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Diophantine Equation Practice

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Here is a diophantine equation for me and Chris and whoever else is interested to practice on:

Here is the solution - but can we reproduce it with minimal cheating?  Maybe Chris can solve it a less conventional way.  He is pretty tricky like that.  :))

Melody  Jul 2, 2015

#4
+20680
+10

(gcd = greatest common divisor)

1. There is only a solution when gcd( 1027, 712 ) = 1 or in other words 1027 and 712 are relatively prime.

2. Using the Euclidian algorithm to calcutate gcd(1027,712)

$$\small{ \begin{array}{|r|r|r|r|} \hline &&&\\ a & b & q& r \\ \hline &&&\\ 1027 & 712 & 1 & 315 \\ 712 & 315 & 2 & 82 \\ 315 & 82 & 3 & 69 \\ 82 & 69 & 1 & 13 \\ 69 & 13 & 5 &4 \\ 13 & 4 &3 & {1} \\ 4 & 1 & 4 & 0 \\ &&&\\ \hline \end{array} }$$

The greatest common divisor gcd(1027,712) $${=1}$$

and we can go on

3. Using Extended Euclidean algorithm to calculate the first solution

$$\small{ \begin{array}{|r|r|r|r||r|r|} \hline &&&&&\\ a & b & q& r &x&y\\ \hline &&&&&\\ 1027 & 712 & 1 & 315 & {-165} & 73 - 1 (-165) = {238}\\ 712 & 315 & 2 & 82 & 73 & -19 - 2\cdot 73 = -165\\ 315 & 82 & 3 & 69 & -19 & 16 -3(-19) = 73\\ 82 & 69 & 1 & 13 & 16 & -3 -1\cdot 16 = -19\\ 69 & 13 & 5 &4 & - 3 & 1 -5(-3) = 16\\ 13 & 4 &3 & 1 & 1 & 0 - 3\cdot 1 = -3 \\ 4 & 1 & 4 & 0 & 0 & 0\cdot 4 + 1 = 1\\ &&&&&\\ \hline \end{array} }$$

$$\small{\text{ The first solution is  (-165,238) \qquad 1027 \cdot ({ -165} ) + 712 \cdot {238} = 1 }}\\\\$$

4. All Solutions:

$$\small{\text{ \begin{array}{lcl} \boxed{ a\cdot x + b \cdot y = 1 }\\\\ \mathrm{First~ solution~~} (x_0,~ y_0)\\ \mathrm{All~ solutions~~} \left(x_0+\dfrac{z\cdot b}{ gcd(a,b) } ,~ y_0 - \dfrac{z\cdot a}{ gcd(a,b) }\right)\\ \end{array} }}\\\\$$

$$\small{\text{ \begin{array}{lcl} \boxed{ 1027\cdot x + 712 \cdot y = 1 }\\\\ \mathrm{First~ solution~~} (x_0=-165,~ y_0=238)\\ \mathrm{All~ solutions~~} \left(-165+\dfrac{z\cdot 712}{ 1 } ,~ 238 - \dfrac{z\cdot 1027}{ 1 }\right)\\ \end{array} }}\\\\$$

we have:

$$\small{\text{ \begin{array}{lcl} \left\{~\left(~ -165 + 712\cdot z,~ 238 - 1027\cdot z\right)~|~z \in Z ~ \right\} \\ \end{array} }}\\\\$$

heureka  Jul 2, 2015
#1
+92763
+10

I'm going to employ maximum cheating on this one.......

Using the Euclidian algorithm, we have

1027  = 1(712) + 315

712 = 2(315) + 82

315 = 3(82) + 69

82 = 1(69) + 13

69 = 5(13) + 4

13 = 3(4) + 1

4 = 4(1) + 0

1        2        3         1        5          3          4

1  0                        1        2        7         9        52     165       712

0  1                        1        3       10       13       75      238     1027

Take the  determinant of          165     712

238   1027        =   -1

So.....

1027(165) - 712(238) = - 1   but...we need a  "+1 "   .....so......multiplying through by -1, we have

1027(-165) + 712(238) = 1

So....x = -165   and y = 238....however.....this is only one solution.....the general solution is given by

x , y   =  [ (-165 + 712k), (238 - 1027k) ]

CPhill  Jul 2, 2015
#2
+94114
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Well aren't you just so clever.  I worked it out Too :))

Melody  Jul 2, 2015
#3
+92763
+5

Well....Ms. SmartyPants......I even provided a general solution.....so.....take that  !!!

CPhill  Jul 2, 2015
#4
+20680
+10

(gcd = greatest common divisor)

1. There is only a solution when gcd( 1027, 712 ) = 1 or in other words 1027 and 712 are relatively prime.

2. Using the Euclidian algorithm to calcutate gcd(1027,712)

$$\small{ \begin{array}{|r|r|r|r|} \hline &&&\\ a & b & q& r \\ \hline &&&\\ 1027 & 712 & 1 & 315 \\ 712 & 315 & 2 & 82 \\ 315 & 82 & 3 & 69 \\ 82 & 69 & 1 & 13 \\ 69 & 13 & 5 &4 \\ 13 & 4 &3 & {1} \\ 4 & 1 & 4 & 0 \\ &&&\\ \hline \end{array} }$$

The greatest common divisor gcd(1027,712) $${=1}$$

and we can go on

3. Using Extended Euclidean algorithm to calculate the first solution

$$\small{ \begin{array}{|r|r|r|r||r|r|} \hline &&&&&\\ a & b & q& r &x&y\\ \hline &&&&&\\ 1027 & 712 & 1 & 315 & {-165} & 73 - 1 (-165) = {238}\\ 712 & 315 & 2 & 82 & 73 & -19 - 2\cdot 73 = -165\\ 315 & 82 & 3 & 69 & -19 & 16 -3(-19) = 73\\ 82 & 69 & 1 & 13 & 16 & -3 -1\cdot 16 = -19\\ 69 & 13 & 5 &4 & - 3 & 1 -5(-3) = 16\\ 13 & 4 &3 & 1 & 1 & 0 - 3\cdot 1 = -3 \\ 4 & 1 & 4 & 0 & 0 & 0\cdot 4 + 1 = 1\\ &&&&&\\ \hline \end{array} }$$

$$\small{\text{ The first solution is  (-165,238) \qquad 1027 \cdot ({ -165} ) + 712 \cdot {238} = 1 }}\\\\$$

4. All Solutions:

$$\small{\text{ \begin{array}{lcl} \boxed{ a\cdot x + b \cdot y = 1 }\\\\ \mathrm{First~ solution~~} (x_0,~ y_0)\\ \mathrm{All~ solutions~~} \left(x_0+\dfrac{z\cdot b}{ gcd(a,b) } ,~ y_0 - \dfrac{z\cdot a}{ gcd(a,b) }\right)\\ \end{array} }}\\\\$$

$$\small{\text{ \begin{array}{lcl} \boxed{ 1027\cdot x + 712 \cdot y = 1 }\\\\ \mathrm{First~ solution~~} (x_0=-165,~ y_0=238)\\ \mathrm{All~ solutions~~} \left(-165+\dfrac{z\cdot 712}{ 1 } ,~ 238 - \dfrac{z\cdot 1027}{ 1 }\right)\\ \end{array} }}\\\\$$

we have:

$$\small{\text{ \begin{array}{lcl} \left\{~\left(~ -165 + 712\cdot z,~ 238 - 1027\cdot z\right)~|~z \in Z ~ \right\} \\ \end{array} }}\\\\$$

heureka  Jul 2, 2015
#5
+94114
0

Thanks for that great answer Heureka :)

Melody  Jul 3, 2015