Square Squares!

The number of divisors, including 1 and the number itself, of a positive integer with prime factorization

$x = p_1^{a_1} \cdot p_2^{a_2} \cdot p_3^{a_3} \cdots p_n^{a_n}$

can be calculated with

$\tau(x) = \left( a_1+1 \right) \cdot \left( a_2+1 \right) \cdot \left( a_3+1 \right) \cdots \left( a_n+1 \right)$

In the factorization of $x^2$ , every exponent $a_i$ will double. Hence, $x^2$ will have

$\tau \left( x^2 \right) = \left( 2a_1+1 \right) \cdot \left( 2a_2+1 \right) \cdots \left( 2a_n+1 \right)$

divisors. What we want, is $(\tau(x))^2 = \tau \left( x^2 \right)$ .

$\begin{aligned} (\tau(x))^2 & = \left( a_1+1 \right)^2 \cdot \left( a_2+1 \right)^2 \cdots \left( a_n+1 \right)^2 \\ & = \left( a_1^2+2a_1+1 \right) \cdot \left( a_2^2+2a_2+1 \right) \cdots \left( a_n^2+2a_n+1 \right) \\ & > \left( 2a_1+1 \right) \cdot \left( 2a_2+1 \right) \cdots \left( 2a_n+1 \right) \\ & = \tau \left( x^2 \right) \end{aligned}$

Therefore, they can never be equal and so the answer is No .

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To be extra clear, because the problem is very ambiguosly stated and it's easy to get confused:

The ${\color{#3D99F6}\text{blue } 2}$ is the number of divisors of the ${\color{#20A900}\text{green } 2}$ .

${\color{#20A900} 2}^2 = {\color{#EC7300} 4}$

and ${\color{#EC7300} 4}$ has ${\color{#D61F06} 3}$ divisors, but it should have

${\color{#3D99F6} 2}^2 = {\color{#3D99F6} 4} \neq {\color{#D61F06} 3}$ ,

so it doesn't satisfy the condition.

We stipulate that x is a positive integer with n factors, and that x^2 has n^2 factors. x^2 is a perfect square and thus has an odd number of factors. But n^2 odd implies that n is odd, which in turn implies that x is also a perfect square.

Repeating this argument reveals that $sqrt(x)$ also has an odd number of factors, and thus all numbers of the form (sqrt(x))^k, for some k, are integers. Since, in general, we will reach a non-integer with repeated square roots, the only such x which satisfies this condition is 1.

I just thought it logical to be No. There isn't any number outside 1 that would retain it's value after been squared and it's factors will be such that when squared are that number squared itself.

For any x with n factors, x^2 will result in 2n-1 factors. The problem asks for x^2 to yield n^2 factors, so 2n-1=n^2 Set up the quadratic equation and solve. n^2 - 2n +1 = 0 = (n-1)^2 Therefore, n=1. There is only one number with only 1 factorisation, x=1, so there can be no other x such that x^2 results in n^2 factors.

I tried every other number and it didn’t work

Positive factors of number 'x' are given by the prime factorization exponents (ei) and

n= product of (ei+1).

For the square 'x^2' of the number 'x' , positive factors will be equal to product of {2*ei+1} and this can not be made equal to n^2, {for single non-zero 'ei' case}:

(ei+1)^2 is not equal to (2ei+1).

Answer= NO

complicated stuff. Just say no. one is special

Let the n factors of x be $a_1,a_2,a_3...a_n$ (include 1 and x), and let the set of them be A .

Select two factors from A , which we will call $a_i, a_j$ ( $a_i$ can be equal to $a_j$ ). Easy to see that $a_i \cdot a_j$ is a factor of $x^2$ . Also, it's easy to prove that any factor of $x^2$ can be constructed in this way.

Since the number of elements in A is n, the number of number pair $(a_i,a_j)$ is then $n^2$ .

Yet some pairs are not necessary. e.g, for (1,a) and (a,1), they give out the same factor of $x^2$ , which is a. For any integer greater than 1, there must be at least one pair of number like this. Therefore, the number of factors of $x^2$ is less than $n^2$

Intuitively the square of a number rises much quicker than the number of factors of a number, you can imagine those two lines plotted on a graph and it is obvious. Checking with the number 2 confirms that even 2² does not have 4 factors and the lines are diverging. Therefore No.

If x^2 has n^2 factors, that means that every factor of x is being multiplied by every factor of x (n factors * n factors). However, this leads to duplications, where x1•x2 = x2•x1. This counts as a single factor of x^2 however, meaning n^2 factors is an overestimation. Hence NO.

Any positive integer x can be written as a prime power product $x=p_1^{m_1} p_2^{m_2} .. p_N^{m_N}=\prod{p_i^{m_i}}$ where $p_i$ is the i-th prime number and $m_i$ is an integer $\geq 0$ . Since $f=\prod{p_i^{f_i}}$ is a factor of x iff $0\leq f_i\leq m_i$ for all i (providing $m_i+1$ options for each i), the number of factors of $x$ is $\sigma_0(x)=(m_1+1)(m_2+1).. (m_N+1)$ Similarly, the number  of factors of $x^2$ is $\sigma_0(x^2)=(2m_1+1)(2m_2+1).. (2m_N+1)$

We are setting $\sigma_0(x)=n$ and $\sigma_0(x^2)=n^2$ , so that we get two expressions for $n^2$ : $A : n^2=((m_1+1)(m_2+1).. (m_N+1))^2$ $B : n^2=(2m_1+1)(2m_2+1).. (2m_N+1)$ Comparing the factors for $m_i$ in both expressions, we see that

• if $m_i=0$ then $(m_i+1)^2=1=2m_i+1$
• If $m_i>0$ then $(m_i+1)^2=m_i^2+2m_i+1> 2m_i+1$ .

The only way to make A and B equal, is to set $m_i=0$ for all $i$ , implying $x=1$ .