# Solve the equation system

• November 19th 2007, 09:13 AM
james_bond
Solve the equation system
Solve the equation system:
$x+y=1$

$x^5+y^5=31$
• November 19th 2007, 09:20 AM
Krizalid
Quote:

Originally Posted by james_bond
Solve the equation system:
$x+y=1$

$x^5+y^5=31$

Power to 5 the first equation:

$(x + y)^5 = x^5 + 5x^4 y + 10x^3 y^2 + 10x^2 y^3 + 5xy^4 + y^5.$

So $\left( {x^5 + y^5 } \right) + \left( {5x^4 y + 5xy^4 } \right) + \left( {10x^3 y^2 + 10x^2 y^3 } \right).$

And $31 + 5xy\left( {x^3 + y^3 } \right) + 10x^2y^2\left( {x+y } \right).$

May be you can go on from there.
• November 19th 2007, 09:30 AM
topsquark
Quote:

Originally Posted by james_bond
Solve the equation system:
$x+y=1$

$x^5+y^5=31$

I'd recommend the brute force approach.

$y = 1 - x$

$x^5 + (1 - x)^5 = 31$

$x^5 + 1 - 5x + 10x^2 -10x^3 + 5x^4 - x^5 = 31$

$5x^4 - 10x^3 + 10x^2 - 5x - 30 = 0$

$x^4 - 2x^3 + 2x^2 - x - 6 = 0$

The rational root theorem gives -1 and 2 as zeros. So do long or synthetic division, which will give you a quadratic to solve at the end.

-Dan
• November 19th 2007, 09:47 AM
ThePerfectHacker
Step 1 (The Krizalid Operation)
$(x+y)^5 = 1$
$x^5+y^5 + 5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5 = 1$
$(x^5+y^5)+(5x^4y+5xy^4)+(10x^3y^3+10x^2y^3)=1$
$31 + 5xy(x^3+y^3)+10x^2y^2(x+y)=1$
$31+5xy(x+y)(x^2-xy+y^2)+10x^2y^2=1$
$5xy(x^2-xy+y^2)+10x^2y^2=-30$
$5x^3y+5x^2y^2+5xy^3=-30$
$x^3y+x^2y^2+xy^3 = -6$ . . . [1]

Step 2 (Emo)
$x^5+y^5= 31$
$(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)=31$
$(x^4+y^4) - (x^3y+xy^3 - x^2y^2) = 31$ . . . (use [1])
$(x^4+y^4) - (-6 - x^2y^2 - x^2y^2) = 31$
$x^4+2x^2y^2+y^4 = 25$
$(x^2+y^2)^2=25$
$x^2+y^2 = \pm 5$ . . . [2]

Step 3 (Cool)
$x+y=1$
$x^2+2xy+y^2 = 1$ . . . (use [2])
$\pm 5 + 2xy = 1$
$xy = 3,2$ . . . [3]

Step 4 (Envy)
$x-y = ?$
$x^2 - 2xy +y^2 = ?^2$ . . . (use [3]) . . . (use 2)
$?^2 = \pm 5 - 3,2 = 2,3,-7,-8$

So you just have to solve,
$x+y=1$
$x-y=c$
Where $c$ is one of those possible values.