An algebra problem by Parth Lohomi

From $2<a^2<3$ ,we have $\lfloor{a^2}\rfloor=2$ that implies, $\langle{a^2}\rangle=a^2-2$ .From $2<a^2<3$ ,we also have that $\sqrt{2}<a<\sqrt{3}$ that implies $a>1$ that implies $\lfloor{\dfrac{1}{a}}\rfloor=0$ that implies $\langle{\dfrac{1}{a}}\rangle=\dfrac{1}{a}$ .That implies $\dfrac{1}{a}=a^2-2$ .That implies $a^3-2a-1=0$ .By observation we have $a=-1$ satisfies the given equation that implies $(a+1)(a^2-a-1)=0$ that implies the solutions are $a=-1$ and the roots of the quadratic $a^2-a-1$ .But $a$ can't be $-1$ as it is given in the question that $a$ is positive.That means $a$ can take two values,which are the roots of the quadratic that we got earlier.The two roots are $:\dfrac{1+\sqrt{5}}{2}$ and $\dfrac{1-\sqrt{5}}{2}$ .But the latter one is negative and that means this is also ruled out,thus, $a=\dfrac{1+\sqrt{5}}{2}$ . Now put the values in the calculator and approximate a bit.You will get $322-89=233$ and done!!

Notice that the golden ratio satisfies the properties for $a$ . Assuming there is a unique such $a$ , $a$ must be the golden ratio. Finally, if $a$ is the golden ratio, then $a^{12}-144 \cdot a^{-1} =233$ .

From $2 < a^2 < 3$ we know that $\lfloor a^2 \rfloor = 2$ .

We can now write out this equation: $a^2 - 2 = \frac 1a$

Multiply through by a to get a off the bottom: $a^3 - 2 a = 1$

Synthetic division, try $a = \pm 1$ and -1 works

$(a + 1)(a^2 -a -1) = a^3 - 2a -1$

Using the quadratic equation the solutions are $\frac {1 \pm \sqrt{5}}{2}$

Since $\frac {1 - \sqrt 5}{2}$ is negative it cannot be correct.

First, $\frac 1a = \frac{2}{\sqrt 5 + 1} * \frac{\sqrt 5 - 1}{\sqrt 5 -1} = \frac{2 (\sqrt 5 - 1) }{5 - 1} = \frac{2(\sqrt 5 - 1)}{4} = \frac{\sqrt 5 - 1}{2}$

Next calculate $a^2, a^4, a^6, a^12$

$a^2 = a + 1$ (from the original quadratic)

$a^2 = \frac{1 + \sqrt 5}{2} + \frac 22 = \frac{3 + \sqrt 5}{2}$

$a^4 = \frac{3 + \sqrt 5}{2} * \frac{3 + \sqrt 5}{2} = \frac{9 + 6\sqrt 5 + 5}{4} = \frac{14 + 6\sqrt 5}{4} = \frac{7 + 3\sqrt 5}{2}$

$a^6 = \frac{7 + 3\sqrt 5}{2} * \frac{3 + \sqrt 5}{2} = \frac{21 + 7\sqrt 5 + 9\sqrt 5 + 15}{4} = \frac{36 + 16\sqrt 5}{4} = 9 + 4\sqrt 5$

$a^12 = (9 + 4\sqrt 5)(9 + 4\sqrt 5) = 81 + 36\sqrt 5 + 36\sqrt 5 + 80 = 161 + 72\sqrt 5$

$161 + 72\sqrt 5 - 144\frac{\sqrt 5 - 1}2 = 161 + 72\sqrt 5 - 72\sqrt 5 +72 = 233$