Prove $\frac{a(a^2+2)}{3}$ is an integer for all integer $a\geqslant 1$

Question

If $\frac{a(a^2+2)}{3}$ then $3\mid{a(a^2+2)}$.

By induction:

Lets define the set, $S=\left\{a\in N:a\geqslant1, 3\mid a(a^2+2) \right\}$

If $a=1$ then, $1\in S$

So we have to prove that if $k(k^2+2)=3m$ then $(k+1)((k+1)^2+2)=3n$ with $m,n\in Z$

If $k(k^2+2)=3m$ then,

$\begin{align*}k(k^2+2)+3(k^2+k+1)=&3m+3(k^2+k+1)\\=&3(m+k^2+k+1)\end{align*}$

Also,

$\begin{align*}k(k^2+2)+3(k^2+k+1)=&k^3+2k+3k^2+3k+3\\=&k^3+2k^2+k^2+2k+3k+3\\=&k^2(k+2)+k(k+2)+3(k+1)\\=&(k+2)(k^2+k)+3(k+1)\\ =&k(k+2)(k+1)+3(k+1)\\=&(k+1)(k(k+2)+3)\\ =&(k+1)(k^2+2k+1+2)\\ =&(k+1)((k+1)^2+2)\\ =&3(m+k^2+k+1) \end{align*}$

where $n=m+k^2+k+1$

Therefore,

$3\mid(k+1)((k+1)^2+2)$

Can I do it simplier using induction?

Answer 1

Yes, we can give a simpler and more conceptual inductive proof. Notice that

$\qquad\qquad a(a^2+2)\, =\, a(a^2\!-\!1 + 3)\, =\, \color{#0a0}{(a-1)a(a+1)} + 3a$

so it suffices to show that one of any $\rm\color{#0a0}{3\ consecutive\ integers}$ is divisible by $3$

This has a simple inductive proof. Note that shifting such a sequence by one simply replaces the old least element $\,\:\color{#C00}n\,$ by the new greatest element $\,\color{#C00}{n+3}$

$$ \begin{array}{} \:\color{#C00}n & n+1 & n+2 \\ \to & n+1 & n+2 & \color{#C00}{n+3} \end{array}$$

Since $ \: \color{#C00}n\equiv \color{#C00}{n+3} \pmod{\! 3},\,$ the shift does not change the set of remainders $\bmod 3$ of the elements. Thus the remainders remain the same as in the base case $ \ 0,1,2\: =\: $ all possible remainders mod $ \,3.\,$ Therefore the sequence has an element with remainder $\,0,\,$ i.e. an element divisible by $ \,3.$

Remark $ $ The same method works to show that a sequence of $d$ consecutive integers contains a multiple of $d.\,$ Alternatively this can be proved by using division with remainder (which has a natural proof by induction), which is closely connected to the above method.

Answer 2

Here is an alternative proof that doesn't use induction - just for fun!

Let $a \in \mathbb{Z}$. Observe that $a^3 - a = a(a^2 - 1) = (a-1)a(a+1)$ is the product of three consecutive integers, hence divisible by three. Adding $3a$ to the initial expression does not alter divisibility by $3$.

In particular, $a^3 - a + 3a = a^3 + 2a = a(a^2 + 2)$ is divisible by $3$, which means that it is still an integer after dividing by $3$. QED

(A ridiculous version of the above: $a(a^2+2)$ is the sum of $a^3 - a$ and $3a$, where the latter addend is a multiple of three by observation, and the former is divisible by three by Fermat's Little Theorem; so, their sum is divisible by three, and its quotient by three is an integer. QED)

Answer 3

I'd suggest writing the proof in a more linear fashion which is easier to read. Below is how I would approach the induction step.

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Let $P(n)$ be the statement $3\mid n(n^2+2)$.

Proving $P(k)\implies P(k+1)$:

$$P(k)\implies 3\mid k(k^2+2)$$ $\implies$ $$3\mid k(k^2+2)+3(k^2+k+1)$$ $\implies$ $$3\mid k^3+3k^2+5k+3$$ $\implies$ $$3\mid(k+1)(k^2+2k+3)$$ $\implies$ $$3\mid(k+1)((k+1)^2+2)$$ $\implies$ $$P(k+1)$$

Answer 4

If a is divisible by 3, then 3 divides a, and therefore a(a^2 + 2).

Otherwise, $a = 3k ± 1$, and 3 divides $(a^2 + 2)$ = $((3k ± 1)^2 + 2)$ = $((9k^2 ± 6k + 1) + 2)$ = $(9k^2 ± 6k + 3)$.

Answer 5

Any number is congruent to $0$, $1$ or $2$ modulo $3$, and upon squaring you get any square is congruent to $0$, $1$ or $1$ modulo $3$. Now take an arbitrary number $a$. If $3$ divides $a$ then you're done, if not $a^2+2$ is, by the above, congruent to $0$ modulo $3$ since $1+2=3$, and so you're good, too.

Answer 6

Work backwards. It's far more natural.

$(k+1)((k+1)^2+2)=$

$(k+1)(k^2+2k+1+2)=$

$(k+1)(k^2+2)+(k+1)(2k+1)=$

$k (k^2+2)+(k^2+2)+(k+1)(2k+1)=$

$3m+(k^2+2)+(2k^2+3k+1)=$

$3m+3k^2+3k+3$.

It is sometimes easier to subtract results.

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

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

$k^3+2k^2+k +k^2+2k+1+2-k^3=$

$3k^2+3k+3$

is a multiple of 3.

But it's probably easiest not to use induction at all.

Let $a \equiv i \mod 3;i=0,\pm 1$

If $i=0$ then $3|a $ so $3|a (a^2+2) $

Otherwise $i=\pm 1$

So $a^2+2\equiv i^2+2\equiv 1+2\equiv 0\mod 3$

So $3|a^2+2$ so $3|a (a^2+2) $.