An algebra problem by Hana Wehbi

If $t$ is divisible by $12$ , then it is divisible by $4=2^2$ .

So it can be written as $t=2^2\times a$ , where $a$ is an integer not divisible by $2$ .

Then $t^2=(2^2a)^2=2^4a^2$ , where $a^2$ is an integer not divisible by $2$ .

$\dfrac {t^2}{2^4}=\dfrac{2^4a^2}{2^4}=a^2$ is an integer.

However $\dfrac {t^2}{2^5}=\dfrac{2^4a^2}{2^5}=\dfrac{a^2}2$ is not an integer.

The answer is $\boxed5$ .

$t = 12n$ for some integer $n$

We must find integer $a$ such that:

$A = \frac{(12n)^2}{2^a}$ $= \frac{(3^2)(2^2)^2(n^2)}{2^a}$ $= 9(2^{4-a})n^2$ is not an integer

For $A = 9(2^{4-a})n^2$ to not be an integer, we must have $\boxed{a = 5}$ , because if so:

$A = 9(2^{4-5})n^2 = \frac{9n^2}{2}$

For $A$ to be an integer again, we need that $n^2 = 2^m$ for some integer $m \gt 0$

If $a \gt 5$ , then $n^2 = 2^{m^2}$ which is obviously larger than $2^m$

Hence, $a = 5$ is the smallest value such that $A \notin \mathbb{Z}$