Can the total number of integers that divide x be express

Solution

• We know that a number in the form of 2k + 1 leaves a remainder of 1 when divided by 2, (that is, a number that leaves a remainder of 1 when divided by 2) is odd.
• So, we are asked to find if the number of factors of x is odd.

• We know that if a number has odd number of factors, it has to be a perfect square. So, we are (indirectly) asked to find if x is a perfect square?

1. √12x is an integer
\(\sqrt{12\mathrm x}=\;\sqrt{2^2\ast3\mathrm x}=\;2\surd3\mathrm x\) is an integer. For √3x to be an integer, x should contain an odd power of 3.

Now, if 3 occurs odd number of times in x, x can’t be a perfect square.

2. The product of √x and √y is an integer, where the total number of factors of y/3 is odd.
Statement-2 tells us that the total number of factors of \(\frac{\mathrm y}3\) is odd i.e. \(\frac{\mathrm y}3\) is a perfect square. Let’s assume \(\frac{\mathrm y}3=\mathrm z^2\), where z is an integer.

\(y = 3z^2\). Since \(z^2\) is always non-negative, y will also be a non-negative integer
So, we know that \(\surd\mathrm x\operatorname{ *}\;\surd3\mathrm z^2\) is an integer i.e. z√3x is an integer.
For √3x to be an integer, x should contain an odd power of 3. If 3 occurs odd number of times in x, x can’t be a perfect square