- #1

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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.

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- #1

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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.

- #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 3. Then n^2 could not be divisible by 6.

So n is divisible by 2 and 3, hence by 6.

- #4

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now if 6 divides n.n, then 6 divides n.

- #5

morphism

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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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6 isn't a prime number, so we cannot apply that theorem directly...

now if 6 divides n.n, then 6 divides n.

- #7

Gokul43201

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Why are folks writing up complete solutions to what appears to be a textbook problem?

- #8

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now if 6 divides n.n, then 3 divides n.n and 3 divides n.6 isn't a prime number, so we cannot apply that theorem directly...murshid_islam said:

now if 6 divides n.n, then 6 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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People, please read and follow the Posting Guidelines.

https://www.physicsforums.com/showthread.php?t=5374

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- #10

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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]

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