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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 [itex]n=3k \Longrightarrow n 3k\left ( n+1 \right )[/itex][itex]\left ( 2n+1 \right )=3k'[/itex]

If [itex]n=3k+1 \Longrightarrow 2n+1=6k+3[/itex] [itex]3k\left ( n+1 \right )\left ( 2k+1 \right )=3k''[/itex]

If [itex]n=3k+2 \Longrightarrow n+1=3k+3[/tex] and [tex]3n\left ( k+1 \right )\left ( 2n+1 \right )=3k'''[/itex]

Thank you very much for your attention

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# Question about arithmetic

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