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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.

Martin Sleziak
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Jack
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    Does it have to be by induction? It seems to be a lot simpler just to compute $n^4-n^2$ modulo 4 for $n=0,1,2,3$. – hmakholm left over Monica Mar 20 '16 at 11:23
  • It easy to do it using strong induction ,but it is required to be done using mathematical conduction for all numbers. – Jack Mar 20 '16 at 11:26
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    $n^{4} - n^{2} = n^{2} (n + 1) (n - 1)$ so you don't have to resort to induction to prove that it is divisible by $3$ and $4$ (for $n \in \mathbb{N}, n \geq 2$): one of $n - 1$, $n$,$n + 1$ must be a multiple of $3$ since they are consecutive integers. If $n$ is odd, then $n - 1$ and $n + 1$ are even. if $n$ is even then $n^{2}$ is divisible by $4$. – eltonjohn Mar 20 '16 at 11:32
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    Just for fun: $n^4-n^2=12 \binom{n}{2} + 36 \binom{n}{3} + 24 \binom{n}{4}$, which is clearly a multiple of $12$. – lhf Mar 20 '16 at 11:58
  • @lhf how did you come up with that? – Airdish Mar 20 '16 at 12:30
  • @TheOddbodNumber, repeated differences and Newton's formula. – lhf Mar 20 '16 at 13:29
  • @TheOddbodNumber Also called the binomial transform – Jean-Claude Arbaut Mar 21 '16 at 12:49

3 Answers3

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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 $\;n+1=0\pmod3\;$)

DonAntonio
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    Thank you, This is what i was looking for. I just missed the fact that either n or n+1 are even. – Jack Mar 20 '16 at 13:53
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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 $4\mid n(n+2)$.

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. ;-)

egreg
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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.

lhf
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  • How did you transition from $$n^3−n$$ to 6(n+1 3) Or how can we prove that the product of 3 consecutive numbers is divisible by 3 – Jack Mar 21 '16 at 18:36
  • @Jack, $\binom{n+1}{3}=\frac{(n+1)n(n-1)}{6}$. – lhf Mar 21 '16 at 18:43