I am trying to show that $n^2 \bmod 3 = 0$ implies $n \bmod 3 = 0$.
This is a part a calculus course and I don't know anything about numbers theory. Any ideas how it can be done? Thanks!
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2The definition of a prime number is that $p|ab\Rightarrow p|a\text{ or }p|b$, which makes this rather tautological in view of 3 being prime... – anon Mar 22 '12 at 15:36
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5What on Earth is this doing in a calculus course? – Peter Taylor Mar 22 '12 at 15:45
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@anon: That is not the usual definition of a prime, is it? – Harald Hanche-Olsen Mar 22 '12 at 16:08
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@Harald: Depending on where you are in number theory, it is (I suppose prime just means irreducible in elementary NT so my comment wasn't really helpful). – anon Mar 22 '12 at 16:16
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@anon It's not quite a tautology since $\rm:q\ |\ n^2:\Rightarrow:q\ |\ n:$ is true iff $\rm:q:$ is squarefree, which is not equivalent to $\rm:q:$ is prime, since, e.g. products of distinct primes are squarefree, but are prime iff the product has $1$ factor. – Bill Dubuque Mar 22 '12 at 17:58
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Hint: Try to show that $n \bmod 3 \ne 0$ does imply $n^2 \bmod 3 \ne 0$. Consider the cases $n \bmod 3 = 1$ and $n \bmod 3 = 2$. If for example $n \bmod 3 = 1$, we can write $n = 3k+1$, what follows for $n^2$?
HTH, AB,
martini
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Hint $\rm\ (1+3k)^2 = 1 + 3\:(2k+3k^2)$
and $\rm\ \ \ (2+3k)^2 = 1 + 3\:(1+4k+3k^2)$
Said mod $3\!:\ (\pm1)^2 \equiv 1\not\equiv 0\ \ $ (note $\rm\: 2\equiv -1$)
Bill Dubuque
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If I could elaborate (since the OP might not be able to connect the dots, even of this good hint)... The contrapositive of the statement $$n^2 \mod 3 = 0 \Rightarrow n \mod 3 = 0$$ is $$n \mod 3 \neq 0 \Rightarrow n^2 \mod 3 \neq 0$$. This is what Bill is showing. There are two options for $n$ if it is not $0$ mod 3... n = 3k + 1 or n = 3k + 2 (here Bill and I are using "n" in different ways...) – The Chaz 2.0 Mar 22 '12 at 15:50
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1@TheChaz Indeed, being an exercise in a calculus book, it may be intended to illustrate proof by contradiction /contrapositive. Probably one cannot assume known any nontrivial number theory. – Bill Dubuque Mar 22 '12 at 15:54
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The natural way to think about the problem is that since $n^2$ is divisible by 3, hence prime factorization of $n^2$ contains at least one 3 in it(since 3 is a prime number). If so is the case, then prime factorization of $n$ must contains 3 in it.
quartz
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