How would you solve the question (in the title)?
Can I apply your approach/solution (for the title question) also for: $13 \cdot a \equiv 1 \pmod {1337}$ and $69 \cdot a \equiv 8 \pmod {8008}$ etc.?
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What methods do you know of? Easiest and fastest may depend on you, and what you're comfortable with and know well, so the more we know about the methods you already know will help us to answer. – Studentmath Feb 06 '15 at 17:26
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To be honest, I'm not interested in the fastest way, but just in a possible way to solve it. If it's easy and fast, that's good :) – Joey Feb 06 '15 at 17:29
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The equation $ax\equiv1\pmod{n}$ has a solution if $\gcd(a,n)=1$ – kingW3 Feb 06 '15 at 17:30
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@kingW3 Could generalize this and say: ax≡ b (mod n) has a solution if gcd(a,n)=b? – Joey Feb 06 '15 at 17:32
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@Joey It has a solution iff $\gcd(a,n)\mid b$. – user26486 Feb 06 '15 at 19:09
3 Answers
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Hint $\ $ By $ $ Gauss's algorithm, $\,\ {\rm mod}\ 1247\!:\,\ \dfrac{1}{11}\equiv \dfrac{1}{11}\dfrac{113}{113}\equiv\dfrac{113}{-4}\equiv\dfrac{113+1247}{-4}\equiv -340$
More generally $\,1247 = 11(\color{#c00}{113})+4\,$ so above is case $\,\color{#c00}{j=113}\,$ of below
$\!\bmod 11\color{#c00}j+4\!:\,\ \dfrac{1}{11}\equiv \dfrac{j}{11j}\equiv \dfrac{12j+4}{-4}\equiv -(3j+1)\ $ again by Gauss and a twiddle.
Beware $\ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.
Bill Dubuque
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Use Bézout: with the Extended Euclidean algorithm, you get: $$-340\cdot 11+3\cdot 1247=1,\quad\mathrm{hence}\quad -340\cdot 11\equiv 1\mod 1247.$$
Bernard
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That's incorrect $,112\cdot 11\equiv -5\pmod{1237}.,$ Also that's not the OP's problem: it is $1247$ not $1237$ – Bill Dubuque Feb 06 '15 at 18:34
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@Bill Dubuque: I know I shouldn't compute all in my head… It's all fixed (and checked). – Bernard Feb 06 '15 at 19:40
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@Bernard Yes, now it agrees with the independent (purely mental!) computation in my answer. – Bill Dubuque Feb 06 '15 at 19:42
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@Bill Dubuque: I discovered Gauß's algorithm on your thread. Fine! I wonder what is its exact link with Bézout. – Bernard Feb 06 '15 at 19:46
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@Bernard One way to view the relationship is that it is the Euclidean algorithm with one argument fixed prime, see the final paragraph of this answer. – Bill Dubuque Feb 06 '15 at 19:49
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By Euclid division algorithm: $D = dq + r, r < d$
$D = 11a$
$d = 1247$
$r = 1$
Diophantine: $11a - 1247 q = 1$
Now, successive divisions for calculating gcd(1247, 11):
$1247 = 11 \cdot 113 + 4$
$11 = 4 \cdot 2 + 3$
$4 = 3 \cdot 1 + 1$
$a = a_0 + 1247 t, q = q_0 + 11t$
We take the remainders above:
$4 = 1247 - 11 \cdot 113 = 1247x + 11y$
$3 = 11 - 4 \cdot 2$
$1 = 4 - 3 \cdot 1 = (1247 - 11 \cdot 113) - (11 - 4 \cdot 2) \cdot 1$, rearrange and exchange $4$:
$1 = 1247 - 11 \cdot 114 + (1247 - 11 \cdot 113) \cdot 2$
$1 = 1247 \cdot 3 - 11 \cdot (114 + 113 \cdot 2)$
$1 = 1247 \cdot 3 - 11 \cdot 340 \Rightarrow a_0 = -340, q_0 = -3$
$11 (-340) - 1247 (-3) = 1$
$a \in \{-340 + 1247 t ; t \in \mathbb{Z} \} = \{..., -1587, -340, 907, 2154, ...\}$
$D = dq + r, r < d$
$D = 69a$
$d = 8008$
$r = 8$
$69a - 8008q = 8$
$8008 = 116 \cdot 69 + 4 \Rightarrow 4 = 8008 - 116 \cdot 69$
$69 = 17 \cdot 4 + 1 \Rightarrow 1 = 69 - 17 \cdot (8008 - 116 \cdot 69) = 69(+1973) - 8008(+17)$
Multiply by $8$:
$8 = 69(+15784) - 8008(+136)$
$a_0 = 15784 ; a = a_0 + 8008t \in \{8008t + 15784\} = \{..., -232, 7776, 15784, ... \}$
Memorize the steps to the test :-)
Vinícius Ferraz
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