The prime factorization of \(20\) is \(2^2\times 5\).
The number \(20\) has \(6\) positive divisors. They are:
\(2^05^0=1\), \(2^05^1=5\), \(2^15^0=2\), \(2^15^1=10\), \(2^25^0=4\), \(2^25^1=20\)
Two of the divisors, \(1\) and \(4\), are perfect squares.
How many positive divisors of \(2025^{2025}\) are perfect squares?
First, let’s look at the prime factorization of four different perfect squares:
\(9=3^2\), \(16=2^4\), \(36=2^2\times 3^2\), \(129\,600=2^6\times 3^4\times 5^2\)
Note that, in each case, the exponent on each of the prime factors is even. In fact, a positive integer is a perfect square exactly when the exponent on each of the prime factors in its prime factorization is an even integer greater than or equal to zero. Now \[\begin{aligned} 2025^{2025}&=(3^4\times 5^2)^{2025}\\ &=(3^4)^{2025}\times (5^2)^{2025}\\ &=3^{8100}\times 5^{4050} \end{aligned}\] All positive divisors of \(2025^{2025}\) will be of the form \(3^k\times 5^n\), \(0\leq k \leq 8100\), and \(0 \leq n \leq 4050\), where \(k\) and \(n\) are each integers.
For \(3^k\times 5^n\) to be a perfect square, \(k\) must be an even integer such that \(0\leq k \leq 8100\) and \(n\) must be an even integer such that \(0 \leq n \leq 4050\).
There are \(8100 \div 2=4050\) even integers from \(1\) to \(8100\). The number \(0\) is also even, so there are \(4051\) possible values of \(k\).
There are \(2025\) even integers from \(1\) to \(4050\). The number \(0\) is also even, so there are \(2026\) possible values of \(n\).
For each of the \(4051\) values of \(k\), there are \(2026\) values of \(n\) so there are \(4051\times 2026=8\,207\,326\) perfect square divisors of \(2025^{2025}\).
Therefore, \(2025^{2025}\) has \(8\,207\,326\) positive divisors that are perfect squares. This is over \(8\) million perfect square divisors!