Divisibility of $n^4 -n^2$ by 4 (induction proof)

Question

We have to show that $$ n^4 -n^2 $$ is divisible by 3 and 4 by mathematical induction
Proving the first case is easy however I do not know how what to do in the inductive step.
Thank you.

Answer 1

Doing it by the book, though you were already told there are simpler/faster ways:

For $\;n=1:\;\;1^4-1^2=0\;\color{green}\checkmark\;$

Suppose it is true for $\;n\;$ and prove for $\;n+1\;$

$$(n+1)^4-(n+1)^2=(n+1)^2\left((n+1)^2-1\right)=(n+1)^2(n+1-1)(n+1+1)=$$

$$(n^2+2n+1)(n^2+2n)=n^4+4n^3+5n^2+2n=n^4-n^2+2n(2n^2+3n+1)=$$

$$=n^4-n^2+2n(n+1)(2n+1)$$

Now observe that the factor $\;2n(n+1)(2n+1)\;$ is always divisible by $\;4\;$ (as either $\;n\;$ or $\;n+1\;$ is even), and also by $\;3\;$ (since either $\;n\;$ is, or $\;n=1\pmod 3\;$ and then $\;2n+1=0\pmod 3\;$ , or $\;n=2\pmod3\;$ and then $\;n+1=0\pmod3\;$)

Answer 2

Let's prove that if $n^4-n^2$ is a multiple of both $3$ and $4$, then so is $(n+1)^4-(n+1)^2$.

Consider $ ((n+1)^4-(n+1)^2)-(n^4-n^2)=4 n^3+6 n^2+2 n $.

Ignoring $6n^2$, we have $4 n^3+2 n=3n^3+3n+n^3-n=3n^3+3n+6\binom{n+1}{3}$, which is always a multiple of $3$.

Ignoring $4n^3$, we have $6 n^2+2 n=4n^2+2n(n+1)=4n^2+4\binom{n+1}{2}$, which is always a multiple of $4$.

We're done.

This problem is definitely one which is much easier not by induction but instead simply by arguing that $n^{4} - n^{2} = n^{2} (n + 1) (n - 1)$, as @eltonjohn suggested.

Answer 3

Since $a_n=n^4-n^2=n^2(n-1)(n+1)$, we can also write $$ a_{n+1}=(n+1)^4-(n+1)^2=(n+1)^2n(n+2) $$ Suppose $a_n=n^4-n^2$ is divisible by $3$; then $3$ divides one among $n$, $n-1$ and $n+1$. Since $n$ and $n+1$ appear in the decomposition of $a_{n+1}$. The only remaining case is when $3$ divides $n-1$; but in this case we can use $n+2=(n-1)+3$ and we are done.

Suppose $4$ divides $a_n$. Then either $2\mid n$ or $2\mid(n+1)$ (and also $n-1$). In the case $2\mid(n+1)$, we have $4\mid(n+1)^2$. In the case $2\mid n$, we have $2\mid(n+2)$, so $4\mid n(n+2)$.

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Using Fermat's little theorem is of course easier; since $n^3\equiv n\pmod{3}$, $$ n^4-n^2\equiv n\cdot n^3-n^2\equiv n^2-n^2\equiv0\pmod{3} $$ Since $n^2\equiv n\pmod{2}$, we have \begin{align} n^2-n&\equiv n-n\equiv0\pmod{2}\\[4px] n^2+n&\equiv n+n\equiv2n\equiv0\pmod{2} \end{align}

Is this a proof by induction? Well, yes! One uses induction for proving Fermat's little theorem. ;-)