Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$. [closed]

Question

Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$.

---

I was thinking of using induction, but wasn't really sure how to do it.

Answer 1

Hint: $$ n^2(n^2+1)(n^2−1)\cong n^2(n^2-4)(n^2−1) = (n-2)(n-1)(n^2)(n+1)(n+2) $$

Answer 2

Hint: Compute $n^2\bmod 5$ for $n=0,1,2,3,4$.

Answer 3

${\rm mod}\ 5\!:\,\ \color{#c00}n^2(\color{#0a0}{n^4-1})\equiv 0\, $ by $\,\color{#c00}{n\equiv 0}\,$ or $\,\color{#0a0}{n^4\equiv 1}\,$ by little Fermat.

Or, directly $\,\color{#c00}{n\equiv 0}\ $ or $\ n\equiv \pm1,\pm2\,\Rightarrow\, n^2\equiv \pm1\,\Rightarrow\, \color{#0a0}{n^4\equiv 1}$

Answer 4

Notice that $(n^2 - 1)(n^2 + 1) = n^4 - 1$. Fermat's little theorem tells us that $n^4 \equiv 1 \pmod 5$. This means that $n^4 - 1$ is a multiple of $5$ if $n$ is not, and therefore $n^2 (n^4 - 1)$ is also a multiple of $5$. For example, if $n = 2$, then $n^4 - 1 = 15$ and $n^2 (n^2 - 1)(n^2 + 1) = 60$.

This leaves us the case where $n$ is a multiple of $5$ to consider. Obviously $n^4 - 1$ is not a multiple of $5$. But $n^2$ is. For example, $n = 5$ gives us $15600$.

Answer 5

$n^2(n^2+1)(n^2-1)=n^2(n^4-1)$ and $n^4\equiv1\mod5$ by FLT for $n\in\{1,2,3,4\}$

Or, as FLT also states that $a^{p-1+k}\equiv a^k \mod p$, and as the equation is $n^6-n^2$, the fact is immediate.

Answer 6

If one wishes to use induction:

The statement holds for $n=0$, so let us assume it's true for $n=k$: $$k^2(k^2+1)(k^2-1)=(k-1)k^2(k+1)(k^2+1)=5m $$ and consider it for $n=k+1$. We have $$(k+1)^2((k+1)^2+1)((k+1)^2-1)=\\ (k+1)^2(k^2+2k+2)(k^2+2k)=\\ k(k+1)^2(k+2)(k^2+2k+2).$$ Suppose this isn't a multiple of $5$. Then none of its factor is, so our inductive hypothesis yields that for some integer $a,b$ either $k-1=5a$ or $k^2+1=5b$, or both:

• if both, then $$(k-1)+(k^2+1)=k(k+1)=5(a+b),$$ contradiction;
• if only the former, we have $$k+4=5(a+1),$$ whence $$ k^2+2k+2=25(a+1)^2-6k-14=25(a+1)^2-2(3k+7)=\\25(a+1)^2-2(3(k-1)+10)=5\left(5(a+1)^2-2\left(\frac{3(k-1)}{5}+2\right)\right),$$ contradiction;
• if only the latter, we also have for some integer $c$ $$k+3=5c,$$ whence $$k^2+2k+2=k^2+2k+7-5=(k^2+1)+2(k+3)-5=5(b+2c)-5=5(b+2c-1),$$ contradiction.

Answer 7

In case $n$ is a multiple of $5$ the $n^2$ on the rhs is also.

If $n=2,3\pmod 5$ then $n^2\equiv4\pmod5$ , hence $n^2+1$ would be a multiple of $5$.

In the remaining cases, i.e.,$n=1,4\pmod 5$ then in $n^2-1= (n+1)(n-1)$ one of the term on the right will be a multiple of $5$.

So $n^2(n^2+1)(n^2-1)$ is always a multiple of $5$.

Answer 8

Fermat's Little Theorem: $p$ prime $\,\Rightarrow\, a^p\equiv a\pmod {p}$.

