# Show that if $n$ is even and larger than 2, then $n^3−4n$ is always divisible by 48.

I have done the following but could not have made the final conclusion!

Base case: $$n=4$$ then $$(4^3-4 \times 4)|48=1$$ that holds true.

Hypothesis: assume $$n=k$$ then $$(k^3-4k)|48$$ always holds true so $$(k^3-4k)=48m$$ where $$m \in \mathbb{Z}$$.

Proof: take $$n=k+1$$ then show that $$((k+1)^3-4(k+1))|48$$ also holds true:

$$(k^3+3k^2+3k+1-4k-1)|48$$

$$((k^3-4k)+3k^2+3k)|48$$ $$(48m+3k^2+3k)|48$$ $$3(16m+k^2+k)|48$$

This is where I am stuck! I know $$16m+k^2+k$$ is already an integer but I cannot conclude the divisibility holds true in the last line. Any advice or clue?

• Since $n$ is even, the induction step is $k+2$, not $k+1$.
– lhf
Commented Oct 9, 2021 at 16:50
• I wonder why was this question downvoted. OP showed his work, it is a good question.
– Mark
Commented Oct 9, 2021 at 16:52
• Just a quick remark concerning your notation: As far as I know the expression $m \mid n$ usually means that $m$ divides $n$, you seem to use it the other way around or to denote a fraction like in $(4^3 - 4 \cdot 4)/48 = 1$. Commented Oct 9, 2021 at 16:54
• The clue is that you forgot that this only holds for even numbers. (For example, for $n=4$ your computation to $n+1=5$ will not work). Commented Oct 9, 2021 at 16:58

Let $$n=2k+2,$$ where $$k$$ is a natural number.

Thus, $$n^3-4n=n(n-2)(n+2)=8k(k+1)(k+2).$$ Now, we see that $$k(k+1)(k+2)$$ is divisible by $$6$$ because one of the numbers $$k$$, $$k+1$$ and $$k+2$$ is divisible by $$3$$ and one of the numbers $$k$$ and $$k+1$$ is divisible by $$2$$, which gives that you wish.

• @bof In the given $n>2$. Commented Oct 9, 2021 at 17:08
• @MichaelRozenberg But he’s right, your substitution assumes $n=2k$, else you’d get $8k(k+1)(k+2)$.
– Lazy
Commented Oct 9, 2021 at 17:09
• @Lazy Thank you! I fixed Commented Oct 9, 2021 at 17:11
• @bof The statement is generally true for any even integer.
– Lazy
Commented Oct 9, 2021 at 17:12
• @MichaelRozenberg how can we see $k(k+1)(k+2)$ is divisible by 6? Also, how come do you take $n=2k+2$ since we already know $k$ ought to be even. Commented Oct 9, 2021 at 18:14

As lhf has stated in the comments you need to make the step to $$k+2$$. Then $$(k+2)^3-4(k+2) - (k^3-4k) = 6k^2+12 k$$ If $$k=2l$$ then this is $$6\cdot 4 l^2+12\cdot 2 l = 48 l$$.

But rather you can prove this directly: If $$n=2k$$, then $$n^3-4n = 8k^3-8k = 8k(k^2-1)$$ Now we know that either $$2|k$$ or $$2|k^2-1$$ (if $$k\equiv 1\mod2$$ then $$k^2\equiv1\mod 2$$) and we know that either $$3|k$$ or $$k^2\equiv 1\mod 3$$, so $$3|k^2-1$$. Thus $$6|k(k^2-1)$$ and thus $$8\cdot6=48|n$$.

By the way: We do not need $$n>2$$. It holds true also for $$n=2$$ and $$n=0$$ and generally for every even integer.

• in the second line, it should be $6k^2+12k+4$, shouldn't it? Commented Oct 9, 2021 at 19:10
• No, as $(k+2)^3-4(k+2) = k^3+6k^2+12k+8-4k-8$, so the constant term cancels out, and $k^3-4k$ is known to be a multiple of $48$ by induction hypothesis.
– Lazy
Commented Oct 9, 2021 at 19:28