Number theory

Question

Why does a perfect square number have odd number of factors?

Comments

Thanks both of you for answering

Answer 1

Suppose we are finding the number of factors of a natural number n

To find it I have written the following pseudo-code

Parameters  i,p,q,c=0,n

for   i =1   to  sqrt(n)

 if ( i  divides  n )

   p= i

   q= n/i

 end  if

 if ( p== q)

   c= c+1

 end if

 else

  c= c+2

 end else

end for

Note that c is the no. of factors of n.

c is odd only when p=q otherwise c is even.

p=q only if n is a perfect square.

Hence c is odd when n is a perfect square.

Hence the no. of factors of a perfect square is odd. (proved)

Answer 2

Let's take an example

$n = 20, \  which \  is \  not \  a \  perfect \  square $

$factors \  of \  20 \  are \ 1, 2, 4, 5, 10, 20$

$1 \ X \ 20 \ \rightarrow 20$

$2 \ X \ 10 \ \rightarrow 20$

$4 \ X \ 5 \ \rightarrow 20$

You can see that every factor comes in pair

$n = 16, \  which \  is \ a \  perfect \  square$

${\color{Red} {factors \ of \ 16 \ are \ 1, 2, 4, 8 , 16 }}$

$1 \ X \ 16 \ \rightarrow 16$

$2 \ X \ 8 \ \rightarrow 16$

Here we can see that $4$ dosen't have any pair.

Take another example

$ n = 36, \  which  \ is  \ a  \ perfect \  square $

${\color{Red} {factors \ of \ 36 \ are \ 1, 2, 3, 4, 6, 9, 12,18,36}}$

$1 \ X \ 36 \ \rightarrow 36$

$2 \ X \ 18 \ \rightarrow 36$

$3 \ X \ 12 \ \rightarrow 36$

$4 \ X \ 9 \ \rightarrow 36$

$6$ doesn;t have any pair

This is the reason for perfect square having odd number of factors