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Within a quite different context, I realized that ($p$ is an odd prime): $$\sum_{i=1}^{m_a}a^i\equiv 0\pmod p$$ where $m_a$ is the multiplicative order of $a\in\mathbb Z/p\mathbb Z\setminus\{1\}$. I'm quite illiterate of (elementary) number theory, so I ask how goes the standard proof of this result, or a reference for it.

Kan't
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    This looks interesting. In what context did you realize this was true? – Zima Feb 04 '25 at 09:20
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    When $a \equiv 0 \pmod{p}$, the congruence is trivially true. Else, the summation on the LHS is equal to $1+ a + a^2 + \cdots + a^{m_a-1} = \frac{a^{m_a}-1}{a-1} \equiv 0 \pmod{p}$ (since $a^{m_a} \equiv 1 \pmod{p}$ and $a-1$ is invertible $\pmod{p}$). – Anuradha N. Feb 04 '25 at 09:41
  • Thanks, @AnuradhaN. Can you provide a reference for the mentioned sum result, please? – Kan't Feb 04 '25 at 09:44
  • @Kan't The result is so basic that it does not require a reference. But maybe you can check out Ireland and Rosen's "A classical introduction to modern number theory". You can find more general results of similar nature there. – Anuradha N. Feb 04 '25 at 09:49
  • @Kan't At least in number theory, this result and its proof is basic. A group-theoretic proof seems unnecessarily complicated. – Anuradha N. Feb 04 '25 at 09:53
  • I believed so, @AnuradhaN., and in fact by "reference" I hoped just a (wiki?) web page. But searching by "sum of powers" I got only "power sum" results, which is a completely different concept. Thanks, though. – Kan't Feb 04 '25 at 09:54
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    @Kan't Search for sums of roots of unity; you may find it on wikipedia. – Anuradha N. Feb 04 '25 at 09:58
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    @Kan't sums of roots of unity in finite fields. I could not edit my comment above as the time limit had lapsed. – Anuradha N. Feb 04 '25 at 10:08
  • See the Lemma and Theorem in the linked dupe. $\ \ $ – Bill Dubuque Feb 04 '25 at 11:11
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    here is another duplicate. – lulu Feb 04 '25 at 21:39
  • Indeed way more appropriate, @lulu. Thanks – Kan't Feb 05 '25 at 04:50
  • Does the formula of the partial sum of the geometric series holds for every field, @lulu? (Not for every ring, I guess, as there inverses set in.) – Kan't Feb 05 '25 at 04:57
  • @Kan't there's no need for geometric series...if the sum is $S$ then $aS\equiv S$ and if $a\not \equiv 1$ then this implies $S\equiv 0$. That's the argument I would look to generalize – lulu Feb 05 '25 at 05:03

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Since $a^k \pmod{p}$ is periodic with period $m_a$ (by definition of the multiplicative order), we can say that

$$ \sum_{i=1}^{m_a} a^i \equiv \sum_{i=0}^{m_a-1} a^i \pmod{p}. $$ It's also well-known that, by the sum of a geometric series, $$ \sum_{i=0}^{m_a-1} a^i = \frac{a^{m_a}-1}{a-1}. $$ So now we're left with $$ \frac{a^{m_a}-1}{a-1} \equiv k \pmod{p}, $$ where we have to show that $k=0$. Since $p$ is prime, every element of $\mathbb{Z}/p\mathbb{Z}$ except $0$ has a modular multiplicative inverse, including $a-1$ (we know that $a-1$ can't be $0$ because $a$ can't be $1$). Let $b$ be this value. Multiplying both sides of the congruence by $b$, we're left with $$ a^{m_a}-1 \equiv kb \pmod{p} $$ By the definition of the multiplicative order, $a^{m_a} \equiv 1 \pmod{p}$, and we're finally left with $kb \equiv 0 \pmod{p}$. We know that $b \not\equiv 0 \pmod{p}$, so the only valid conclusion is that $k \equiv 0 \pmod{p}$.

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