Find all the integral solutions to the equation $323x+391y+437z=10473$

Question

Find all the integral solutions to the equation $323x+391y+437z=10473$.

I know how to find integer solutions in two variables using Diophantine Equations.

But I am stuck here because it involves 3 variables.

Can I get a hint?

Answer 1

Recall that - by linearity - the general solution of a non-homogeneous linear equation is obtained by $\color{#0a0}{\text{adding}}$ any particular solution $\rm P$ to the general solution $\rm H$ of the associated homogeneous equation. We can use this to reduce the solution of a trivariate linear Diophantine equation to the well-known bivariate case as below. Here I have followed the presentation that is implicit in the (unproved) formula applied in Robert's answer, and also appended a complete proof of that (unproven) formula.

Homogeneous solution: $\ 323 x + 391 y = - 437 z\ $ is solved as below:

$\gcd(323,391) = 17\mid 437z\,\Rightarrow\, 17\mid z,\ $ so $\ z = 17 m\,$ for $\,m\in\Bbb Z$

Cancelling $\,17\,$ above yields: $\ \,19x\, +\, 23 y\, = -437m.\, $ Recursively solving this bivariate case:

$\ \ \ $ Particular solution: $\bmod 19\!:\ 4y\equiv 0\iff y\equiv 0,\ $ so $\ x = {\large \frac{-437m}{19}} = -23m$

$\ \ \ $ Homogeneous solution: $\ 19x+23y = 0\iff {\large \frac{y}x =\, \frac{\!\!-19}{23}}\!$ $ \iff$ $(x,y) = (23k,-19k)$

$\ \ \ \ \color{#0a0}{\text{Adding}}\rm\ P\!+\!H\!:$ $\ (x,y,z)\, =\, (23k,\, -19(m\!+\!k),\, 17m) = $ general homogeneous solution.

Particular solution $\ (x,y,z) = (8,9,10)\ $ is obtained as follows:

$10473 = 437z + 17(\color{#c00}{19x\!+\!23y}) =: 437z + 17\,\color{#c00}T $

$\!\bmod 17\!:\ z \equiv {\large \frac{10473 }{437}\equiv \frac{1}{12} \equiv \frac{18}2\frac{18}6}\equiv 9\cdot 3\equiv 10\ $ so $\ \color{#c00}T = {\large \frac{10473-437(10)}{17}} = \color{#c00}{359}$

$\color{#c00}{19x\!+\!23y = 359}\ $ $\Rightarrow \bmod 19\!:\ \begin{align}4y&\equiv -2\\ 2y&\equiv -1\equiv 18\end{align}\!\!\iff y\equiv 9\ $ so $\,x = {\large \frac{359-23(9)}{19}} = 8$

$\color{#0a0}{\text{Adding}}\rm\ P\!+\!H\!:$ $\,\ \bbox[5px,border:1px solid #c00]{(x,y,z) = (8\!+\!23k,\, 9\!-\!19(m\!+\!k),\, 10\!+\!17m)}\:$ is a general solution.

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Below is a complete proof of the cited formula - proved exactly as above.

Theorem $ $ Let $\,a,b,c\in\Bbb Z\,$ with gcd $\,(a,b,c) = 1,\,$ let gcd $\, (a,b) =: g,\,$ and $\,a' = a/g,\ b' = b/g.$

Let $\ \ z_0,\, t_0\in\Bbb Z\,\ $ be any solution of $\, \ c\,z\:\! +\ g\,t\, =\, d$
and $\ \color{#90f}{u_0,v_0}\in\Bbb Z\ $ be any solution of $\ \, a'u + b'v\, =\, c$
and $\ x_0, y_0\in\Bbb Z\ $ be any solution of $\ \ a'x + b'y =\:\! t_0$.

