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Let $p$ be an odd prime. Then, by Euler's theorem, I know that $$ 2^{k}\equiv 2^{k\pmod{p-1}}\pmod p. $$ From Lagrange I also know that for the order of $2$ modulo $p$ we have $ ord_p(2)|p-1 \Leftrightarrow \exists m\in\mathbb Z: m\cdot ord_p(2) = p-1.$

But how can I deduce that $$ 2^k \equiv 2^{k\pmod{p-1}} \equiv 2^{k\pmod{ord_p(2)}}\pmod p? $$

I have no idea how to reduce via the product $ m\cdot ord_p(2)$.

Bill Dubuque
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mathquester
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  • Euler's theorem is just Lagrange's theorem applied to the group $(\mathbb{Z}/p\mathbb{Z})^\times$. Where is the confusion, and what is the context for this question? – student91 Mar 01 '23 at 16:23
  • @student91 Let's say I want to solve $2^{17}\pmod{7}.$ By Euler I know $2^{17}\equiv 2^{17 \mod 6} \equiv 2^5\pmod 7$. But I could also calculate $2^{17}\equiv 2^{17 \mod 3}\equiv 2^2\pmod 7$, because $3$ is the order. But I don't know why I may do that. – mathquester Mar 01 '23 at 16:27
  • Then after reducing $17\equiv5\pmod{6}$ like you did, you only need to do $2^5\equiv32=28+4\equiv 4\pmod{7}$. – student91 Mar 01 '23 at 16:30
  • @student91 I am struggling to proof this in general, that I may reduce modulo order. – mathquester Mar 01 '23 at 16:36
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    See mod order reduction in the linked dupe. – Bill Dubuque Mar 01 '23 at 19:32

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@student91 Let's say I want to solve $2^{17}\pmod{7}$. By Euler I know $2^{17}\equiv2^{17\pmod{6}}\equiv2^5\pmod{7}$. But I could also calculate $2^{17}\equiv2^{17\pmod{3}}\equiv2^2\pmod{7}$, because $3$ is the order. But I don't know why I may do that.

Given a finite group $G$ (Say $G=(\mathbb{Z}/17\mathbb{Z})^\times$) and an element $x\in G$ (Say $x=2$), and a number $m$ which can be written as $m = k\cdot \mathrm{ord}_G(x)+r$, with $k,r,m\ge0$, we have that $$x^m=x^{k\cdot\mathrm{ord}_G(x)+r}=(x^{\mathrm{ord}_G(x)})^k\cdot x^r=1^k\cdot x^r=x^r.$$

student91
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