0
$\begingroup$

I started with Fermat's Little Theorem taking $2^{12}\equiv1\pmod{13}$ and raising it to power $5$, we got $2^{60}\equiv1\pmod{13}$, then multiplied the congruence $2^{10}\equiv(-3)\pmod{13}$.

I did the same with $5$ and I get $2^{70}+5^{70}\equiv(-4)\pmod{13}$.

Is another method to find it

$\endgroup$
  • 2
    $\begingroup$ I agree with you. $2^{70}+5^{70}\equiv 9 \pmod {13}$. $\endgroup$ – lulu Jun 1 '18 at 17:28
  • 1
    $\begingroup$ Everything you said is correct. What you're trying to prove is not true. $13$ does not divide $2^{70} + 5^{70}$. $\endgroup$ – JGA Jun 1 '18 at 17:29
  • $\begingroup$ Why did you roll back to a wrong edition of your question? $\endgroup$ – egreg Jun 1 '18 at 17:34
  • 3
    $\begingroup$ Possible duplicate of Show that 13 divides $2^{70}+3^{70}$ $\endgroup$ – Arnaud Mortier Jun 1 '18 at 21:27
  • $\begingroup$ A correct formulation is already on the site. $\endgroup$ – Arnaud Mortier Jun 1 '18 at 21:27
3
$\begingroup$

You're doing quite well: $2^{12}\equiv1\pmod{13}$, so $$ 2^{70}=(2^{12})^5\cdot2^{10}\equiv2^{10}\pmod{13} $$ Now note that $2\cdot7\equiv1\pmod{13}$, so $2^{10}\equiv2^{12}\cdot7^{2}\equiv7^2\pmod{13}$ and finally $$ 2^{70}\equiv49\equiv10\pmod{13} $$ With similar computations, noting that $5\cdot8\equiv1\pmod{13}$ we have $$ 5^{70}\equiv(5^{12})^6\cdot8^2\equiv64\equiv12\pmod{13} $$ This shows $$ 2^{70}+5^{70}\equiv10+12\equiv9\not\equiv0\pmod{13} $$

$\endgroup$

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.