Then $\,ax + by + cz = d\,$ has the general solution $\,\ \begin{align} x &= x_0 + b'k - u_0 m\\ y &= y_0 - a'k - v_0 m\\ z &= z_0 + gm\end{align}\,\ $ for any $\,k,m\in\Bbb Z$

Proof: $ $ Homogeneous solution: $\ a x + b y = -c z\ $ is solved as below:

$(a,b)\!=\! g\mid cz\overset{(g,\,c)=1}\Rightarrow\! g\mid z,\ $ so $\ z = g m,\,m\in\Bbb Z,\,$ by $(g,c)\! =\! (a,b,c)\!=\!1\,$ & $ $ Euclid's Lemma.

Cancel $\,g\,$ in $\,ax+by = -cz\,$ $\Rightarrow \,a'x\, +\,b' y\, = -cm.\, $ Recursively solving this bivariate case:

$\ \ \ $ Particular solution: $\ (x,y) = (-u_0m,-v_0m)\ $ by $\,\color{#90f}{u_0,v_0}\,$ hypothesis scaled by $\,-m$.

$\ \ \ $ Homogeneous solution: $\ a'x+b'y = 0\iff {\large \frac{y}x =\, \frac{\!\!-a'}{b'}}\!$ $ \iff$ $(x,y) = (b'k,-a'k)$

$\ \ \ \ \color{#0a0}{\text{Adding}}\rm\ P\!+\!H\!:$ $\ (x,y,z)\, =\, (b'k\!-\!u_0m,\, -a'k\!-\!v_0m,\, gm) = $ general homogeneous solution.

Particular solution $\,\ (x,y,z) = (x_0,y_0,z_0)\ $ is obtained as follows:

$d = cz + g(\color{#c00}{a'x\!+\!b'y}) =: cz + g\,\color{#c00}t\ $ has solution $\,(z,\color{#c00}t) = (z_0,\color{#c00}{t_0})\,$ by hypothesis.

and also: $\, \ \ \ \color{#c00}{a'x\!+\!b'y = t_0}\,\ $ has $ $ as $ $ a $ $ solution: $\ \, (x,y) = (x_0,y_0)\,$ by hypothesis.

Therefore $\,(x,y,z) = (x_0,y_0,z_0)\,$ is a particular solution.

$\color{#0a0}{\text{Adding}}\,$ the Particular and Homogeneous solutions yields the claimed general solution.

Remark $ $ If $\, e := (a,b,c) > 1\,$ then $\,e\mid d\,$ so cancelling $e$ in the equation reduces to the above case.

Answer 2

Hint. Reducing $323x+391y+437z=10473\ $ modulo $\,17\,$ yields $\,12z\equiv1\pmod{\!17}$. Similar constraints can be found on $x$ by reducing $\!\bmod 23\,$ and $y$ by reducing $\!\bmod 19$.

Answer 3

Since $\gcd(323,391,437)=1$ divide $10473$ we are supposed to find infinite solutions.

Hint. First find a solution $u_0$, $v_0$ of $$19u + 23 v = 437$$ where $19=323/17$ and $23=391/17$ with $17=\gcd(323,391)$. Then let $t_0$, $z_0$ be a solution of $$17t+ 437z=10473$$ and $x_0$, $y_0$ be a solution of $$19x + 23y = t_0.$$ Then $(x_0,y_0,z_0)$ is a particular solution of $323x+391y+437z=10473$, whereas the general solution is given by $$\begin{cases} x = x_0 - 23k - u_0j\\ y = y_0 + 19k - v_0j\\ z = z_0 + 17j \end{cases}$$ with $j,k\in\mathbb{Z}$.

Then compare your result given by Script.

P.S. Finally, I got the general solution: $$\begin{cases} x = 8 - 23k -23j\\ y = 9 + 19k\\ z = 10 + 17j \end{cases}\tag{*}$$ with $j,k\in\mathbb{Z}$.

