high power congruences finding $x$

Question

trying to solve:

$$x^{13} \equiv 11 \pmod{135}$$

I came to the fact that $x = 11^{59}$ but its in mod $72$ and needs to be converted to mod $135$

any suggestions? I'm not sure how to change it to mod $135$ with such a large number

Answer 1

$135 = 3^3 \cdot 5$, so $\varphi(135) = 72$. Using Euclid, you find that $$ 1 = 13 \cdot (-11) + 72 \cdot 2 = 13 \cdot 61 + 72 \cdot (-11), $$ so $$ x \equiv (x^{13})^{61} \equiv 11^{61} \pmod{135}. $$

PS Once again, apologies for the initial mistake in this late-night post (never again!), and thanks to Gerry Myerson and user62340 for calling my attention to it.

To finish the calculation by hand, it is probably safer to compute first $$ 11^{11} = 11 \cdot 11^2 \cdot 11^{8} \equiv 41 \pmod{135}, $$ and then the inverse $56$ of $41$ modulo $135$, which is the required solution.

Answer 2

If $\rm\: x^{13}\equiv 11\,\ (mod\ 5\cdot27)\:$ then the same congruence holds mod $5$ and $27$, and we can use CRT (Chinese Remainder Theorem) to recombine the two solutions. It turns out to be very simple:

$\rm\quad\ mod\,\ 5\!:\ x^4\equiv 1,\:$ so $\rm\,x \equiv x (x^4)^3 \equiv x^{13}\equiv 11\equiv \color{#C00}1.\ \ $ Next $\ \phi(27) = 18,\:$ hence

$\rm\quad\ mod\ 27\!:\ 1 \equiv x^{18}\equiv x^{13} x^5 \equiv 11 x^5\:$ so $\rm\,x^5 \equiv \dfrac{1}{11}\equiv \dfrac{55}{11}\equiv 5\equiv 2^5\Rightarrow\:x\equiv \color{#0A0}2\ $ (unique, see Note)

$\rm\Rightarrow mod\ 5\cdot 27\!:\,\ x \equiv \color{#0A0}2 + 27\,\left[\dfrac{\color{#C00}1\!-\!\color{#0A0}2}{27}\ mod\ 5\right]\! \equiv 56,\ $ by $\rm\ mod\ 5\!:\, \dfrac{-1}{27}\equiv \dfrac{4}2 \equiv 2,\ $ using Easy CRT.

Note $\,\ 5$'th roots are unique $\rm\,mod\ 27\:$ since $\rm\, (5,\phi(27)) = (5,18) = 1,\:$ so $\rm\:n\equiv 1/5\ mod\ 18\:$ exists, hence $\rm\,x^5 = y^5\:\Rightarrow x\equiv x^{5n}\equiv y^{5n}\equiv y\,$ since $\rm\,5n\equiv 1\,\ (mod\ 18).$

Answer 3

$x^{13}\equiv11\pmod{135}\implies x^{13}\equiv11\pmod5$ and $x^{13}\equiv11\pmod{27}$

Using Euler's Totient Theorem, $$a^{\phi(m)}\equiv1\pmod m \text{ where }(a,m)=1$$

So, if $\phi(m)\mid(r-s),a^r\equiv a^s\pmod m$

$\implies x^{13}\equiv x\pmod5$ as $\phi(5)=5-1=4\implies x^{13}\equiv11\pmod5\iff x\equiv1\pmod5--->(1)$

Using this, if $a$ is primitive roots of prime $p$ and $p^2,a$ will be primitive roots of $p^n$ for $n\ge1$

$2^1\equiv-1\pmod3,2^2\equiv1\implies2$ is a primitive root of $3$

Now, $2^1\equiv2\pmod9,2^2\equiv4,2^3\equiv-1\implies ord_92=6=\phi(9)\implies2$ is a primitive root of $9$

So, $2$ is a primitive root of $3^n$ for $n\ge1$

Applying Discrete Logarithm with respect to base $2$ on $x^{13}\equiv11\pmod{27},$

$13\cdot ind_2x\equiv ind_211\pmod{18}$ as $\phi(27)=3^2(3-1)=18$

Now, $2^3\equiv8\pmod{27},2^4\equiv16\equiv-11\pmod{27}$

Again as $2$ is a primitive root of $27,2^{\frac{\phi(27)}2}\equiv-1\pmod{27}$ i.e., $2^9\equiv-1$

So, $11\equiv2^4\cdot2^9\equiv2^{13}\implies ind_211=13$

So, $13\cdot ind_2x\equiv 13\pmod{18}\implies ind_2x\equiv1\pmod{18}$ as $(13,18)=1$

So, $x\equiv2^1\pmod{27}\equiv2--->(2)$

Applying well known CRT on $(1),(2)$ we get $x\equiv2b_1\frac{27\cdot5}{27}+1\cdot b_2\frac{27\cdot5}5\pmod{27\cdot5}\equiv 10b_1+27b_2\pmod{135}$

where $b_1\frac{27\cdot5}{27}\equiv1\pmod {27}\iff 5b_1\equiv1$

and $b_2\frac{27\cdot5}5\equiv1\pmod{27}\iff 27b_2\equiv1$

Expressing $\frac{27}5$ as continued fraction, $\frac{27}5=5+\frac25=5+\frac1{\frac52}=5+\frac1{2+\frac12}$

So, the last convergent of $\frac{27}5$ is $5+\frac12=\frac{11}2$

Using convergent property of continued fraction, $27\cdot2-5\cdot11=-1$

$\implies 5\cdot11\equiv1\pmod{27}\implies 5^{-1}\equiv11\implies b_1\equiv11 $

and $27\cdot2\equiv-1\pmod5\implies (27)^{-1}\equiv-2\equiv3\implies b_2\equiv3$

So, $x\equiv27\cdot3+10\cdot11\pmod{135}\equiv191\equiv56$