Why perfect square has odd number of factors can someone please describe me why only the perfect square has odd number of factors.why does other number not has odd numbers of factors? I understand it but don't find any mathmetical proof.Please help me
 A: You can always list the factors of a number, N, into pairs $(a_i,b_i)$ where $a_i \le \sqrt N \le b_i$. This means that a number will always have an even number of factors, unless the number is a perfect square, in which case one pair will consists of the same two numbers. The two examples below should demonstrate why.
\begin{align}
    \text{factors} &\;  \text{of $36 = 6^2$} \\
    \hline
    1 &,\, 36 \\
    2 &,\, 18 \\
    3 &,\, 12 \\
    4 &,\, 9 \\
    6 &,\, 6 & \text{A total of $9$ factors} \\
    \hline
\end{align}
\begin{align}
    \text{factors} &\;  \text{of $12$ } \\
    \hline
    1 &, 12 \\
    2 &, 6 \\
    3 &, 4 & \text{A total of $6$ factors} \\
    \hline
\end{align}
A: I agree that it is counter-intuitive. The catch is that when "Factors" is written, generally what is actually meant is "unique factors". Therefor if a number is a perfect square, it will have an even number of total factors, but odd number of UNIQUE factors, which is what is truly meant. 
E.g. 36: 1x36, 2x18, 3x12, 4x9, 6x6 - 10 total numbers but only 9 unique ones. For fun we just decide to ignore the second '6'.
Hope that helps.
A: For a given number $n$ we can group its divisors in pairs $(d,\frac nd)$, except that if $n=m^2$ this would pair $m$ with itself.
A: Well, without the square root, the square no. would be even. Since the square root is multiplied by itself, then there is only 1 more factor, not 2.
Example: 64. Has 6 factors (excluding 8 (squared)). With 8, there is now 7 factors... but not eight because the root (in this case, 8) will be repeated, and there is simply no point in repeating it.
I hope your question is answered! :)