Proof: Clearly $0^p\equiv 0$. Assume $a^{p}\equiv a$. Then $$(a+1)^p\,\stackrel{\text{BT}}\equiv \,a^p+\binom{p}{1}a^{p-1}+\binom{p}{2}a^{p-2}+\cdots+\binom{p}{p-1}a+1\equiv a^p+1\stackrel{\text{ind. hyp.}}\equiv a+1$$

$\text{BT}$ refers to binomial theorem. I used $\binom{p}{i}\equiv 0,\,\forall i\in\{1,2,\ldots,p-1\}$.

It follows from $\binom{p}{i}=\frac{p!}{(p-i)!i!}$ and $p\mid p!,\ \ p\nmid (p-i)!i!$.

$5$ prime $\,\Rightarrow\, n^2(n^2+1)(n^2-1)\equiv n^2(n^4-1)\equiv n(n^5-n)\equiv n\cdot 0\equiv 0\pmod {5}$.

Answer 9

Here's a proof based on the fact that among any five consecutive numbers, one of them must be divisible by $5$.

Suppose that $n^2(n^2+1)(n^2-1)$ is not divisible by $5$. Then either $n^2+2$ or $n^2-2$ is divisible by $5$. Let's write this as $n^2+2\sigma=5k$, where $|\sigma|=1$.

But if $n^2(n^2+1)(n^2-1)$ is not divisible by $5$, then neither is $n(n^2-1)=n(n+1)(n-1)$, and therefore $5$ divides either $n-2$ or $n+2$. Let's write this as $n+2\tau=5h$ with $|\tau|=1$.

Combining these cleverly, we can conclude $n^2-\sigma\tau n=5(k-\sigma\tau h)$, which implies $5$ divides $n(n-\sigma\tau)$, where $|\sigma\tau|=1$. But that implies $5$ divides $n(n-1)(n+1)$, which is a contradiction. Therefore $5$ must divide $n^2(n^2+1)(n^2-1)$.

Answer 10

If $n$ is divisible by five, then $n^2(n^2+1)(n^2-1)$ is divisible by $5$, since $n^2$ is a factor.

Now suppose that $n$ is not divisble by $5$. Then $n=5m+r$ for some $r=1,2,3,4$. Note that $(n^2 + 1)(n^2-1) = n^4-1$.

$$n^4 = (5m+r)^4 = (5m)^4 + 4 (5m)^3r + 6 (5m)^2 r^2 + 4 (5m) r^3 + r^4$$ by the binomial theorem. We see then that $n^4 = 5\cdot M + r^4$ for some $M$. Now we just need to show that $r^4-1$ is divisible by $5$ to establish that $n^4-1$ is divisible by $5$.

This does follow by Fermat's little theorem. If that sounds foreign to you, it's easy enough to check by hand. $r$ can take four different values, $1,2,3,4$ (remember we already assumed that $r\neq 0$ since we are assuming $n$ is not divisble by $5$).

$$1^4 -1 = 0$$ $$2^4 -1= 15$$ $$3^4 - 1 = 80$$ $$4^4 - 1 = 255$$

Thus we see that, if $n$ is not divisible by $5$, then $$n^4 -1 = 5 M + r^4 - 1$$ and both $5M$ and $r^4 - 1$ are divisible by five. Therefore, $n^4-1$ is divisible by five, and $$n^2(n^2+1)(n^2-1) = n^2(n^4-1)$$ is divisible by $5$.

Answer 11

Lemma:

For any polynomial of integer coefficients $P(n)$, $n=0,1,2,3,4$ exhaust the possible values of $P(n)\bmod 5$.

Indeed for any $a$, $(n+a)\bmod 5=(n\bmod5+a)\bmod5$, and $(na)\bmod5=((n\bmod5)a)\bmod5$, so that $P(n)\bmod5=P(n\bmod 5)\bmod5$. QED.

Now, $$0^2(0^4-1)\bmod5=0\\ 1^2(1^4-1)\bmod5=1\cdot0\bmod5=0\\ 2^2(2^4-1)\bmod5=4\cdot15\bmod5=0\\ 3^2(3^4-1)\bmod5=9\cdot80\bmod5=0\\ 4^2(4^4-1)\bmod5=16\cdot255\bmod5=0\\$$