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I've tried solving this equation for my assignment by replacing $n$ by any of the following possible integers

  • a number divisible by $3 (3k$)
  • $1 \mod 3 (3k+1)$
  • $2 \mod 3 (3k+2)$

However, I found that none of them are divisible by $3$ this way. Is there something I am doing wrong?

In my assignment, b, c, and d are equally impossible for me to prove. enter image description here

Any help is appreciated

  • 0
    "*However, I found that none of them are divisible by $3$ this way. Is there something I am doing wrong?*" Yes, apparently there is. $n\equiv 0\pmod{3}$ implies $n$ is divisible by three. $n\equiv 1\pmod{3}$ implies $2n+1$ is divisible by three. $n\equiv 2\pmod{3}$ implies $n+1$ is divisible by three... – 2017-02-22
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    Please ask separate questions if you wish to receive feedback about (c) and (d). – 2017-02-22
  • 1
    The end result though is that all four of the parts to this question are proven in exactly the same way. – 2017-02-22
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    If n = 3k then 3|n. If n = 3k +1 then 3|2n + 1 = 2(3k+1) + 1 = 6k + 3. If n= 3k + 2 then 3|n + 1 = (3k +2) + 1 = 3k+3. So... try harder. – 2017-02-22

4 Answers 4

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if $$n=3m$$ then $3$ divides $$3m$$ which is clear, if $$n=3k+1$$ then $$2n+1=6k+2+1$$ is divisible by $3$ if $$n=3k+2$$ then $$n+1=3k+3$$ is divisible by $3$

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Here is a strategy that works whenever one wants to prove that at least one outcomes from of a list must take place: Suppose all but one of them fail, and show that the final outcome holds.

Consider the example at hand. If it happens that $3$ divides either $n$ or $n+1$, then there's nothing to show. We got one of the three desired outcomes by assumption.

Now, suppose that $3$ does not divide $n$ and $3$ does not divide $n+1$. We need to show that $3$ divides $2n+1$, or else our entire list of outcomes is false. Based on our assumption, we can conclude that $3$ divides both $n-1$ and $n+2$ (why?), so $3$ divides $(n-1)+(n+2)$, which is $2n+1$.

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There are three possibilities.

$n \equiv 0 \mod 3$ and therefore $3|n$ and $3 \not \mid n+1$ and $3 \not \mid 2n+1$.

$n \equiv 1 \mod 3$ and therefore $3|2n +1$ and $3 \not \mid n+1$ and $3 \not \mid n$.

Or $n \equiv 2 \mod 3$ and therefore $3|n+1$ and $3 \not \mid n$ and $3 \not \mid 2n + 1$.

But there is no way to know which without knowing more information about $n$.

Note that that although one of the three is divisible by $0$ it does imply that one of the three is congruent to $1$ and the other to $2$ mod $3$.

If $n \equiv 0 \mod 3$ then $n+1 \equiv 2n + 1 \equiv 1 \mod 3$.

If $n \equiv 1 \mod 3$ ten $n+1 \equiv 2 \mod 3$ and $2n+1 \equiv 0 \mod 3$.

If $n \equiv 2 \mod 3$ then $n+1 \equiv 0 \mod 3$ and $2n + 1 \equiv 1 \mod 3$.

So $n, n+1, 2n+1$ is not necessarily a reduced modulo class.

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If you want to get slick. Let $n = 3k + i$ where $i = 1, 0$ or $-1$.

Then $n+1 = 3k + i + 1 = 3k + j$ where $j = 2, 1$ or $0$ if $i = 1, 0 , -1$.

And $2n + 1 = 2(3k + i) = 3(2k) + 2k + i = 3N + l$ where $N=2k$ and $l = 0, 1, -1$ if $i = 1, 0, -1$.

So $3$ divides $2n+1, n$ or $n+1$ if $i = 1, 0$ or $-1$.

You can to the rest just as easily.

a) $n^3 - n = (3k + i)^3 - (3k + i) = 27k^3 + 27k^2i + 9ki^2 + i^3 - 3k - i = 3(9k^3 + 9k^2i + 3ki^2 - 3) + (i^3 - 1)$. So $1^3 = 1; 0^3 = 0; (-1)^3 = -1$, $i^3 - i = 0$.

c) $n + 2 = 3k + 2 +i$ and $2+i = 3, 2, 1$ while $n + 4 = 3k + 4 + i$ and $4+i = 5, 4, 3$.

d) $2n + 1 = 6k + 2i + 1$ and $2i + 1 = 3,1,-1$ and $2n -1 = 6k +2i - 1 = 1, -1, -3$.

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Regarding question d:

$3$ will divide either $a-1$, $a$, or $a+1$. It follows that $3$ will divide either $2n -1$, $2n$ or $2n+1$ and since if $3$ divides $2n$ it divides $n$ it is that $3$ will either divide $2n -1$, $n$ or $2n+1$