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
but i dont know what can i do from now
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
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..!
There are only 3 possible remainders for m/3: 0,1 or 2
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
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.