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Powering (Posted on 2013-10-05) Difficulty: 3 of 5
Given that a,b,x and y are real numbers such that

a+b=23
ax+by=79
ax2+by2=217
ax3+by3=691

Determine ax4+by4

No Solution Yet Submitted by Danish Ahmed Khan    
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this does it | Comment 3 of 5 |
I got bogged down in the algebra of substitution too but had a fresh look last night.

Tag the LHS of the 4 equations L1,L2,L3,L4 top to bottom.

L4 = (x+y)*L3 - (xy)*L2
L3 = (x+y)*L2 - (xy)*L1

691 = (x+y)*217- (xy)*79
217 = (x+y)*79 - (xy)*23

Solve to get (x+y) = 1, xy = -6.

Then ax^4 + by^4 = (x+y)*L4 - (xy)*L3 = 1*691 - (-6)*217 = 1993. 

  Posted by xdog on 2013-10-07 12:21:31
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