Once, I was playing with my calculator at school during the break. I was doing many things with the number 16, when I decided to calculate the square root of it. The result, of course, was 4. However, I pressed, by accident, the square root button again, and the result that was being shown was 2. I then thought of what had just happened: I had a number, then I calculated the square root of it, and the square root of the result was an integer, which was not a perfect square. In other words, the result of the square root of a perfect square (16) was also a perfect square (4). Let the initial number be \(x\), and the perfect square results be \(x_{1}, x_{2}, x_{3}, ... ,x_{m}\), where \(x_{m}\) is the last possible result, which is not a perfect square. As an explicit example, I used \(x = 16\) that day at school, so \(m\) was \(2\) and \(x_{m} = 2\). Consider that \(n\) and \(k\) are positive integers. In order to maximize the value of \(m\), \(x\) should be of the form:
###### Image credit: Like Cool.

**Note**: the exponent of \(x_m\) must be minimized.

×

Problem Loading...

Note Loading...

Set Loading...