What is the remainder when $12^{39} + 14^{39}$ is divided by $676$?

Question

I tried following but then I got stuck

$676 = 26*26$

$12^{39} + 14^{39}$ is divisible $26$ for sure since $a^n + b^n$ is divisible by $(a+b)$ when $n$ is odd. But what to do next?

You might have a look at this question and some other questions posted there among listed questions. — Martin Sleziak, Sep 03 '14 at 08:12

lab bhattacharjee · Accepted Answer · 2014-09-03T06:00:39.037

3

Clearly, $$12^{39}+14^{39}$$ is divisible by $4$

Again, $$12^{39}=(13-1)^{39}\equiv-1+\binom{39}113\pmod{169}\equiv-1$$

$$14^{39}=(13+1)^{39}\equiv1+\binom{39}113\pmod{169}\equiv1$$

More generally if $\displaystyle a\equiv b\pmod p, a^p\equiv b^p\pmod{p^2}$

Here $\displaystyle p=13,12\equiv-1\pmod{13}\implies12^3\equiv(-1)^3\equiv-1$

Similarly, $\displaystyle14\equiv1\pmod{13}\implies14^3\equiv(1)^3$

edited Sep 03 '14 at 06:00

answered Sep 03 '14 at 05:48

lab bhattacharjee

Could you please elaborate the second part How (mod 169) is 0? – user3710832 Sep 03 '14 at 05:52
@user3710832, Have you understood how $12^{39}\equiv-1\pmod{169}?$ – lab bhattacharjee Sep 03 '14 at 05:53
No. Please elaborate a little. – user3710832 Sep 03 '14 at 05:54
@user3710832, Do you understand binomial theorem ? – lab bhattacharjee Sep 03 '14 at 05:55
Yes. But how is that applied here? – user3710832 Sep 03 '14 at 06:00
@user3710832, $$(13-1)^{39}=13^{39}+\cdots+\binom{39}113-1\equiv39\cdot13-1\equiv-1\pmod{169}$$ as the other terms contain higher powers of $13,$ hence divisible by $169=13^2$ – lab bhattacharjee Sep 03 '14 at 06:02

score 0 · Answer 2 · edited Sep 03 '14 at 09:14

0

$$\left((39-12)^{39}+(39-14)^{39}\right) \div 26 \times 26$$ so $$\left(27^{39}+25^{39}\right) \div 26$$ so $$1^{39}+(-1)^{39}$$

edited Sep 03 '14 at 09:14

DanielV

answered Sep 03 '14 at 08:42

jyothika1

2 Answers2