Verification that (*) are ALL the solutions of the given linear Diophantine equation. It is easy to check that the particular solution $(x_0,y_0,z_0)=(8,9,10)$ works. Moreover, the related homogeneous equation is $$323(x-x_0)+391(y-y_0)+437(z-z_0)\\=17\cdot 19 (x-x_0)+17\cdot 23(y-y_0)+19\cdot 23 (z-z_0)=0$$ and it follows that $z-z_0$ is a multiple of $17$, i.e. $z = z_0 + 17j$, $y-y_0$ is a multiple of $19$, i.e. $y = y_0 + 19k$, and therefore $$x=x_0-\frac{391(y-y_0)+437(z-z_0)}{323}=x_0-\frac{(17\cdot 19)\cdot 23 k+(19\cdot 23) \cdot 17j}{17\cdot 19}\\=x_0-23k-23j$$ and we are done.

Note that along the same lines, you may show that the method outlined above works in general.

Answer 4

Applying the Extended Euclidean Algorithm

The Extended Euclidean Algorithm is usually applied to a pair of numbers, but by combining the results from the pairs $(391,323)$, $(437,391)$, and $(437,323)$, we can get a similar result for the triple $(437,391,323)$.

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Applying the Extended Euclidean Algorithm as implemented in this answer to $391$ and $323$ $$ \begin{array}{r} &&1&4&1&3\\\hline 1&0&1&-4&5&-19\\ 0&1&-1&5&-6&23\\ 391&323&68&51&17&0\\ \end{array} $$ we get $\gcd(391,323)=17$ and $$ \begin{align} 5\cdot391-6\cdot323&=17\tag{1a}\\ 19\cdot391-23\cdot323&=0\tag{1b} \end{align} $$ Applying the Extended Euclidean Algorithm to $437$ and $391$ $$ \begin{array}{r} &&1&8&2\\\hline 1&0&1&-8&17\\ 0&1&-1&9&-19\\ 437&391&46&23&0\\ \end{array} $$ we get $\gcd(437,391)=23$ and $$ \begin{align} 9\cdot391-8\cdot437&=23\tag{2a}\\ 19\cdot391-17\cdot437&=0\tag{2b} \end{align} $$ Applying the Extended Euclidean Algorithm to $437$ and $323$ $$ \begin{array}{r} &&1&2&1&5\\\hline 1&0&1&-2&3&-17\\ 0&1&-1&3&-4&23\\ 437&323&114&95&19&0\\ \end{array} $$ we get $\gcd(437,323)=19$ and $$ \begin{align} 3\cdot437-4\cdot323&=19\tag{3a}\\ 17\cdot437-23\cdot323&=0\tag{3b} \end{align} $$

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Writing $\bf{1}$ as a Linear Combination of $\bf{323}$, $\bf{391}$, and $\bf{437}$

Since $17$, $19$, and $23$ share no common factors, we can write $1$ as a linear combination of $323$, $391$, and $437$.

We start by applying the Extended Euclidean Algorithm to $23$ and $17$, the gcds in $\text{(2a)}$ and $\text{(1a)}$: $$ \begin{array}{r} &&1&2&1&5\\\hline 1&0&1&-2&3&-17\\ 0&1&-1&3&-4&23\\ 23&17&6&5&1&0\\ \end{array} $$ We get that $\gcd(23,17)=1$ and $$ 3\cdot23-4\cdot17=1\tag4 $$ Applying $\text{(1a)}$ and $\text{(2a)}$ to $(4)$ yields $$ \begin{align} 1 &=3\cdot\overbrace{23}^\text{(2a)}-4\cdot\overbrace{17}^\text{(1a)}\\ &=3(9\cdot391-8\cdot437)-4(5\cdot391-6\cdot323)\\ &=24\cdot323+7\cdot391-24\cdot437\tag5 \end{align} $$ Equation $(5)$ shows how to write $1$ as a linear combination of $323$, $391$, and $437$. Using $\text{(3b)}$, $(5)$ can be reduced to $$ 1=1\cdot323+7\cdot391-7\cdot437\tag6 $$

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Writing $\bf{10473}$ as a Linear Combination of $\bf{323}$, $\bf{391}$, and $\bf{437}$

