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AleksandrR [38]
3 years ago
7

Let n be a natural number. Show that 3 | n3 −n

Mathematics
1 answer:
Vladimir [108]3 years ago
4 0
Let n=1. Then n^3-n=1^3-1=0. By convention, every non-zero integer n divides 0, so 3\vert n^3-n.

Suppose this relation holds for n=k, i.e. 3\vert k^3-k. We then hope to show it must also hold for n=k+1.

You have

(k+1)^3-(k+1)=(k^3+3k^2+3k+1)-(k+1)=(k^3-k)+3(k^2+k)

We assumed that 3\vert k^3-k, and it's clear that 3\vert 3(k^2+k) because 3(k^2+k) is a multiple of 3. This means the remainder upon divides (k+1)^3-(k+1) must be 0, and therefore the relation holds for n=k+1. This proves the statement.
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