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Let n and a be positive integers. Prove that if $a^{170} ≡ 1 \text{ (mod n) and }a^{111} ≡ 1 \text{ (mod n)} \text{ then } a ≡ 1 \text{ (mod n)}$

I know that if $a ≡ 1 \text{ (mod n)}$ then $a^{170} ≡ 1 \text{ (mod n)}$ and $a^{111} ≡ 1 \text{ (mod n)}$, but I am not sure how I can show this in the opposite direction.

Carlisle Manson
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  • so I get that $gcd(a^{170}-1, a^{111}-1) = a^{gcd(111,170)} - 1 = a - 1$, but how do I show that $n | a-1$? – Carlisle Manson Mar 08 '20 at 16:52
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    @JaredL: If $n$ divides $a^{170}-1$ and $a^{111}-1$ then $n$ divides $\gcd(a^{170}-1,a^{117}-1)=a-1$ – J. W. Tanner Mar 08 '20 at 20:16
  • $a^{111}\equiv 1\pmod n$ then $a^k \equiv a^k1 \equiv a^ka^{111}\equiv a^{k+111}\pmod n$ and in particular $a^{59}\equiv a^{111+59}\equiv a^{170}\equiv 1 \pmod n$. Which means $a^k\equiv a^{59+k}\pmod n$. Inparticular $a^{52}\equiv a^{52+59}\equiv a^{111}\equiv 1 \pmod n$ ... and so on – fleablood Mar 08 '20 at 23:06

5 Answers5

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Here is a hint:

$32\times170-49\times111=1$

J. W. Tanner
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$ a^{\large \color{#c00}1}\! \equiv 1,\,$ by $\,a\,$ has order $\,\color{#c00}{k\!=\!1},\,$ by $\,k\,$ divides coprimes $170,111,\,$ by the Order Theorem.

Remark $\ $ This is a multiplicative analog of the well-known fact that a fraction writable with coprime denominators is an integer (by its least denominator ("order") divides coprimes so it must be $\color{#c00}1).\,$ See here for further discussion of this analogy (and denominator and order ideal generalizations).

Bill Dubuque
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  • How do you know they’re coprime? – J. W. Tanner Mar 08 '20 at 19:40
  • @J.W.T Of course there are many simple ways to see that they're coprime without arduously calculating the Bezout equation, e.g. $111$ is clearly coprime to $10$ & $17.\ \ \ $ – Bill Dubuque Mar 08 '20 at 19:53
  • Since $ord_n(a)|170$ and $ord_n(a)|111$, therefore $ord_n(a)|(170,111)=1$. So $ord_n(a)=1 $ and hence $a^1\equiv 1 (mod n) $ – Lai Oct 17 '21 at 16:18
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You can indirectly use the Euclidean algorithm on the exponents to derive the result, as follows:

  • $a^{170} = a^{111}\cdot a^{59} \equiv 1$, and $a^{111} \equiv 1$. Thus, $a^{59} \equiv 1$.
  • $a^{111} = a^{59}\cdot a^{52} \equiv 1$, and $a^{59} \equiv 1$. Thus, $a^{52} \equiv 1$.
  • $a^{59} = a^{52}\cdot a^{7} \equiv 1$, and $a^{52} \equiv 1$. Thus, $a^{7} \equiv 1$.
  • $a^{52} = (a^{7})^{7}\cdot a^{3} \equiv 1$, and $a^{7} \equiv 1$. Thus, $a^{3} \equiv 1$.
  • $a^{7} = (a^{3})^{2}\cdot a^{1} \equiv 1$, and $a^{3} \equiv 1$. Thus, $a^{1} = a \equiv 1$.

Applying this concept in general leads to the theorem

$$a^x \equiv a^y \equiv 1\quad(\text{mod } n)\quad\implies\quad a^{\mathrm{gcd}(x,y)} \equiv 1\quad(\text{mod }n).$$

Jivan Pal
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$1 \equiv a^{170}=a^{111}a^{59}\equiv 1*a^{59}\equiv a^{59}\pmod n$.

And $1 \equiv a^{111}= a^{59}a^{52}\equiv 1a^{52}\equiv a^{52}\pmod n$ and .... keep going....

This is akin to using Euclid's algarithm to show $\gcd(111,170) = 1$.

....

$1 \equiv a^{59} = a^{52}a^{7}\equiv 1*a^7\equiv a^7\pmod n$.

$1 \equiv a^{52}\equiv (a^7)^7a^3 \equiv 1^7a^3\equiv a^3 \pmod n$

$1 \equiv a^7 \equiv (a^3)^2*a \equiv 1^2a \equiv a \pmod n$.

That's that.

Alternatively we can use EA and Bezouts lemma to show that

$1 = 7- 2*3 = 7- 2(52- 7*7)$

$= 15*7 - 2*52 = 15*(59-52)- 2*52 = 15*59 - 17*52=$

$=15*59- 17(111-59) = 32*59 - 17*111 =$

$ 32*(170-111) - 17*111=32*170 - 49*111$

So $5440 = 32*170$ and $5540= 5439+1 = 49*111 + 1$.

And so....

So $a^{5440}=(a^{170})^{32} \equiv 1^{32} \equiv 1 \pmod n$.

But also $a^{5440}= a*a^{5439}=a*(a^{111})^{49}\equiv a*1^{49}\equiv a \pmod n$.

So transitively, $a\equiv a^{5440}\equiv 1 \pmod{n}$.

fleablood
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Since $170$ and $111$ are relatively prime, there exist, by Bezout's theorem, $x,y$ such that $x170+y111=1$. But then $1\cong {(a^{170})}^x\cdot {(a^{111})}^y\cong a^{x170+y111}\cong a\pmod n$.

Of course, this generalises to any coprime exponents.