1. May 17, 2012

### inverse

Hello,

The formulation of the question says:

Show, using congruences and disjunction of cases, which,for all natural n є Ninteger, the integern (n+1) (2n+1) is a multiple of 6.

Simultaneously, themultiple of 3 is a multiple of 2 (number), then it is divisible by 6

3 cases are possible by applying the congruence module 3:

If $n=3k \Longrightarrow n 3k\left ( n+1 \right )$$\left ( 2n+1 \right )=3k'$

If $n=3k+1 \Longrightarrow 2n+1=6k+3$ $3k\left ( n+1 \right )\left ( 2k+1 \right )=3k''$

If $n=3k+2 \Longrightarrow n+1=3k+3[/tex] and $$3n\left ( k+1 \right )\left ( 2n+1 \right )=3k'''$ Thank you very much for your attention 2. May 17, 2012 ### DonAntonio You should be more careful before you send over a post, and that's what the "preview post" button exists for. You want to know for what positive integers n, the number $\,(n+1)(2n+1)$ is a multiple of 6. Your division in cases is fine, but for example: [tex]n=3k\Longrightarrow (n+1)(2n+1)=(3k+1)(6k+1)$$ which is never a multiple of 6 (I just can't understand what you did in this case)

Simmilarly for the next case, $$n=3k+1\Longrightarrow (n+1)(2n+1)=(3k+2)(6k+3)$$ which is multiple of 6 iff $\,k\,$ is even (why?).

Now try to do correctly the last case by yourself.

DonAntonio

3. May 17, 2012

### inverse

I cannot understand what are you asking me, can you ask the question again?
thanks

4. May 17, 2012

### ramsey2879

Don asked you to check your first post for errors and typos using the preview button. Looks like you should have corrected the typos as he only saw the last two factors since you left out a space between "integers" and "n". By the way you did not show that the product is divisible by 2, so your work is incomplete.