Compute the number of ordered triples of integers $$(x, y, z)$$, $$1729 < x, y, z < 1999$$ which satisfy: $$ x^2 + xy + y^2 = y^3 - x^3 \qquad \text{and} \qquad yz + 1 = y^2 + z. $$

Guest Jun 5, 2018

#2**+3 **

We first use \(y^3-x^3=(y−x)(x^2+xy+y^2)\)

\(x^2 + xy + y^2 = y^3 - x^3 \\ x^2 + xy + y^2=(y-x)(x^2+xy+y^2)\\ y=x+1 \)

From the second equation, we get:

\(yz + 1 = y^2 + z\\ (x+1)z+1=(x+1)^2+z\\ xz+z+1=x^2+2x+1+z\\ x^2+2x-xz=0\\ x^2+x=zx-x\\ x(x+1)=x(z-1)\\ x+1=z-1\\ y=z-1 \)

Concluding:

\(z=y+1\\ y=x+1.\\ \therefore z>y>x\)

They are also consecutive integers.

z | y | x |

1998 | 1997 | 1996 |

1997 | 1996 | 1995 |

. | . | . |

. | . | . |

. | . | . |

1732 | 1731 | 1730 |

In this list, there are \(1998-1732+1\) rows, and therefore, 267 ordered triples.

I hope this helped,

Gavin

GYanggg Jun 6, 2018