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I want to use Euler's theorem to calculate the following:

$14^{(2019^{2019})} \mod 60$

$a^{\phi(n)} \equiv 1 \mod n \iff \gcd(a,n) = 1$

Now I start with the outer one first:

$14^{(2019)}\mod 60 \iff \gcd(14,60) = 2$

Now I don't know how can I solve it ...

Can anyone suggest me any hints/further calculations?

Thank you in advance,

Kind Regards.

Bernard
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John M
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  • Hint: your large number is clearly divisible by $4$, so you can park that and achieve a radical simplification. – Mark Bennet Apr 30 '19 at 11:42
  • @Mark Don't know how you mean that - will the congruence get easier then, after a radical simplification? – John M Apr 30 '19 at 11:54
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    Indeed it will because $60/4=15$ and $14\equiv -1 \bmod 15$ and you can raise $-1$ to any power quite easily (see lhf's answer) – Mark Bennet Apr 30 '19 at 12:33

5 Answers5

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You can't use Euler's theorem because $\gcd(14,60) > 1$.

But you can use the Chinese remainder theorem, using $60 = 4 \cdot 15$: $$ \begin{align} 14^{(2019^{2019})} & \equiv \hphantom{-} 0 \bmod 4 \\ 14^{(2019^{2019})} & \equiv -1 \bmod 15 \end{align} $$ because $$ \begin{align} 14^2 & \equiv \hphantom{-} 0 \bmod 4 \\ 14\hphantom{^2} & \equiv -1 \bmod 15 \end{align} $$

lhf
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You can calculate separately $2^{2019^{2019}}$ and $7^{2019^{2019}}$ first:

  • It is not long to check that, if $n\ge 2$, $$2^n\equiv\begin{cases} 16&\text{ if }n\equiv 0\mod 4,\\ 32&\text{ if }n\equiv 1\mod 4,\\ 4&\text{ if }n\equiv 2\mod 4,\\ 8&\text{ if }n\equiv 3\mod 4. \end{cases}$$ So we have to compute the exponent mod. $4$: $$2019^{2019}\equiv (-1)^{2019}=-1\equiv3\mod 4\quad\text{and thus }\quad 2^{2019^{2019}}\equiv8\mod 60.$$
  • As it is coprime to $60$, $7$ has order dividing $\varphi(60)=16$. Actually, $\;7^2=49\equiv -11\mod 60$, so $\;7^4\equiv (-11)^2=121\equiv 1$: $7$ has order $4$, so that $$7^{2019^{2019}}\equiv 7^{2019^{2019}\bmod4}\equiv 7^3\equiv-77\equiv -17\mod 60.$$
Bernard
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  • I see, seems odd! But how do we get to the mod 4 part? Thanks in advance! – John M Apr 30 '19 at 12:34
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    The mod 4 part for $2$ or for $7$? – Bernard Apr 30 '19 at 12:35
  • for 2.. - at 7 just using Euler's Theorem? – John M Apr 30 '19 at 12:38
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    Yes, for $7$ it's jst Euler's theorem to begin with, then checking for the exact order. For $2$ I simply computed the first successive powers of $2$ modulo $60$, until the same value reappeared: for exponent $n=1,2,3,3,4,5,6$, i obtained the sequence $2,4,8,16,32, 4\pmod{60}$. From there on, one knows the sequence is cyclic mod. $60$, with period $4$.. – Bernard Apr 30 '19 at 12:47
  • It is very hard for me to understand how we got so fast from $2019^{2019}$. If we don't find the step with $\mod 4$, we would have calculate $2019^{2019} \mod 60$ right? How to do this in an example case, if I didn't find a simplification? – John M Apr 30 '19 at 13:09
  • But if a number is not coprime to ,the modulus, it always happens its powers are ultimately cyclic (the cycle may be $0, 0, 0,\dots$). Now to compute high powers fast, you also have the fast exponentiation algorithm, based on repeated squaring. – Bernard Apr 30 '19 at 13:57
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Like How to find last two digits of $2^{2016}$,

Get the last two digits of $16^{100}$ and $17^{100}$

what are the last two digits of $2016^{2017}$?

last two digits of $14^{5532}$?,

As $(14^n,60)=2^2$ for $n\ge2$

and as $14\equiv-1\pmod{15}\implies14^m\equiv(-1)^m$

$\implies F_{m+2}=14^{m+2}=14^2\cdot14^m\equiv14^2(-1)^m\pmod{15\cdot14^2}$

$F_{m+2}\equiv14^2(-1)^m\pmod{15\cdot4}$ as $15\cdot4$ divides $15\cdot14^2$

$F_{m+2}\equiv16(-1)^m\pmod{60}$

If $m$ is odd like here $2019^{2019}-2,$ $$F_{m+2}\equiv16(-1)\pmod{60}\equiv-16+60$$

lab bhattacharjee
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Here's a way:

  • Get rid of the gcd,
  • then attempt by Euler's theorem, then reformulate mod 60

$$gcd(14,60)=2; gcd(14,15)=1; 14^{2019^{2019}}\equiv 14^{3^3}\equiv -1^3\equiv -1\bmod 15$$ which adding back a factor of 4 coincides with $44 \bmod 60$

  • Hm, this answer seems to be pretty nice. Can you give a few instructions from step 1 to step 3? How do I exactly get rid of the gcd? In which cases can I do that? Would be awesome! Thank you in advance! – John M May 01 '19 at 10:39
  • Euclidean gcd. all cases. otherwise you can't directly use Euler.. –  May 01 '19 at 11:04
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$n\ge 2\,\Rightarrow\, 4\mid 14^n\Rightarrow\,14^n\bmod 60 = 4 (14^n/4 \bmod 15) = 4((-1)^n 4) = \left\{\begin{align} &16\ \ {\rm if}\ \ 2\mid n\\ &44\ \ {\rm if}\ \ 2\nmid n\end{align}\right. $

Bill Dubuque
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