How can I disprove that $13\mid 2^{70}+5^{70}$

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

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

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 $$2^{70}+5^{70}\equiv10+12\equiv9\not\equiv0\pmod{13}$$