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 Integer Solutions (Posted on 2006-03-30)
Determine all integer solutions (x,y,z) for the system of equations

x²z + y²z + 4xy = 40,

x² + y² + xyz = 20

 See The Solution Submitted by Bractals Rating: 2.5000 (6 votes)

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 kiss- corrected and reinforced | Comment 15 of 16 |

LET

X^2+Y^2=W  AND XY=T

WE GET:      WZ+4T=40             (1)

AND             W+TZ=20            ( 2)  multiply by z and

SUBTRACT   (1)-(2)*Z   ====>  T=20/(2+Z)

CHECK FOR  Z =  0,2,3,8 AND 18   to get integer T and then evaluate W.  ONLY Z=3 AND Z=18 QUALIFY.

The rest follows easily.T=4 ===>x=y +2 or -2

T=1           x=y +1 or -1

remark:

There is much shorter   way to solve- by reasoning that symmetry implies x=y (writing x instead of y and vice versa leaves the set of equations unchanged...)

So  ( 2x^2)*z+4*x^2=40

the other equation becomes redundant, - both imply

x^2=20/(z+2  )  and we throw away the non-integer solutions

Edited on April 1, 2006, 4:04 am
 Posted by Ady TZIDON on 2006-04-01 03:49:27

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