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

Guest Jun 5, 2018
 #1
avatar+28 
0

Help plz

Max0815  Jun 5, 2018
 #2
avatar+797 
+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. 

 

zyx
199819971996
199719961995
 ...
...
...
173217311730

 

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

 

I hope this helped,

 

Gavin

GYanggg  Jun 6, 2018
edited by GYanggg  Jun 6, 2018
edited by GYanggg  Jun 6, 2018
 #3
avatar+28 
+1

Thank you very much. Your help is appreciated! :)

Max0815  Jun 7, 2018

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