<s>[INST] Proving a congrent modular How do I show that $3^{1974}+5^{1974}\equiv 0 \pmod {13}$? I have tried feeding the values onto a calculator but they are so big to be computed. What is the best approach? 
 [/INST] since Fermat's little theorem, implies $3^{12}\equiv5^{12}\equiv 1\pmod{13}$ we have: 
$$
3^{1974}+5^{1974}=3^{12*164+6}+5^{12*164+6}=(3^{12})^{164}3^6+(5^{12})^{164}5^6\equiv 3^6+5^6 \pmod{13}
$$ 
Now, you can easily see that:
$$3^3=27\equiv 1\pmod{13}\Rightarrow 3^6=1\pmod{13}$$
and:
$$5^2=25\equiv -1\pmod{13}\Rightarrow5^6=-1\pmod{13}$$
thus, summing the above:
$$
3^{1974}+5^{1974}\equiv 3^6+5^6 \equiv 1-1 =0 \pmod{13}
$$
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