We can simply multiply $(6)$ by $10473$ and reduce using $\text{(1b)}$ and $\text{(2b)}$: $$ \begin{align} 10473 &=10473\cdot323+73311\cdot391-73311\cdot437\\ &+455\,(19\cdot391-23\cdot323)\tag{7a}\\ &=8\cdot323+81956\cdot391-73311\cdot437\\ &-4313\,(19\cdot391-17\cdot437)\tag{7b}\\ &=8\cdot323+9\cdot391+10\cdot437\tag{7c} \end{align} $$ Explanation:
$\text{(7a)}$: reduce the coefficient of $323$ using $\text{(1b)}$
$\text{(7b)}$: reduce the coefficient of $391$ using $\text{(2b)}$
$\text{(7c)}$: a reduced linear combination

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The General Solution

The difference of two solutions to $323x+391y+437z=10473$ is a solution to the homogeneous equation $$ 323x+391y+437z=0\tag8 $$ Consequences of $(8)$:
Since $\gcd(323,437)=23$, we have $23\mid x$, so WLOG let $x=23a$.
Since $\gcd(391,437)=19$, we have $19\mid y$, so WLOG let $y=19b$.
Since $\gcd(323,391)=17$, we have $17\mid z$.
Note that $323(23a)+391(19b)=437(17a+17b)$, so we need $z=-17a-17b$.

Therefore, the general solution to $(8)$ is $$ 323(23a)+391(19b)-437(17a+17b)=0\tag9 $$ Thus, combining $\text{(7c)}$ and $(9)$, the general solution to $323x+391y+437z=10473$ is $$ \bbox[5px,border:2px solid #C0A000]{10473=(8+23a)\,323+(9+19b)\,391+(10-17a-17b)\,437}\tag{10} $$ Looking at $(10)$, it appears that $\text{(7c)}$ is the only solution with all positive coefficients.

Answer 5

Below we show how it can be solved using more general methods for solving systems of Diophantine equations by reducing them to Hermite / Smith triangular / diagonal and related normal forms. If you search on those keywords you should find expositions on these general methods.

Below is one simple way to do so, via this method, except here we need to keep track of the $3$ current rows, which are the current row plus the two notated at row end, e.g. rows $\color{#c00}{[\![5]\!]}$ snd $[\![4,2]\!]$ below.

$\ \ \ \ \begin{array}{rrrrrl} [\![1]\!] & 437 & 1 & 0 & 0 \\ [\![2]\!] & 391 & 0 & 1 & 0 \\ [\![3]\!] & 323 & 0 & 0 & 1 \\ [\![1]\!]-1\,[\![2]\!]\, \rightarrow\, [\![4]\!] & 46 & 1 & -1 & 0 &[\![3,2]\!]\\ [\![3]\!]-7\,[\![4]\!]\, \rightarrow\, \color{#c00}{[\![5]\!]} &\color{#c00}1 & \color{#c00}{{-}7} & \color{#c00}7 & \color{#c00}1& [\![4,2]\!]\ \ \ \ \ \ \smash{\overbrace{\color{#c00}1 = \color{#c00}7\cdot 437 \color{#c00}{-7}\cdot 391+\color{#c00}1\cdot 323}^{\large \color{#c00}{\text{Bezout}}\text{ Identity}}}\\ [\![2]\!]-8\,[\![4]\!]\, \rightarrow\, [\![6]\!] & 23 & {-}8 & 9 & 0& [\![5,4]\!]\\ [\![4]\!]-2\,[\![6]\!]\, \rightarrow\, \color{#0a0}{[\![7]\!]} & \color{#0a0}0 & \color{#0a0}{17} & \color{#0a0}{{-}19} & \color{#0a0}0&[\![6,5]\!]\ \ \ \ \ \ \color{#0a0}{\text{Null$_{\:\!1}$}}\\ [\![6]\!]\!\!-23[\![5]\!]\, \rightarrow\,[\![8]\!] & 0 &\!\! 153 & \!\!{-}152 &\! {-}23&[\![7,5]\!]\\ [\![8]\!]-9\,[\![7]\!]\, \rightarrow\, \color{#90f}{[\![9]\!]} & \color{#90f}0 & \color{#90f}0 & \color{#90f}{19} & \!\color{#90f}{{-}23} &[\![7,5]\!]\ \ \ \ \ \ \color{#90f}{\text{Null$_{\:\!2}$}}\\ \end{array}$

