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For $207^{321} \pmod{7},$ I got $$ 207^{321} = 207^{6\cdot 53+3}$$ and $$207^{\Phi(7)} \equiv 207^6 \equiv 1 \pmod{7}$$ by Euler's Theorem. Then $$207^3 \equiv 4^3 \equiv 1 \pmod{7} $$

Is there any simpler way? I'm also not sure about the format of module symbol.Should there be only one (mod 7) written on the right of the equation so as to avoid redundancy ?

I have also seen equation like this 26 mod 5=1,rather than $26\equiv 1 \mod{5}$. What's the difference?

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  • $\begingroup$ use \pmod, not \Mod. $\endgroup$ – Ben Grossmann Oct 5 '15 at 17:15
  • $\begingroup$ @Omnomnomnom: What's the difference between pmod and bmod, by the way? $\endgroup$ – Brian Tung Oct 5 '15 at 17:17
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    $\begingroup$ @BrianTung \bmod is meant for use as a binary operator. \pmod puts parentheses around the second argument (and spaces things differently). \mod uses the spacing of \pmod but not the parentheses. $\endgroup$ – Ben Grossmann Oct 5 '15 at 17:19
  • $\begingroup$ Also $207^{6^{53}}$ is not the same as $\left(207^6\right)^{53}$ Which has been already corrected, as I can see. $\endgroup$ – I want to make games Oct 5 '15 at 17:22
  • $\begingroup$ @Omnomnomnom: Excellent, thanks; I've added it to my own answer. $\endgroup$ – Brian Tung Oct 5 '15 at 17:26
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I don't know, your line of reasoning seemed pretty quick and easy to me. I suppose you could observe that (modulo $7$)

$$ 207^{321} \equiv (207 \bmod 7)^{(321 \bmod 3)} \equiv 4^0 = 1 $$

Is that simpler, by your lights?

The difference in notation, incidentally, is that when you write $26 \bmod 5 = 1$, the mod is treated as an operation—essentially the remainder left when you divide $26$ by $1$. When you write $26 \equiv 1 \pmod 5$, you mean that $26$ and $1$ fall into the same equivalence class, modulo $5$. The two formulations are equivalent under ordinary circumstances.

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  • $\begingroup$ That should be $321 \bmod 6 = 3$ $\endgroup$ – Ben Grossmann Oct 5 '15 at 17:29
  • $\begingroup$ I had that first (well, I had that second, after I fixed an earlier mistake). But isn't it the case that powers of $4$ have periodicity $3$ modulo $7$? $\endgroup$ – Brian Tung Oct 5 '15 at 17:31
  • $\begingroup$ oh, you're right; I didn't notice. $\endgroup$ – Ben Grossmann Oct 5 '15 at 17:35
  • $\begingroup$ @BrianTung - good solution. (+1). how do we know that powers of $4$ have periodicity $3$ mod $7$? what about other powers and other mods, and how can this be shown? $\endgroup$ – Hypergeometricx Oct 5 '15 at 17:45
  • $\begingroup$ Thank you.But I didn't got your method.Why did you take mod 3 on the power? How can I use 'powers of 4 have periodicity 3 mod 7'? $\endgroup$ – Charlotte Gu Oct 5 '15 at 17:50
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Note that $207^{321}=(207^{6})^{53} \cdot 207^3=1\cdot 4^3=1$ (mod. $7$).

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