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I'm stuck on this basic problem,

Show that $39$ divides $17^{48}-5^{24}$

I am familiar with the basic arithmetic properties of modular. But I have absolutely no idea how I'm supposed to apply them to solve this type of question. Some advice on the pattern I'm supposed to be looking for would be very helpful.

Bobby B
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1 Answers1

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Hint:

You only have to prove that $3$ and $13$ both divide $17^{48}-5^{24}$, i.e. that $2^{48}-2^{24}\equiv 0\mod 3$, and that $\:4^{48}-5^{24}\equiv 0\mod 13$.

Remember that Fermat's theorem can be written as $$ a^k\equiv a^{k\bmod p-1}\mod p\quad\text{ for any $a$ coprime to }p. $$

Bernard
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  • My textbook does not cover Fermat's theorem nor any of his other work with modular until later. I have to solve this problem without using that theorem. The only properties of modular I've learned about have been additivity, multiplicativity, symmetry, and transitivity, – Bobby B Feb 26 '21 at 23:40
  • @BobbyB Then simply calculate small powers e.g. $\bmod 13!:\ 5^2\equiv -1\Rightarrow 5^4\equiv 1,,$ and $,16\equiv 4,\ 4^2\equiv 3,\ 3^3\equiv 1\Rightarrow 4^6\equiv 1,$ then use the linked order reduction to conclude that $,5^{4k}\equiv 1\equiv 16^{6n},$ by congruence arithmetic rules – Bill Dubuque Feb 27 '21 at 00:18