Problem of the year 2016

Algebra Level 5

$\begin{aligned} ax + by & = 3 \\ a{ x }^{ 2 } + b{ y }^{ 2 } & = 7 \\ a{ x }^{ 3 } + b{ y }^{ 3 } & = 16 \\ a{ x }^{ 4 } + b{ y }^{ 4 } & = 42 \end{aligned}$ .

If $a$ , $b$ , $x$ , and $y$ are real numbers satisfying the system of equations above, find $a{ x }^{ 5 } + b{ y }^{ 5 }$ .

The answer is 20.

1 solution

Priyanshu Mishra
Sep 7, 2016

For $n = 2; n = 3$ , the identity

$\large\ \left( a{ x }^{ n }+b{ y }^{ n } \right) (x+y)-\left( a{ x }^{ n-1 }+b{ y }^{ n-1 } \right) (xy)=a{ x }^{ n+1 }+b{ y }^{ n+1 }$

leads to the equations

$7{(x+y)} - 3xy = 16$ and $16{(x+y)} - 7xy = 42$ .

Solving these two equations simultaneously yields

$x + y = -14$ and $xy = -38$ .

Applying recurrence identity for $n = 4$ gives

$a{ x }^{ 5 } + b{ y }^{ 5 } = (42)(-14) - (16)(-38) = \boxed{20}$ .

Problem of the year 2016

The answer is 20.

1 solution

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