$\begin{array}{r r r r r l} 10473\color{#c00}{[\![5]\!]}\, \rightarrow\, [\![a]\!] & 10473 &\!\!\!-73311 &\!\! 73311 &\!\! 10473&\ \ 10473\times \color{#c00}{\text{Bezout}}\text{ Identity}\\ [\![a]\!]\!-\!4313\color{#0a0}{[\![7]\!]}\, \rightarrow\, [\![b]\!] & 10473 &\!\!\!10 &\!\!\! -8636 &\!\! 10473&\ \ \text{use }\color{#0a0}{{\text {Null}}_{\:\!1}}\text{ to reduce coef of }437\\ [\![b]\!]\:\!+\:\!455\color{#90f}{[\![9]\!]}\, \rightarrow\, [\![c]\!] & 10473 &10 & 9 & 8& \ \ \text{use }\color{#90f}{{\text{Null}}_{\:\!2}}\text{ to reduce coef of }391\end{array}$

A particular solution is: $\ 10473 = 10\cdot 437 + 9 \cdot 391 + 8\cdot 323\ $ from the prior row, and

the null space is $\ (z,y,x) = (\color{#0a0}{17,-19,0})m - (\color{#90f}{0,19,-23})k = (17m,-19(m\!+\!k),23k)$

Compare robjohn's answer, which is similar, but does not explicitly use Hermite row reduction.

Answer 6

Three variables doesn't really make as much of a difference as you'd think.

$323 =17*19$ and $391 = 17*23$ and $437 = 19*23$

Let's suppose that $(x,y,z)$ and $(x+a, y+b, z+c)$ are two solutions to $323x + 391y + 437z = 10473$.

Then $17*19(a) + 17*23(b) + 19*23(c)= 0$ and so

$a\equiv 0 \pmod 23$, and $b\equiv 0 \pmod 19$ and $c\equiv 0 \pmod 17$

So suppose $a= 23j; b=19k; c=17m$

then we must have $j+k+m = 0$ and any such combination is possible.

So if $(x,y,z)$ is a solution then $(x + 23j,y+19k,z-17(j+k))$ will be a solution and that generates all solutions.

Now by Bezout we can solve $323A + 391B= 17$ and $391C + 437D = 23$ and $17M+23N = 10473$

so $(323A + 391B)M + (391C+437D)N = 10473$

and $323AM + 391(BM+CN) + 437DN = 10473$ is a solution.

So the solution set is $\{(AM+23j, BM+CN+19k, DN+17m| j+k + m = 0\}$.

..... .....

Now was the question supposed to be that $x,y,z$ must all be positive?

If so:

$323A+391B =17$

$19A + 23B = 1$

(Argh, I really hate doing this but...)

$23 - 19 = 4$

$3 = 19-4*4 = 19 - 4(23-19) = 19*5 - 4*23$

$1 = 4-3 = (23-19)-(19*5 - 4*23) = 5*23- 6*19$

So we can let $A=-6$ and $B=5$.

$391C + 437D = 23$

$17C + 19D = 1$ so

$19 - 17 = 2$

$1=17 - 8*2 = 17- 8(19-17) = 9*17-8*19$

So we can let $C= 9$ and $D = -8$.

And for

$17M+23N = 10473$

$23 -17 = 6$

$1=3*6 - 17=3(23-17) - 17=3*23 - 4*17$

So we can let $M=-4(10473)$ and $N=3(10473)$.

