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## #1 2005-11-04 21:21:44

Ikuku
Guest

### Division of m

m is a number.
i need to prove that if m^2 is divided by 3, then m is divided by 3 too.

i did something like this
m^2=3k

but i dont know what can i do from now

## #2 2005-11-04 22:09:43

MathsIsFun

Offline

### Re: Division of m

So, you are asking: "m is an integer (whole number), and if m² is divisible by 3, then m is also divisible by 3", is that right?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

Ikuku
Guest

Yes...

## #4 2005-11-05 07:16:54

Ikuku
Guest

### Re: Division of m

can someone help me??
I need to show that if
m^2 can be divided by 3 then m can be divided by 3 too.

how can i do that?

Induction can be no use because i think it must be true for every m not for only specific ones.

anyway, if  someone has an idea then thnx..!

## #5 2005-11-05 09:09:32

MathsIsFun

Offline

### Re: Division of m

There are only 3 possible remainders for m/3: 0,1 or 2

For example:
m      m/3       whole and remainder
1     0.333..       0 R1
2     0.666..       0 R2
3     1               1 R0
4     1.333..       1 R1
etc...

Let us call these answers n, n+1/3, and n+2/3

Now, you want to square "m" then divide that by 3 and see if there is a remainder.

We know that "m/3" is going to be either (n, n+1/3, and n+2/3), and (m/3)²= m²/9, so we can calculate what m²/3 will be:

(m/3)² = (n, n+1/3, or n+2/3)²
m²/9 = (n, n+1/3, or n+2/3)²
m²/3 = (n, n+1/3, or n+2/3)² × 3

Let us test each possibility:

• n² × 3 = 3n² (this has no fractional part and is therefore divisible by 3)
• (n+1/3)² x 3 = (n² + 2n(1/3) + (1/3)²) x 3 = 3n² + 2n + 1/3 (has a fractional part)
• (n+2/3)² x 3 = (n² + 2n(2/3) + (2/3)²) x 3 = 3n² + 4n + 4/3 (has a fractional part)

I think I have gone about this the long way, but I have shown that whenever "m/3" has a fractional part, then "m²/3" ALSO has a fractional part.

Enjoy!

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman