# 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

Note by Mashrur Fazla
4 years, 7 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Suppose we have a factor $$n$$ of the number $$N$$. Then $$N/n$$ is also a factor of the number $$N$$.

These factors are different, unless $$n = N/n$$, or $$N=n^2$$.

So unless $$N$$ is a square, every factor $$n$$ can be paired with $$N/n$$ and thus there is an even number of factors.

If $$N=n^2$$, then every factor $$m$$ ($$m \neq n$$) can be paired with $$N/m$$. Adding the factor $$n$$ that can't be paired with a different factor, we have an odd number of factors.

- 4 years, 7 months ago

Hai Mashrur.

Let N be a perfect square.

Then the prime factorization of N is a product of primes with even powers. Thus the total number of factors is (even+1)(even+1)....(even+1), which is odd.

( I dont know how to format power terms here, otherwise i would have given you a nice complete solution. Hope this works.)

- 4 years, 7 months ago

Just look at the following examples: $$24$$ is not a square number but $$36$$ is a perfect square number.

$$24=1 \times 24$$

$$24=2 \times 12$$

$$24=3 \times 8$$

$$24=4 \times 6$$

$$24=6 \times 4$$

$$24=8 \times 3$$

$$24=12 \times 2$$

$$24=24 \times 1$$

and.........................................................

$$36=1 \times 36$$

$$36=2 \times 18$$

$$36=3 \times 12$$

$$36=4 \times 9$$

$$36=6 \times 6$$

$$36=9 \times 4$$

$$36=12 \times 3$$

$$36=18 \times 2$$

$$36=36 \times 1$$

I did factorizing this way so that the number on the left side (or right side) of the multiplication sign is a factor of the number. It can be easily seen that 24 has even number of factors. You see, at one step of factorizing 36, we come across a term with same numbers on both sides of the multiplication sign ($$6 \times 6$$) which when reversed makes no difference. Hence, you can see that 36, a square number, has odd number of factors.

- 4 years, 7 months ago

Gaussian pairing solves this problem before one can spell 'brilliant'

- 4 years, 7 months ago

Lets have an example,

9=3x3 if you are doing a factor tree, you cant write the same number twice.

- 4 years, 7 months ago