"]26
[/TD]
[/TR]
[TR]
[TD="bgcolor: transparent"]5852
[/TD]
[TD="bgcolor: transparent"]5068
[/TD]
[/TR]
[TR]
[TD="bgcolor: transparent"]1135290
[/TD]
[TD="bgcolor: transparent"]983190
[/TD]
[/TR]
[TR]
[TD="bgcolor: transparent"]220240440
[/TD]
[TD="bgcolor: transparent"]190733816
[/TD]
[/TR]
[TR]
[TD="bgcolor: transparent"]42725510102
[/TD]
[TD="bgcolor: transparent"]37001377138
[/TD]
[/TR]
[/TABLE]