Pell Equation

The first such triple is 8 = \(2^2+2^2\),9 = \(3^3+0^2\),10=\(3^2+1^2\), which suggests we consider triples \(x^2-1,x^2,x^2+1\).Since \(x^2-2y^2=1\) has infinitely many positive solutions (x,y), we see that \(x^2-1=y^2+y^2,x^2=x^2+0^2\)and \(x^2+1\) satisfy the requiment and there are infinitely many such triples.

The first such triple is 8 = 22+22,9 = 33+02,10=32+12, which suggests we consider triples x2−1,x2,x2+1.Since x2−2y2=1 has infinitely many positive solutions (x,y), we see that x2−1=y2+y2,x2=x2+02and x2+1 satisfy the requiment and there are infinitely many such triples.

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