Why do perfect squares have a odd amount of factors I cannot find out why perfect squares have an odd amount of factors
 A: A factor is a number that can be multiplied with another number to get a specific product. For most (not perfect square numbers), we can think of factors as coming in pairs. 
For example, let's look at the factors of 12. The following multiplication facts/pairs can get a product of 12: 1X12, 2X6, and 3X4. Therefore, the factors are 1, 2, 3, 4, 6, 12. This has an even number of factors because each factor has another factor paired with it to get to the desired product. 
However, the case of perfect squares are different because for one of the factors, the paired factor is itself. Let's look at 16 as an example. The following multiplication facts/pairs can get a product of 36: 1x36, 2x18, 3x12, 4x9, 6x6. Note that 6 shows up twice, but when we write the list of factors, 6 need only be mentioned once. Therefore, the factors are 1, 2, 3, 4, 6, 9, 12, 18, and 36. Here, there is an odd number of factors because the square root of the perfect square (in this case 6) does not have a pair. 
Therefore, perfect squares have an odd number of factors because the square root of the perfect square does not have a pair. 
A: First 'factorize' a positive integer into real numbers as:
$N = \sqrt N \times \sqrt N$. We want integer factors, so let us change the two (equal) 'factors' appearing above. If one of them is increased the other has to be decreased so that the product remains the same $N$.
In any factorization $n=a\times b$ we can take $a$ as the smaller and $b$ the bigger. So the factors occurs in pairs. 
In case $N$ is a perfect square then $\sqrt N$ is a whole number and $N = \sqrt N \times \sqrt N$ leads to only a single new factor.
So overall we have  odd number of factors for a perfect square.
A: If $d$ is a factor of $n$, then so is $n/d$. Also $d<\sqrt n$ iff $n/d>\sqrt n$.
So each $n$ has the same number of factors $<\sqrt n$ as $>\sqrt n$ so
$n$ has an even number of factors $\ne \sqrt n$.
When $n$ is a square also $\sqrt n$ is a factor, so $n$ has an odd number of
factors. Otherwise $n$ has an even number of factors.
A: Let $n=m^2$ be a perfect square.
Write the prime factorisation of $m$ as $\prod_{i=1}^k p_i^{a_i}$, 
then we know that the prime factorisation of $n$ equals $\prod_{i=1}^k p_i^{2a_i}$
Any divisor $d$ of $n$ is uniquely determined by picking some $b_i \in \{0,\ldots,2a_k\}$ for each $i$, and is of the form $\prod_{i=1}^k p_i^{b_i}$ (as every prime factor of $d$ must occur among those of $n$, at most $2a_i$ times).
So the number of those divisors is $\prod_{i=1}^k (2a_i+1)$ which is a product of odd numbers, hence odd.
Or as an alternative note that $f: d \to \frac{n}{d}$ is an involution on the set of divisors of $n$ with a unique fixed point $\sqrt{n}$. So that set has an odd number of members.
A: Because if $k$ is a factor of $n$ then $k$ is matched up in one to one correspondence with the factor $\frac nk$.  
If we assume $k \ne \frac nk$ then we can list all the factors of $n$ in pairs.  For every $k_i$ being a factor, $\frac n{k_i}$ is a matching factor.
But if we have a case where $k = \frac nk$ then you have an odd singleton sticking out.
That happens if, and only if,  $n = k^2$.  Non perfect squares never have this odd singleton and perfect squares always have this odd singleton.
So the factors of a non perfect square will but the sets of $(k_i, \frac n{k_i})$ pairs and that's an even number of factors.  The factors of a perfect square will have  $(k_i, \frac n{k_i})$ pairs plus the one odd singleton   $k = \sqrt n = \frac n{\sqrt{n}} = \frac nk$.  That's an odd number of factors.
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Another way of putting it.  Suppose $n$ has $m$ factors that are strictly less than $\sqrt n$.  (We can label them as $1,k_2, ....,k_m$).  Then there are cooresponding factors $m_i$ so that $k_i*m_i = n$ and as $k_i < \sqrt n$ then $m_i = \frac n{k_i} >\frac n{\sqrt n} = \sqrt n$.  
And vice versa  $k_i< \sqrt n$ and $k_i*m_i;k_i,m_i\in \mathbb N$ if and only if $m_i > \sqrt n$.
So $n$ will have exactly $2m$ factors that are not equal to $\sqrt n$.  That is true of all numbers.
If $n$ is not a perfect square that is all the factors.  If $n$ is a perfect square then $k=\sqrt n$ will be one more final factor.
