0

This is a problem in a test stands as a simulation of the NCEE (the college entrance examination of China), it starts by giving a definition of the discrete log:

Let $p$ be a prime number, and let $X=\lbrace 1,2,...,p-1\rbrace$, define $u\otimes v$ to be $uv\bmod{p}$ for $u,v\in X$, and define $u^{m,\otimes}$ to be $u^m\bmod{p}$. Now let $a\in X$ and $1, a, a^{2,\otimes},...,a^{p-2,\otimes}$ are pairwise different, if $a^{n,\otimes}=b(n\in\lbrace0,1,...,p-2\rbrace)$, we say that $n$ is the discrete logarithm of $a$ to the base $b$, written $n=\log(p)_a b$.

Then it asks to prove:

  • Let $m_1,m_2\in\lbrace0,1,...,p-2\rbrace$, define $m_1\oplus m_2=(m_1+m_2)\bmod{(p-1)}$, prove that $\log(p)_a(b\otimes c)=\log(p)_a b\oplus \log(p)_a c$.
  • Let $n=\log(p)_a b$, for $x\in X$ and $k\in\lbrace 1,2,...,p-2\rbrace$, let $y_1=a^{k,\otimes}$, $y_2=x\otimes b^{k,\otimes}$, proves that $x=y_2\otimes y_1^{n(p-2), \otimes}$

My problem: Is there a proof of this problem using algebraic approach, not only number theory.

Background: I apologize for the non-standard and confusing notation here, but this is literally how it looks on the test paper. This is at first glance a problem of number theory on which I don't have any background (In fact any students should not be supposed to equipped such background in a normal senior school here because they never appear on the textbook), since I do have some algebra background and to me it has a strong algebraic feel, I tried to prove it using some group theory but failed because of the lack of knowledge from the number theory part. I check the answer and it seems that if you are of number theory or cryptography background, then this problem will become rather basic. But the answer used a pure number-theoretic approach, I'm wondering is there a way to prove it by leveraging the tools from algebras?

Bill Dubuque
  • 282,220
Dylech30th
  • 151
  • 6
  • I would definitely have not made it into college if this was an example question from an entrance exam. – abiessu Jan 19 '24 at 16:36
  • @abissu Well it is pretty weird because in here, number theory was never a part of the math education from junior to senior school, it just never appear, this is the 19th problem of the test paper, and normally the test paper would contain 3 extra problems which are sequences, calculus, and analytic geometry, respectively, but in this test the paper stops in this problem, so I guess the creator of this paper admits its difficulty and weirdness :) – Dylech30th Jan 19 '24 at 16:42
  • The notation is just awfully non standard, even though I am familiar with the above it took me a non zero time to decipher what is meant :/. I probably also wouldn't have passed such an entrance exam after high school. –  Jan 19 '24 at 16:51
  • Well first time I saw it I thought it was some kind of algebraic sum and product, until I find it seems to be irrelevant... – Dylech30th Jan 19 '24 at 16:53
  • Working in the ring $\Bbb Z/p\Bbb Z$ is algebra and number theory at the same time. There is no difference. So "not only number theory" is not possible. $p$ is a prime, so this is definitely number theory. But an identity or congruence is algebra. – Dietrich Burde Jan 19 '24 at 17:05
  • @DietrichBurde field :P –  Jan 19 '24 at 17:06
  • @eulersgroupie It is a ring (every field is a ring). – Dietrich Burde Jan 19 '24 at 17:06
  • @DietrichBurde I know but imo it should say field in your comment. –  Jan 19 '24 at 17:07
  • @eulersgroupie No, the point is that algebra treats this as rings first, because usually it is more generally $\Bbb Z/n\Bbb Z$ for all natural numbers $n$ (and then it is not always a field). But because of $n=p_1^{e_1}\cdots p_n^{e_n}$ as prime powers, it is very much number theory as well. – Dietrich Burde Jan 19 '24 at 17:08
  • As I recall, you can prove this using generators. I don't think you can escape using some variation of Fermat's Little Theorem. Here, there's a lot of overlap between Group Theory and Number Theory. – TurlocTheRed Jan 19 '24 at 17:55
  • Are you familiar with groups? This boils down to showing that an element of multiplicative order $,m, (= p!-!1,$ here) generates a cyclic group isomorphic to the additive group of $,\Bbb Z_m =$ integers $!\bmod m.\ \ $ – Bill Dubuque Jan 19 '24 at 18:08
  • For a simple worked example see this answer. To learn more see discrete logarithm and index calculus. $\ \ $ – Bill Dubuque Jan 19 '24 at 18:20
  • @BillDubuque Thanks for the inspiration! However could you please provide some more detail because I really haven't face any number theory problems or definitions, properties, theorems before which makes me have some problem in translating the original problem properly. – Dylech30th Jan 20 '24 at 05:30

0 Answers0