perplexus.info :: Just Math : Powering

Powering (Posted on 2013-10-05)

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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