The question is to find all solutions of each equation: $$444x = 148 \in \Bbb Z_{964}$$
$$39x = 37 \in \Bbb Z_{53}$$
I've googled the answer and got no results. I think I'm supposed to use extended gcd algorithm but I'm not sure how
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Googled... the answer ? It is internet, not black magic! What've you tried so far? – DonAntonio Nov 08 '13 at 19:38
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I've tried finding a method of answering this question online (googled it) and I didn't find anything. – Miller.Steve Nov 08 '13 at 19:40
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@observeRReacts: Maybe you should consult a software like WA or Mathematica. – Mikasa Nov 08 '13 at 19:48
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Should it be $44x$ or $444x$? – Cameron Buie Nov 08 '13 at 19:58
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@Cameron Buie 444x – Miller.Steve Nov 08 '13 at 20:05
3 Answers
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I'll work the second one first. Note that $$53=39+14\\39=2\cdot14+11\\14=11+3\\11=3\cdot 3+2\\3=2+1\\2=2\cdot 1+0,$$ so $$1=3-2\\2=11-3\cdot 3\\3=14-11\\11=39-2\cdot14\\14=53-39.$$ Hence, by repeated substitution, we have $$\begin{align}1 &= 3-2\\ &= 3-(11-3\cdot 3)\\ &= 4\cdot 3-11\\ &= 4\cdot (14-11)-11\\ &= 4\cdot14-5\cdot11\\ &= 4\cdot14-5(39-2\cdot14)\\ &= 14\cdot14-5\cdot39\\ &= 14(53-39)-5\cdot39\\ &= 14\cdot53-19\cdot39.\end{align}$$ This means that in $\Bbb Z_{53},$ we have $1=39\cdot-19,$ so multiplying both sides of $39x=37\pmod{53}$ by $-19,$ we have $$\begin{align}x &= -19\cdot 37\pmod{53}\\ &= -703\pmod{53}\\ &= 14\cdot53-703\pmod{53}\\ &= 742-703\pmod{53}\\ &= 39\pmod{53},\end{align}$$ and so $x=39+53n$ for some $n\in\Bbb Z$.
For the first one, note that $$964=2\cdot 444+76\\444=5\cdot76+64\\76=64+12\\64=5\cdot12+4\\12=3\cdot4+0.$$ Hence, $4$ is the greatest common divisor of $964$ and $444$. In particular, working our way back through, we end up with $$4 = 76\cdot444-35\cdot964.$$ So multiplying both sides of $444x=148\pmod{964}$ by $76,$ we have $$\begin{align}4x &= 22\cdot 148\pmod{964}\\ &= 11248\pmod{964}\\ &= -11\cdot964+11248\pmod{964}\\ &= -10604+11248\pmod{964}\\ &= 644\pmod{964},\end{align}$$ so $4x=644+964n$ for some $n\in\Bbb Z,$ and so $x=161+241n$ for some $n\in\Bbb Z$.
Cameron Buie
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You are the man, THANK YOU. I just needed an example that broke it down like this so that I could do more. – Miller.Steve Nov 08 '13 at 20:16
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1Well, since you're probably assuming for the first one that $0\le x\le 963,$ then we can restrict our attention to $x=161+241n$ for $n=0,1,2,3.$ That is, $x=161,402,643,884.$ Does that match up? Note that (for example) when $n=4,$ we get $1125,$ but $$\begin{align}1125 &= -964+1125\pmod{964}\ &= 161\pmod{964},\end{align}$$ so we didn't really get anything new. – Cameron Buie Nov 08 '13 at 20:25
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On the first one, did you mean x= 39 + 53n? And here, do you mean 161, 402, 643, 884? THanks again. – Miller.Steve Nov 08 '13 at 20:26
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Yes, on the first one, I meant $x=39+53n$. That was a typo. And here, I did mean that. If you refresh the page, you should see that I corrected that comment. – Cameron Buie Nov 08 '13 at 20:28
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1Hint: Don't forget that $964=0\pmod{964}.$ We will do the same sort of thing that we did to get $-19\cdot37=39\pmod{53}.$ – Cameron Buie Nov 09 '13 at 23:43
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1You need to work your way through the repeated substitution, as in the first part, where we started with $$\begin{align}1 &= 3-2\ &= 3-(11-3\cdot3)\ &\vdots\end{align}$$ – Cameron Buie Nov 10 '13 at 01:05
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Here is a start:
The Euclid-Wallis Algorithm is an implementation of the Extended Euclidean Algorithm. $$ \begin{array}{r} &&21&1&10\\\hline 1&0&1&-1&11\\ 0&1&-21&22&-241\\ 964&44&40&4&0\\ \end{array} $$ So we get that $$ (22)44+(-1)964=4 $$ Multiplying by $37$ and adding the homogenous solution, we get $$ (814-241k)44+(-37+11k)964=148 $$
Even if the numbers are changed, the process is the same: $$ \begin{array}{r} &&2&5&1&5&3\\\hline 1&0&1&-5&6&-35&111\\ 0&1&-2&11&-13&76&-241\\ 964&444&76&64&12&4&0\\ \end{array} $$ So we get that $$ (76)444+(-35)964=4 $$ Multiplying by $37$ and adding the homogenous solution, we get $$ (2812-241k)444+(-1295+111k)964=148 $$ We can adjust $k$ to get a smaller particular solution: $$ (161-241k)444+(-74+111k)964=148 $$
