We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
244
3
avatar

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
 #1
avatar+194 
+1

Help plz

 Jun 5, 2018
 #2
avatar+974 
+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

 Jun 6, 2018
edited by GYanggg  Jun 6, 2018
edited by GYanggg  Jun 6, 2018
 #3
avatar+194 
+1

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

 Jun 7, 2018

10 Online Users

avatar
avatar
avatar