Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Real Numbers.

Problem on Real Numbers – AIME I, 1990


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

ax+by=3

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

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

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

  • is 107
  • is 20
  • is 634
  • cannot be determined from the given information

Key Concepts


Integers

Equations

Algebra

Check the Answer


But try the problem first…

Answer: is 20.

Source
Suggested Reading

AIME I, 1990, Question 15

Elementary Algebra by Hall and Knight

Try with Hints


First hint

Let S=x+y, P=xy

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

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

Second Hint

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

Final Step

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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