What is the remainder when the expression $$\prod_{n=1}^{100}(1- n^{2} +n^{4})$$ is divided by $101$?

If $\zeta=\dfrac{-1+\sqrt{-3}}{2}$, then $$1-n^2+n^4=(1-n+n^2)(1+n+n^2)=(-\zeta-n)(-\bar{\zeta}-n)(\zeta-n)(\bar{\zeta}-n).$$ We then have $$\prod_{n=1}^{100}(1-n^2+n^4)\equiv \prod_{n=1}^{100}\big((-\zeta-n)(-\bar{\zeta}-n)(\zeta-n)(\bar{\zeta}-n)\big)\pmod{101}\,.$$ Since $$\prod_{n=1}^{100}(x-n)\equiv x^{100}-1\pmod{101},$$ we obtain $$\prod_{n=1}^{100}(1-n^2+n^4)\equiv\big((-\zeta)^{100}-1\big)\big((-\bar\zeta)^{100}-1\big)\big(\zeta^{100}-1\big)\big(\bar{\zeta}^{100}-1\big)\pmod{101}\,.$$ Since $\zeta^3=1$ and $\bar{\zeta}^3=1$, we get $$(-\zeta)^{100}=\zeta^{100}=\zeta\text{ and }(-\bar\zeta)^{100}=\bar\zeta^{100}=\bar\zeta\,.$$ Therefore, $$\prod_{n=1}^{100}(1-n^2+n^4)\equiv (\zeta-1)^2(\bar{\zeta}-1)^2=\big((1-\zeta)(1-\bar{\zeta})\big)^2\pmod{101}\,.$$ As $$(x-\zeta)(x-\bar{\zeta})=x^2+x+1\,,$$ we get $$\prod_{n=1}^{100}(1-n^2+n^4)\equiv (1^2+1+1)^2=9\pmod{101}\,.$$ Are there other solutions? How do we solve this problem without resorting to complex numbers?

  • 1
    $\begingroup$ According to WolframAlpha, the remainder is $9$. $\endgroup$
    – an4s
    Jul 9, 2020 at 5:45
  • $\begingroup$ If $f(n) \equiv 1 - n^2 + n^4 \pmod {101}$, then $f(n) = f(-n) = f(101-n)$ since there are only even powers of $n$. $\endgroup$
    – Toby Mak
    Jul 9, 2020 at 5:52
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    $\begingroup$ Have you seen the answers on AoPS? You should always use Approach0 to search if there are any duplicates, instead of us having to find them for you. $\endgroup$
    – Toby Mak
    Jul 9, 2020 at 5:55
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    $\begingroup$ Might be able to use: $n^4-n^2+1=\frac{n^6+1}{n^2+1}$ $\endgroup$ Jul 9, 2020 at 5:59
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    $\begingroup$ Since $100$ is not divisible by $3,$ the values of $n^2+1$ and the values of $n^6+1$ are the same set of values, modulo $101.$ That means we only need to consider the entries when $n^2+1$ is divisible by $101.$ These are $n=10,91.$ This means the remainder is the same as the remainder of $(10^2-10+1)(91^2-91+1)$ which has remainder $1.$ $\endgroup$ Jul 9, 2020 at 6:08

1 Answer 1


Modulo $101$, the set of values $0^3, 1^3,\dots,100^3$ is a permutation of $0,1,2,\dots,100.$ This is because $101$ is prime and $3$ is not a divisor of $100.$

But $$n^4-n^2+1=\frac{n^6+1}{n^2+1}$$

Now, if $n=10,91$ then $n^2+1$ is divisible by $101.$ The other terms are a permutation, so:

$$\begin{align}\prod_{n=1}^{100} (n^4-n^2+1)&=(10^4-10^2+1)(91^4-91+1)\prod_{n\neq 10,91}\frac{n^6+1}{n^2+1}\\ &\equiv (10^4-10^2+1)((-10)^4-(-10)^2+1)\pmod{101}\\ &\equiv 3\cdot 3=9\pmod{101} \end{align}$$

This works more generally if $p\equiv 5\pmod {12}:$

$$\prod_{n=1}^{p-1}\left(n^4-n^2+1\right)\equiv 9\pmod p$$

If $p\equiv 11\pmod{12},$ the remainder is $1.$

I think when $p\equiv 1\pmod{12},$ the remainder is $0.$

Not sure about $p\equiv 7\pmod{12}.$

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    $\begingroup$ Nice! Too bad the question had appeared in AOPS earlier. $\endgroup$ Jul 9, 2020 at 6:31
  • $\begingroup$ The question and the answer, it appears. @JyrkiLahtonen Well, the main hint in the first part here. Oh well, it was fun to solve. $\endgroup$ Jul 9, 2020 at 6:34

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