Question

Given a nonnegative real number $x,$ let $\langle x\rangle$ denote the fractional part of $x;$ that is, $\langle x\rangle = x - \lfloor x\rfloor,$ where $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x.$ Suppose that $a$ is positive, $\langle a^{-1}\rangle = \langle a^2\rangle,$ and $2 \lt a^2 \lt 3.$ Find the value of $a^{12} - 144a^{-1}.$

Answer

Solution:

From $2 \lt a^2 \lt 3$ we get $\sqrt{2} \lt a \lt \sqrt{3},$ so $0 \lt a^{-1} \lt 1$ and $\langle a^{-1}\rangle = a^{-1},$ while $\langle a^2\rangle = a^2 - 2.$ The condition becomes $a^{-1} = a^2 - 2,$ i.e. $a^3 - 2a - 1 = 0,$ which factors as $(a + 1)(a^2 - a - 1) = 0.$ Since $a \gt 0,$ we get $a = \frac{1 + \sqrt{5}}{2},$ the golden ratio, and indeed $a^2 = a + 1 \approx 2.618$ lies in $(2, 3).$

Using $a^2 = a + 1$ repeatedly: $a^4 = (a+1)^2 = 3a + 2,$ $a^8 = (3a+2)^2 = 9(a+1) + 12a + 4 = 21a + 13,$ and $a^{12} = a^8 a^4 = (21a + 13)(3a + 2) = 63(a+1) + 81a + 26 = 144a + 89.$ Also $a^{-1} = a - 1$ from $a^2 = a + 1.$

Therefore $a^{12} - 144a^{-1} = 144a + 89 - 144(a - 1) = 89 + 144 = 233.$

1997 AIME Problem 9

Solution:

Problem 9 in Other Years