A very interesting problem, leading to a result that seems counter intuitive till you prove it. The difficulty is that intuition doesn't encourage one to try and prove it.
Consider any number n.
It has prime factors having varying powers, and those powers may be odd or even
After squaring, all the powers will become even. Let the power be p, therefore there are p + 1 ways to put that prime
in our factor prime^(0, 1, 2, ... p). Therefore the number of factors of any number are:
(p1 + 1)(p2 + 1)... so on
But since p is always even, p + 1 is guaranteed to be odd.
Since a product of odd numbers is always odd, the number of factors is odd.
Therefore, a perfect square will always have odd number of factors.
Consider any number n.
It has prime factors having varying powers, and those powers may be odd or even
After squaring, all the powers will become even. Let the power be p, therefore there are p + 1 ways to put that prime
in our factor prime^(0, 1, 2, ... p). Therefore the number of factors of any number are:
(p1 + 1)(p2 + 1)... so on
But since p is always even, p + 1 is guaranteed to be odd.
Since a product of odd numbers is always odd, the number of factors is odd.
Therefore, a perfect square will always have odd number of factors.
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