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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

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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.

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please explain briefly – user100315 Oct 14 '13 at 17:58
    
@FRM, I strongly recommend you write out examples. For this sort of mathematics, thinking about examples and non-examples is highly illuminating and will generally lead you to understand the proof. – Ted Shifrin Oct 14 '13 at 21:00

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} \\ \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}

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