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# Intermediate number theory

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

Jun 5, 2018

### 3+0 Answers

#1
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Help plz

Jun 5, 2018
#2
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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

Jun 6, 2018
edited by GYanggg  Jun 6, 2018
edited by GYanggg  Jun 6, 2018
#3
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Thank you very much. Your help is appreciated! :)

Jun 7, 2018