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I've tried many times via Euler's Theorem.

$ \varphi( 2019) = 2\times 672 = 1344 $

$a^{\varphi(2019)} \equiv a^{1344}\equiv 1 \mod 2019$

then , $a^{2019} \equiv a^{675} \mod 2019$ , obviously $1$ is 1 of 4 for answers and then how we can find another integer.

p.s. according to wolframalpha they've shown $3$ equation for integer solutions

I really appreciate for your help .

Bernard
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ABCDEFG user157844
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2 Answers2

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As $2019=3\cdot673$

From the given condition

$a^{2019}\equiv1\pmod{673}$

But $2019\equiv3\pmod{\phi(673)}$

So, $a^3\equiv1\pmod{673}$

which has exactly three solutions

Trivially, $a\equiv1$

Else $a^2+a+1\equiv0\pmod{673}$

$\iff(2a+1)^2\equiv-3$ which is solvable as Prove that $-3$ is a quadratic residue mod an odd prime $>3$ if and only if $p$ is of the form of $6n+1$

Similarly $$1\equiv a^{2019}\equiv a^1\pmod3$$ as $\phi(3)=2$

So using http://mathworld.wolfram.com/ChineseRemainderTheorem.html, we should have exactly three solutions $\pmod{3\cdot673}$

lab bhattacharjee
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  • it's torture with large prime to find $x^2 (mod 673) = - 3$ by hands ,there are $\frac{p-1}{2}$ distinct residue and I don't know what number between 1 to $\frac{p-1}{2}$ give $-3$ as residue $(mod 673)$ . – ABCDEFG user157844 Jun 09 '19 at 22:29
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    @ABCDEFGuser157844 We don't need to compute the nontrivial cube root. We only need to prove that it exists - which is easy by CRT (see the omment on my answer). – Bill Dubuque Jun 09 '19 at 22:39
  • @ABC I just realized there is an ambiguity: do you seek the sum mod $2019$ or the sum as integers? – Bill Dubuque Jun 09 '19 at 22:55
  • I don't know how to describe in english , it's like you finally got equation that for $a^{2019} \ mod(2019) = 1 \ $ we can represent $(a)$ by

    $a = 2019m + 1$ , $a = 2019m + 928$ , $a = 2019m +1090$ , which $m \in \mathbb{Z}$

    and the first four integer $(a)$ is $1,928,1090,2020,.......$ (arrange from minimum value)

     – ABCDEFG user157844 Jun 09 '19 at 23:03
  • and that's a problem that I cannot find that 3 equation to represent (a) because I don't know the way how to get them. – ABCDEFG user157844 Jun 09 '19 at 23:14
  • @ABC Even if we seek the sum as integers (vs. mod $2019)$ we still don't need to know the actual values of the 2nd & 3rd terms $,a,, a^2,\bmod 2019,,$ since I showed $, S := 1+a+a^2\equiv 0\pmod{!2019}$ and $, a,a^2\le 2018,$ when reduced mod $2019,,$ thus $,1\le S < 2\cdot 2019,$ so $,S = 2019,$ since $,2019\mid S.\ \ $ – Bill Dubuque Jun 09 '19 at 23:21
  • I will do my best try to understand your solution ,although it's hard to understand for me because I'm still a 10th grade high school student :) . – ABCDEFG user157844 Jun 09 '19 at 23:28
  • @ABC Do you know CRT = Chinese Remainder Theorem, and primitive roots?. Where does the problem come from, i.e. what was taught before it was posed? – Bill Dubuque Jun 09 '19 at 23:47
  • I'm familiar with CRT but primitive roots am not, sorry no taught in the class is 3x harder for my school math subject. i just curious want to learning math out of my class. searching many knowledge on internet by myself. – ABCDEFG user157844 Jun 10 '19 at 00:18
  • @ABC I updated my answer to show how to avoid using any knowledge of primitive roots. – Bill Dubuque Jun 10 '19 at 17:57
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Hint $ $ Since $\ a^n\equiv 1\equiv a^k\,\Rightarrow a^{\gcd(n,k)}\equiv 1,\,$ we deduce $\,a^3\equiv 1$ so the sum of the first three roots is $\equiv 1+a+a^2\equiv 0,\,$ so the sum of the first four roots = fourth root = least root $(=1) + 2019$

Update $ $ Primitive roots are not known so we give a direct proof that if $p$ is prime and $3\mid p-1\,$ then $\,x^3-1\,$ has $\,3\,$ roots in the field $\,\Bbb Z_p = $ integers $\!\bmod p.\,$ Note $\,3\mid p-1\,\Rightarrow\, x^3-1\mid x^{p-1}-1\,$ so $\,x^{p-1}-1 = (x^3-1)f(x)\,$ for a polynomial $\,f(x)\,$ with integer coef's. Since a polynomial over a field has no more roots than its degree, if $\,x^3-1\,$ had less than $\,3\,$ roots then $\,(x^3-1)f(x)\,$ would have less than $\,p-1\,$ roots, contra it equals $\,x^{p-1}-1\,$ with $\,p-1\,$ roots, all $\,x\not\equiv 0,\,$ by little Fermat.

Bill Dubuque
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    Further hint: verify via CRT, i.e. combine results mod $3$ and mod $,p=673$. We get a cube root $,a\not\equiv 1\pmod{!3p}$ from a CRT solution of $,a\equiv 1\pmod{!3},$ and $,a\equiv \zeta_3\pmod{!p},,$ where $,\zeta_3,$ is a primitive cube root of $1$ mod $p,,$ e.g. $,\zeta_3 = g^{(p-1)/3},$ for $g$ a primitive root mod $p.,$ And $,1+a+a^2\equiv 0\pmod{!3p}$ since it holds mod $3$ and mod $p$. – Bill Dubuque Jun 09 '19 at 22:35
  • Thank you for further hint. – ABCDEFG user157844 Jun 09 '19 at 22:46