3
$\begingroup$

Problem: Use Euler's Theorem to Find a Number $x$ between $0$ and $28$ such that $x^{85} \equiv6\pmod{35}$

Hi all,

I have narrowed this problem down to $x^{13} \equiv6 \pmod{35}$. I got here by splitting up $x^{85}$ into $x^{24}\cdot x^{24}\cdot x^{24}\cdot x^{13}$, and then cancelling out the $x^{24}$'s to just $1$'s (using Euler's theorem which states that $x^{\phi(n)} \equiv1 \pmod{n}$; $\phi(35) = 24$).

I'm stuck on how to proceed from here. Any tips?

$\endgroup$
3
$\begingroup$

$a^{24} = 1 \mod(35)$ so $a^{12} = \pm 1 \mod(35)$

So $a^{85} = a^{13} = a^{12}a = \pm a = 6 \mod(35)$. So $a = \{6, -6\}\mod 35$. As $6^2 = 1 \mod (35)$, $(-6)^{13} = -6 \mod 35$, so $a = 6\mod 35$, so a = 6.

=== oh, we were give a is between 0 and 28. So 29 = 35 -6 was out by premise.

$\endgroup$
0
$\begingroup$

$x^{85}\equiv6\pmod{35}\ \ \ \ (1)$

As $35=5\cdot7$ with $(5,7)=1$

$(1)\implies x^{85}\equiv6\pmod5\equiv1$

As $\phi(5)=4,85\equiv1\pmod4; x\equiv1\pmod5\ \ \ \ (2)$

$(2)\implies x^{85}\equiv6\pmod7$

As $\phi(7)=6,85\equiv1\pmod6; x\equiv6\pmod7\ \ \ \ (3)$

Apply CRT on $(2),(3)$ or by observation $x\equiv6\pmod{35}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.