$7^{-1}\pmod{11}$
the above can be found by
$7x\pmod{11}\equiv 1$ and $x=8$
now i am confused on how to find $7^{-2}\pmod{11}$ and $7^{-3}\pmod{11}$ .
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oh so its $7^{-2}\pmod {11}\equiv 9$ ? – R K Feb 26 '15 at 15:39
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1Indeed. It took me more than it should have to verify $64\equiv 9$ mod $11$ :) – Jack Yoon Feb 26 '15 at 15:41
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Hint $\,\ 7^{-2} \equiv (7^{-1})^2\ $ since $\ 7x\equiv 1\,\Rightarrow\, 7^2 x^2\equiv 1.\,$
Bill Dubuque
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so can i substitute like $7^{-2}\equiv (7^{-1})^2\equiv(8)^2\\equiv 64 \equiv 9 \pmod {11}$ ? – R K Feb 26 '15 at 15:41
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1@RK Yes, $\ \color{#c00}{7^{-1}\equiv 8},\Rightarrow, (\color{#c00}{7^{-1}})^2\equiv \color{#c00}8^2\ $ by the $ $ Congruence Power Rule. $ $ Similarly for $\ 7^{-3}\ \ $ – Bill Dubuque Feb 26 '15 at 15:46
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and similarly for $7^{-3} \equiv 7^{-2}\cdot 7^{-1}\ \equiv 9\times 8 \equiv 72 \equiv 6 \pmod {11}$ – R K Feb 26 '15 at 15:49
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