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Let $p$ be a prime of the form $p = 8k + 5$, and let $g$ be a primitive root modulo $p$.

Prove that: $$ (g^4 + 1)(g^8 + 1)(g^{12} + 1)\cdots(g^{4k} + 1) \equiv g^{k(k+1)} \pmod{p}. $$

Here's what I found:

Since $x^{(p-1)/4} - 1 = (x - g^0)(x - g^4)(x - g^8) \cdots (x - g^{8k})$ in $\mathbb{F}_p$, substituting $x = -1$ gives $$ (g^4 + 1)(g^8 + 1) \cdots (g^{8k} + 1) \equiv 1 \pmod{p}. $$

Now multiply both sides by $g^4 \cdot g^8 \cdots g^{4k}$. That gives: $$ \left[(g^4 + 1)(g^8 + 1) \cdots (g^{4k} + 1)\right]^2 \equiv g^{4 + 8 + \cdots + 4k} = g^{2k(k+1)} \pmod{p}. $$

So we get: $$ (g^4 + 1)(g^8 + 1) \cdots (g^{4k} + 1) \equiv \pm g^{k(k+1)} \pmod{p}. $$

However, I’m still stuck on how to determine the sign.

Bill Dubuque
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HarryD
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    $(g^4+1)(g^8+1)(g^{12}+1)\cdots(g^{4k}+1)\equiv \pm g^{2+4+\cdots+2k}$ is equivalent to $(g^2+g^{p-3})(g^4+g^{p-5})\cdots(g^{2k}+g^{p-1-2k})\equiv\pm1$, and if $\zeta=\exp(2\pi i/(p-1))$ then $(\zeta^2+\zeta^{p-3})(\zeta^4+\zeta^{p-5})\cdots(\zeta^{2k}+\zeta^{p-1-2k})=1$ not $-1$, because all the factors are positive real numbers – J. W. Tanner Jul 08 '25 at 14:53
  • Thanks! I'm guessing instead of modulo, we map it to rotation around the unit circle for easier evaluation right? This is the first time I've seen this technique, do you know where I can read more on this and is this practical for math olympiads? @J.W.Tanner – HarryD Jul 08 '25 at 18:37
  • That was just some thought off the top of my head. I haven't participated in math olympiads in decades – J. W. Tanner Jul 08 '25 at 19:04
  • Ah a nice primitive roots question +1 – mick Jul 21 '25 at 23:01

1 Answers1

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Not an answer, but (maybe?) some progress: from observation for low $k$s it seems that the polynomial $(x^4+1)(x^8+1)\cdots (x^{4k}+1) -x^{k(k+1)}$ can be factored such that

Edit $$ \tag1 (x^4+1)(x^8+1)\cdots (x^{4k}+1) -x^{k(k+1)}= \Phi_{4k+2}(x^2)P_k(x^2)= \Phi_{8k+4}(x)P_k(x^2),$$ where $\Phi_n$ is the $n^{th}$ cyclotomic polynomial and $P_k$ is a polynomial with integer coefficients. This would entail the expected outcome, since when $g$ is a primitive root modulo $p=8k+5$ it is easy to see that $\Phi_{8k+4}(g)\equiv 0 \pmod p$.

But Equation $(1)$ needs a proof...

Second Edit ... which is given below in the comment of @mathematician123.

Third Edit ...and reproduced hereafter as suggested by @J.W.Tanner. $$ (g^4 + 1)(g^8 + 1) \cdots (g^{4k} + 1)\equiv \pm g^{k(k+1)} \pmod{p}. $$ is equivalent to $$ (g^2 + g^{-2})(g^4 + g^{-4}) \cdots (g^{2k} + g^{-2k}) \equiv \pm 1 \pmod{p}. $$ Let $\zeta=\mathbb{exp}\frac{2\pi i}{p-1}$ a primitive $(p-1)^{th}$ root of unity, we also have $$ (\zeta^2 + \zeta^{-2})(\zeta^4 + \zeta^{-4}) \cdots (\zeta^{2k} + \zeta^{-2k}) = \pm 1, $$ because the multiplicative group of $\mathbb{Z}/p\mathbb{Z}$ is isomorphic to the cyclic group of order $p-1$. So we have $$\prod_{j=1}^{k}2\cdot\mathbb{cos}\frac{j \pi}{2k+1}= \pm 1$$ and we see that it cannot be $-1$, because all factors in the product are positive. This establishes that $\zeta$ is a root of the polynomial $ Q(x)= (x^4+1)(x^8+1)\cdots (x^{4k}+1) -x^{k(k+1)}$. From field theory it is known that any algebraic number is a root of its minimal polynomial. The minimal polynomial of $\zeta$ is $\Phi_{8k+4}$ and any polynomial of $\mathbb{Q}[x]$ that vanishes on $\zeta$ is a multiple $\Phi_{8k+4}$. Then $\frac{Q(x)}{\Phi_{8k+4}(x)} \in \mathbb{Q}[x]$. And it follows from Gauss's lemma that all the coefficients of $\frac{Q(x)}{\Phi_{8k+4}(x)}$ are integers.

René Gy
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    I think the comment of J. W. Tanner helps here. We already know that $\zeta=\exp(2\pi i/(p-1))$ is a root of $(x^4+1)(x^8+1)\cdots (x^{4k}+1) -x^{k(k+1)}$. It remains to show that $\zeta=\exp(2\pi it/(p-1))$ is a root of this polynomial if $\gcd(t, p - 1) = 1$. And I guess this is true because $ \prod_{j = 1}^k (\zeta^{2j} + \zeta^{-2j}) = \prod_{j = 1}^k 2\cos(\frac{4\pi tj}{p - 1})$ contains an even number of negative numbers. At least, I proved this for $t = 3$. – mathematician123 Jul 13 '25 at 21:06
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    Everything is much simpler. We already know that $\zeta=\exp(2\pi i/(p-1))$ is a root of $P(x) = (x^4+1)(x^8+1)\cdots (x^{4k}+1) -x^{k(k+1)}$. From field theory it is known that any algebraic number is a root of a minimal polynomial). $\Phi_{8k + 4}$ is a minimal polynomial of $\zeta$. Any polynomial that vanishes on $\zeta$ is a multiple $\Phi_{8k + 4}$. So $\frac{P(x)}{\Phi_{8k + 4}} \in Q[x]$. And it follows from Gauss's lemma that $\frac{P(x)}{\Phi_{8k + 4}} \in Z[x]$. – mathematician123 Jul 14 '25 at 11:20
  • You should put the proof in the answer. Comments could be deleted – J. W. Tanner Jul 20 '25 at 03:42