Diophantine Equation with 3 Variables example

Question

$15x+72y+21z=9 (1)$

$15+72y = gcd(15,72).r1$

(1) can be written as :

15x+72y=gcd(15,72).r1    (2)
gcd(15,72).r1 + 21z = 9  (3)

$15x+72y=gcd(15,72).r1$

Calculating $GCD(15,72)$ gives:

$72 = 4*15 + 12$

$15 = 1*12 + 3$

$12 = 4*3 + 0$

from (2) and (3) $15x+72y=3.r1 (4)$

$3.r1 + 21z = 9$ (5)

Then applying the Extended Euclidean Algorithm:

$3 = (1 * 15) + (-1 * 12) = (-1 * 72) + (5 * 15)$

A particular solution is $x,y = (5,-1)$ multiply by r1 the $(5r1,-r1)$

general solution $(x,y)=(5r1+24m, -r1-5m)$

$3.r1 + 21z = 9 gcd (3,21) = 3 | 9$ so the is solution $(r1,z) = (3+7n, -n)$

$x=5r1+24m = 5(3+7n)+24m=15+35n+24m$

$y=-r1-5m=-3-7n-5m$

$z=-n$

$(x,y,z) = (15+35n+24m, -3-7n-5m, -n)$

I provide this exercise as an example for diophantine equations with 3 variables however please somebody check for mistakes

Please use mathJax: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference — person, Dec 14 '20 at 04:45
Substituting your $x,y,z$ back into the original equation, there's nothing to cancel the appearance of $m$ on the left side. — Gerry Myerson, Dec 14 '20 at 04:51
I am very new and I do not know how to use it yet. I will try net time — gav, Dec 14 '20 at 04:52
error corrected (x,y,z) = (15+35n+24m, -3-7n-5m, -n) is the solution — gav, Dec 14 '20 at 12:49
That looks better. Have you tried plugging it in to check that it works? — Gerry Myerson, Dec 15 '20 at 06:39
I checked my final solution and it is correct. However, there are many other ways to come to the same result — gav, Dec 18 '20 at 20:14

poetasis · Answer 1 · 2020-12-15T01:46:03.670

I'm not sure what you are after with modular arithmetic but the equation has integer solutions that can be found by other means... assuming a solution in integers is what you seek.

Experimenting with values in a spreadsheet, we can let $x$ be any integer and notice that $y,z$ must jump in respective multiples of $7$ and $24$ if the equation is to sum to a consistent value. Further experimentation locates the constants in their separate equations so that the original sums to zero.

$$15x+72y+21z=9\implies 5x+24y+7z-3=0$$

$$\text{Let}\qquad y = 7 n + 3 x + 1\qquad z = -24 n - 11 x - 3\qquad n,x \in Z$$ Here are some example triples of $(x,y,z)$ as $n$ and $x$ vary from $-2$ to $2$

For $n=-2$ $$\cdots\quad (-2,-19,67)\quad (-1,-16,56)\quad (0,-13,45)\quad (1,-10,34)\quad (2,-7,23)\quad\cdots$$

For $n=-1$ $$ \cdots\quad (-2,-12,43)\quad (-1,-9,32)\quad (0,-6,21)\quad (1,-3,10)\quad (2,0,-1)\quad\cdots$$

For $n=0$ $$\cdots\quad (-2,-5,19)\quad (-1,-2,8)\quad (0,1,-3)\quad (1,4,-14)\quad (2,7,-25)\quad\cdots$$

For $n=1$ $$ \cdots\quad (-2,2,-5)\quad (-1,5,-16)\quad (0,8,-27)\quad (1,11,-38)\quad (2,14,-49)\quad\cdots $$

For $n=2$

$$\cdots\quad (-2,9,-29)\quad (-1,12,-40)\quad (0,15,-51)\quad (1,18,-62)\quad (2,21,-73)\quad\cdots$$

Diophantine Equation with 3 Variables example

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