Problem on Real Numbers | AIME I, 1990| Question 15

Find \(ax^{5}+by^{5}\) if real numbers a,b,x,y satisfy the equations

ax+by=3

\(ax^{2}+by^{2}=7\)

\(ax^{3}+by^{3}=16\)

\(ax^{4}+by^{4}=42\)

Let S=x+y, P=xy

\((ax^{n}+by^{n})(x+y)\)

\(=(ax^{n+1}+by^{n+1})+(xy)(ax^{n-1}+by^{n-1})\)

or,\( (ax^{2}+by^{2})(x+y)=(ax^{3}+by^{3})+(xy)(ax+by)\) which is first equation

or,\( (ax^{3}+by^{3})(x+y)=(ax^{4}+by^{4})+(xy)(ax^{2}+by^{2})\) which is second equation

or, 7S=16+3P

16S=42+7P

or, S=-14, P=-38

or, \((ax^{4}+by^{4})(x+y)=(ax^{5}+by^{5})+(xy)(ax^{3}+by^{3})\)

or, \(42S=(ax^{5}+by^{5})+P(16)\)

or, \(42(-14)=(ax^{5}+by^{5})+(-38)(16)\)

or, \(ax^{5}+by^{5}=20\).

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