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I am seeking to prove the following:

If $n \in \mathbb{Z}$, then exactly one of the following is true: $\frac{n}{3} \in \mathbb{Z}, \frac{n + 1}{3} \in \mathbb{Z}, \frac{n + 2}{3} \in \mathbb{Z}$.

I believe this is pretty self evident, and hence have no idea where to start. Any tips would be greatly appreciated!

Robert Soupe
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Will Pike
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  • Hint; Induction for non-negative $n$, then the negative values follow. – Thomas Andrews Mar 04 '15 at 02:40
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    Or you could break it into cases. If $n$ is a multiple of $3$, you're done. Otherwise, $n = 3k + 1$, or $n = 3k + 2$. Hopefully the ability to write $n$ like this is just as obvious, and you don't feel you need to prove that too! – pjs36 Mar 04 '15 at 02:43

5 Answers5

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By the Division Algorithm $\, n+2 = 3q+r,\,$ for some $\, r\in\{0,1,2\}$

Therefore we conclude that $\ n+\! \underbrace{r'}_{\large 2\,-\,r}\!\! = 3q,\, $ for some $\,r'\in \{0,1,2\}$

Bill Dubuque
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Induction proof:

$0$ is divisible by $3$.

If one of $n,n+1,n+2$ is divisible by $3$ then one of $(n+1)+2,(n+1)+1,(n+1)$ is divisble by $3$.

Prove the negative case from the positive case, or prove it be induction, too.

The more general theorem is that if $a,b$ are integers with $b\neq 0$ then there exists $q,r$ so that:

$$a=bq+r,\,r<|b|$$

This is called the division algorithm.

Thomas Andrews
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You can use the Division Algorithm to write $$ n=3k+r $$ where $0\leq r<3$. If $r=0$ then $n$ is divisible by $3$. If $r=1$, then $n+2$ is divisible by $3$. If $r=2$, then $n+1$ is divisible by $3$.

J126
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Since $Z$/3 has 3 elements,they are 0,1,2.
That means every integer has one of the form as following: 3k , 3k+1 , 3k+2, k is an integer
so you can divide this problem into three cases.
(1)if n=3k, then n/3=k;
(2)if n=3k+1, then (n+2)/3=k+1;
(3)if n=3k+2, then (n+1)/3=k+1;
Q.E.D

You can also try to prove 3 divides $n(n+1)(n+2)$
That implies 3|n or 3|(n+1) or 3|(n+2)

yakaqi
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    The statement that $\mathbb Z/3$ has three elements is a more complicated statement that the above theorem. Your second approach is good, because $3$ is prime, but it wouldn't general to $6$ dividing one of $n,n+1,\cdots,n+5$. – Thomas Andrews Mar 04 '15 at 03:01
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If $n \in \mathbb{Z} \to$ $n = 3k$ or $3k+1$ or $3k+2$, hence one of the statements must be true.

user642796
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DeepSea
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    You basically assert a theorem which is equivalent. – Thomas Andrews Mar 04 '15 at 02:46
  • @ThomasAndrews Do you think so? I find it much easier to convince myself of this fact, than the assertion that one of the numbers is divisible by $3$. Perhaps because I'm used to the $2k$, or $2k + 1$ way of thinking of even/odd numbers. – pjs36 Mar 04 '15 at 02:49
  • It's still an assertion, and it is an equivalent statement. That's all I said. – Thomas Andrews Mar 04 '15 at 02:50