Proof by induction that $(n-1)^3 + (n+1)^3$ is divisible by 4 for all $n \in \mathbb{N}^+$

Question

I want to prove that $(n-1)^3 + (n+1)^3$ is divisible by 4 for all $n \in \mathbb{N}^+$. I have looked up solutions for similar problems but cannot make it work for this one. The expression $(n-1)^3 + (n+1)^3$ can be simplified to $2n^3 + 6n$ which, if assumed to be divisible by $4$ (induction hypothesis), could be written as $4m$ (m is a positive integer). So we get: $$2n^3 + 6n = 4m$$ Now, assuming that $(n-1)^3 + (n+1)^3$ divides 4 for n, we want to prove that it also applies to $n+1$: So we get: $$2(n+1)^3+6(n+1) = 2n^3 + 6n+6n^2+6n+8 = [2n^3 + 6n = 4m] = 4m + 8 + 6n^2 + 6n$$

This is where I get stuck because we have two $6$ coefficients for $n^2$ and $n$ in the expression. I can tell that if I write it as $2(3n^2 + 3n)$ then its clear to me that if n is an even number then the expression would be divisible by $4$ and if it's an odd number then the sum of two odd numbers would give an even number (and since it's multiplied by $2$) which would also be divisible by $2$. But this doesn't satisfy me and I'm sure there should be more convincing proof of this.

Answer 1

I think you have confusion in inductive step.
Inductive step: Suppose inductively that $(n-1)^3+(n+1)^3$ is divisible by $4$ for some $n \in \mathbb N$, i.e. $(n-1)^3+(n+1)^3=4q$ for some integer $q$. Which is equivalent to $2n^3+6n=4q$. Then
$$n^3+(n+2)^3=2n^3+6n^2+12n+8 \\ =4q +12n+8 \\ =4(q+3n+2)$$ Since $q+3n+2 \in \mathbb Z$. Thus we conclude that the statement is true for $n+1$. This closes the induction.

Answer 2

Things might get a bit messy with induction so let's see another approach:

We check the cases, $n$ is even and $n$ is odd.

Let $n=2k+1, k\in \mathbb{N}$.

Then we have $(n-1)^3+(n+1)^3=(2k)^3+(2k+2)^3=8k^3+8(k+1)^3=4(2k^3+2(k+1)^3)=4m$ and we are done.

Let $n=2k, k\in \mathbb{N}$.

Now we have $(n-1)^3+(n+1)^3=(2k+1+2k-1)(\text{some terms})=4k(\text{some terms})=4km$ and again we are done.

Answer 3

If $f(m)=(m-1)^{3}+(m+1)^{3}$ is divisible by $4$, it is easy to show that $f(m+2)$ is also divisible by $4$ because:

$$ \begin{aligned} f(m+2)-f(m)&=(m+3)^{3}-(m-1)^{3}\\ &=12m^{2}+24m+28 \end{aligned} $$

$f(1)$ and $f(2)$ are divisible by $4$ and we're done.

Answer 4

We will continue from where you left off. You deduced $f(n) = (n - 1)^3 + (n + 1)^3$ simplifies to $f(n) = 2n^3 + 6n$. We note the base case $f(1)$ is divisible by $4$. Applying induction we get:

$$ f(n + 1) = 2(n + 1)^3 + 6n + 6 = f(n) + 8 + 6n^2 + 6n $$

which implies $f(n + 1)$ is divisible by $4$ if $g(n) = 6n^2 + 6n$ is divisible by $4$. We show this using induction:

$$ g(n + 1) = 6(n + 1)^2 + 6(n + 1) = g(n) + 12(n + 1) $$

and so $g(n + 1)$ is divisible by $4$ if $g(n)$ is divisible by $4$. Furthermore the base case $g(1)$ is divisible by $4$ and so we conclude $g(n)$ is divisible by $4$ and therefore that $f(n)$ is divisible by $4$ for all positive naturals.