Divisibility of n by 6

  • #1
I need to prove that if n is a natural number and n^2 is divisible by 6, then n is divisible by 6.

I know that I knew how to do this at one point fairly recently, if you could refresh my memory I would greatly appreciate it.
 

Answers and Replies

  • #2
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Write n out in terms of it's prime factors, and it should be obvious.
 
  • #3
CRGreathouse
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Suppose n was not divisible by 2. Then n^2 could not be divisible by 6.
Suppose n was not divisible by 3. Then n^2 could not be divisible by 6.
So n is divisible by 2 and 3, hence by 6.
 
  • #4
we know that if p divides ab, then p divides a or p divides b.
now if 6 divides n.n, then 6 divides n.
 
  • #5
morphism
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Here's another way to look at it.

n(n+1)(2n+1)/6 is the sum of the first n squares, and thus must be an integer. Its numerator is 2n^3 + 3n^2 + n. So if n^2 is divisible by 6, n must be as well.
 
  • #6
Hurkyl
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we know that if p divides ab, then p divides a or p divides b.
now if 6 divides n.n, then 6 divides n.
6 isn't a prime number, so we cannot apply that theorem directly...
 
  • #7
Gokul43201
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Why are folks writing up complete solutions to what appears to be a textbook problem?
 
  • #8
murshid_islam said:
we know that if p divides ab, then p divides a or p divides b.
now if 6 divides n.n, then 6 divides n.
6 isn't a prime number, so we cannot apply that theorem directly...
now if 6 divides n.n, then 3 divides n.n and 3 divides n.
again, if 6 divides n.n, then 2 divides n.n and 2 divides n.
so then 3 divides n and 2 divides n. therefore, 6 divides n.
 
Last edited:
  • #9
Gokul43201
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Great! That makes 3 complete solutions for the OP to copy down. Anyone else?

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  • #10
Hello, my math skills are crap, but can't we simply do that?
If n is divisible by 6, then it can be writed as 6x where x is an integer.

[tex]\left(6x\right)^{2}=6^{2}x^{2}=36x^{2}=6\left(6x^{2}\right)[/tex]
 
  • #11
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You could do that... if the intention was to show "if n is divisible by 6, then n^2 is too". But the original question was the other way around: "if true for n^2, then it's true for n".
 

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