Advanced arithmetic question

[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 \(\displaystyle n=3k\)\(\displaystyle \Longrightarrow n\)\(\displaystyle \left 3k (n+1 \right)\)\(\displaystyle \left ( 2n+1 \right )=3k'\)

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

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

Thank you very much for your attention

 

[Deleted member 4993]

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

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

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

Thank you very much for your attention

You have shown that n(n+1)(2n+1) is divisible by 3.

You know that n(n+1) is divisible by 2.

Where are you stuck?

 

[guitarguy]

I find this problem interesting and I agree with the result.

There are some terms and notation I am unfamiliar with.

What do congruence and disjunction mean?

What is module three?

And, what do k', k'' and k''' stand for?

Thanks.