So the solution set is:

$\{(AM+23j, BM+CN+19k, DN+17m| j+k + m = 0\}=$

$\{(24*10473+23j, 7*10473+19k, -24*10473+17m| j+k + m = 0\}$

To make these monster managable:

Since $24*10473, 7*10473, -24*10473$ is a solution then so is

$24*10473 - 23*2*10473, 7*10473, -24*10473 + 17*2*10473 = -22*10473,7*10473, 10*10473$ and so is

$10473, 7*10473 -19*10473, 10*10473= 10473,-125676, 104730 $ and so is

$10473, -125676+19*6615, 104730-17*6615=10473,9,-7725$ and so is

$ 10473-455*23,9, -7725+455*17= 8,9,10$

And ... that's the only positive solution. To have $j + k +m=0$ then one of $j,k,m \le -1$ and $8-23,9-19, 10-17$ are all negative.

Answer 7

$$323x+391y+437z=10473$$

My version of EEA is as follows.

$\begin{array}{r|r|rrr|l} & & 437 & 391 & 323 \\ \hline & 437 & 1 & 0 & 0 & 437-1\cdot323 = 114\\ & 391 & 0 & 1 & 0 & 391-1\cdot323 = 68\\ & 323 & 0 & 0 & 1 & \\ \hline & 323 & 0 & 0 & 1 & 323-4\cdot 68 = 51\\ & 114 & 1 & 0 & -1 & 114 - 1\cdot 68 = 46\\ & 68 & 0 & 1 & -1 & \\ \hline & 68 & 0 & 1 & -1 & 68-1\cdot 46 = 22\\ & 51 & 0 & -4 & 5 & 51 - 1 \cdot 46 = 5\\ & 46 & 0 & -1 & 1 & \\ \hline & 46 & 1 & -1 & 0 & 46 - 9\cdot 5 = 1\\ & 22 & -1 & 2 & -1 & 22 - 4\cdot 5 = 2\\ & 5 & -1 & -3 & 5 & \\ \hline & 5 & -1 & -3 & 5 & 5-5\cdot 1 = 0\\ & 2 & 3 & 14 & -21 & 2-2\cdot 1 = 0\\ & 1 & 10 & 26 & -45 & \\ \hline & 1 & 10 & 26 & -45 & \\ & 0 & -51 & -133 & 230 & \text{See comments below.}\\ & 0 & -17 & -38 & 69 & \\ \hline & 1 & 10 & 26 & -45 & \\ & 0 & 17 & -19 & 0 & \text{See comments below.}\\ & 0 & 0 & 19 & -23 \\ \hline \end{array}$

Comments. The "null space" generated by the algorithm tends to be "ugly". Looking at the basis elements pairwise will give a much prettier null space.

So

$$323(-45) + 391(26) + 437(10) = 1$$

and

$$323(-471285) + 391(272298) + 437(104730) = 10473.$$

Hence

$$(x,y,z) = (-471285-23t, 272298-19s+19t, 104730+17s)$$

Answer 8

We are looking for all integer solutions of

$$\tag 1 323x+391y+437z=10473$$

Using prime factorization this is equivalent to

$$ (17\cdot 19) x+(17\cdot 23)y+(19\cdot 23)z=10473$$

Applying the

$$ \pmod{23} / \pmod{19} /\pmod{17} $$

operators and solving linear congruence equations we get, respectively, that there exist integers $s, t, u \in \mathbb Z$ so that

$$ x = 8 + 23s$$
$$ y = 9 + 19t$$
$$ z = 10 + 17u$$

But the equation

$$ 323 (8 + 23 s) + 391 (9 + 19 t) + 437 (10 + 17 u) = 10473 $$

is equivalent to

$$ 7429 s + 7429 t + 7429 u = 0 $$

or

$$ s + t + u = 0 $$

or

$$ u = -s -t $$

and so we can choose any $(s,t) \in \mathbb Z$ to determine $u$ and a solution to $\text{(1)}$.

General solution:

$$ \{ \bigr( 8+23s, 9 + 19t, 10 -17(s+t) \bigr ) \mid (s,t) \in \mathbb Z \}$$