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I'm trying to find $10^{11} \mod 101$. Here's what I did:

$\pmod{101}$

However, the calculator shows that $10^{11}\equiv91\pmod{101}$

Where did I go wrong?

blz
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  • Fractionary exponents are not defined in $\mathbf Z/n\mathbf Z$. They only make sense in $\mathbf R$. – Bernard Nov 09 '16 at 09:49
  • @Bernard They don't really make sense for $\Bbb R$ either. They only make sense for $\Bbb R_{\geq 0}$. – Arthur Nov 09 '16 at 10:05
  • As a concept, they make sense for $\mathbf R$, but they exist only for non-negative real numbers. – Bernard Nov 09 '16 at 10:28
  • @Bernard No, in $\Bbb R$, they do not make sense, any more than they do for $\Bbb Z/n\Bbb Z$. For instance, $25^{1/2}$ is both $5$ and $-5$. Only when you restrict yourself to non-negative reals do fractional exponents make sense. – Arthur Nov 09 '16 at 10:56
  • Fractionary exponents denote the positive square root, conventionally. – Bernard Nov 09 '16 at 11:04
  • @Bernard Which is the same as saying that when defining fractional exponents, we do it within $\Bbb R_{\geq0}$, not all of $\Bbb R$. – Arthur Nov 09 '16 at 11:40

2 Answers2

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You should've gone with $10^{11} = 100^5\cdot 10$ to make it clearer. However, now that you've come this far, note that the square roots of $100$ are $\pm 10$. You chose the wrong square root in your last step.

Arthur
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${\rm mod}\,\ 101\!:\ 10^{\large 11}\!\equiv 10(\color{#c00}{10^{\large 2}})^{\large 5}\!\equiv 10(\color{#c00}{\bf-1})^{\large 5} \equiv -10 \equiv 91$

Remark $\ $ The problem with your method is that you tried to use $\, n\equiv (n^{\large 2})^{\large 1/2},\, $ which is not true here, e.g. for $\, n\equiv 91\,$ your method gives $\ 91\equiv {-}10\equiv ((-10)^{\large 2})^{\large 1/2}\equiv 100^{\large 1/2}\equiv 10.\,$ Generally you can use that identity only in domains where squaring is injective $(1$-$1)$.

Bill Dubuque
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