(x,y)=(0,0) is obviously a solution. Furthermore, the equation can be written as: (2y+1)2=4y(y+1)+1=x2+x+1, for x>0,x2+x+1, strictly between x2 and (x+1)2, cannot be a perfect square.
Source: NuminaMath-1.5,
licensed Apache-2.0.
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