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**Ikuku****Guest**

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

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,608

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?

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**Ikuku****Guest**

Yes...

**Ikuku****Guest**

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..!

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,608

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!

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