diophantine equations in three variables

8/11/2019 Diophantine Equations in Three Variables

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Diophantine Equations in Three Variables

Date: 11/17/2004 at 08:19:26

From: Kazim

Subject: Diophantine Equations

I need to know how to get positive integer solutions of twoDiophantine equations having three variables. For example:

2x + 3y + 7z = 32 ; 3x + 4y - z = 19

(give the positive set of triples for the above equations)

Date: 11/17/2004 at 19:34:32

From: Doctor Vogler

Subject: Re: Diophantine Equations

Hi Kazim,

First of all, there are two ways to interpret your two equations, and

I'm not sure which one you want. Either you want positive integer

triples (x, y, z) that satisfy *both* equations

2x + 3y + 7z = 32

3x + 4y - z = 19,

or you want all positive integer triples that satisfy the equation

2x + 3y + 7z = 32

as well as all positive integer triples that satisfy the equation

3x + 4y - z = 19.

If the first case is what you want, then you just solve for the

variable z in the second equation, substitute into the first equation,

and that gives you a normal linear Diophantine equation in two

variables. (If there weren't a variable with a coefficient of 1, then

you could make one of the coefficients 1--leaving the rest integers--

by adding and subtracting multiples of one equation from the other,

back and forth.)

Here's an answer for our archives that discusses this case:

Systems with More Variables than Equations

http://mathforum.org/library/drmath/view/61825.html

If the second case is what you want, then you are asking how to solve

a linear Diophantine equation in three variables. The following

answer discusses this case:

Diophantine Equations, Step by Step

http://mathforum.org/library/drmath/view/61325.html

However, I would like to take this opportunity to describe a different

technique for solving three-variable linear Diophantine equations that

gives an answer more like the formula for all integer solutions totwo-variable linear Diophantine equations.
http://mathforum.org/library/drmath/view/61825.htmlhttp://mathforum.org/library/drmath/view/61325.htmlhttp://mathforum.org/library/drmath/view/61325.htmlhttp://mathforum.org/library/drmath/view/61825.html


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So I will use this technique to find all integer solutions to an

equation, and then mention how you use this to find all *positive*

integer solutions.

First let's solve the equation

6x + 10y + 15z = 0.

We know that z has to be even, right? And we know that y has to a

multiple of 3, right? So suppose that

z = 2a

and

y = 3b.

Then we solve for x:

6x + 10(3b) + 15(2a) = 0

6x = -30b - 30a

x = -5(a+b).

Then that means that all solutions are

x = -5(a+b)

y = 3b

z = 2a

for any integers a and b.

You can describe all of the solutions in different ways, as well.

Basically, x has to be divisible by 5, y by 3, and x by 2. Then you

can pick any two of those variables to be anything (satisfying the

divisibility condition) and the third variable is determined from the

equation.

Let's try another. Let's solve your equation

2x + 3y + 7z = 32.

As in solving a two-variable linear Diophantine equation, the first

thing to do is find any *one* integer solution. In three variables,

the easiest way to do this is often to pick any number for one of thevariables and then use your usual techniques (such as modular inverses

or the Euclidean algorithm) to find solutions in the others. Here's

an easy solution for this equation: Take x=16 to cancel the 32 on the

right, so you have

3y + 7z = 0,

and then take y = 7 and z = -3. So now we have one solution,

(16, 7, -3)

and we want to find all solutions. Well, let's suppose that

(x, y, z)


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is another solution. Then what happens if we subtract our known

solution from it?

2(x - 16) + 3(y - 7) + 7(z + 3) = (2x + 3y + 7z) - 32 = 0.

So that means that if

a = x - 16

b = y - 7

c = z + 3,

then (a, b, c) satisfy the equation

2a + 3b + 7c = 0.

Now you see why I started with an equation that equaled zero. In the

previous case, there were common factors between any two of the

coefficients, and that allowed us to divide by those common factors.

If there are not, as here, then you don't divide by anything. (I.e.

you divide by the common factor 1.)

But we still can't solve for a until we know that

3b + 7c

is even. So we have to make it even. We do this by noticing that if

3b + 7c

is even, then so is b - c. So we let

b = c + 2r,

and now we know that

2a = -(3b + 7c) = -(3c + 6r + 7c) = -(10c + 6r)

is even, so we can solve for a:

a = -(5c + 3r).

And c can be anything. So we have

a = -5s - 3r

b = s + 2r

c = s

for some integers r and s. That is, every solution to

2a + 3b + 7c = 0

has that form for some r and s. Furthermore, for any integers r and

s, those three numbers are solutions. If we recall that

a = x - 16

b = y - 7

c = z + 3,

then we can also say about all solutions to


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2x + 3y + 7z = 32

that they are given by

x = 16 - 5s - 3r

y = 7 + s + 2r

z = -3 + s.

Finally, if we want to find only positive solutions, then we just have

to find all (s, r) pairs that satisfy the three inequalities

x = 16 - 5s - 3r > 0

y = 7 + s + 2r > 0

z = -3 + s > 0.

The easiest way to solve these inequalities might be to graph them (as

lines) on a plane and count all of the lattice points (integer

points).

Now, let me summarize (and generalize). Suppose we want to solve an

equation of the form

Ax + By + Cz = N

for some (known) integers A, B, C, and N. We may assume that A, B,

and C have no common factor (of all three), since if such a common

factor divides N, then we can divide the whole equation by that

number. If that common factor does not divide N, then there are no

solutions. First we find a single particular solution by choosing any

number x. If gcd(B, C) > 1, then our x must satisfy

Ax = N (mod gcd(B, C)).

Then we find some solution for y and z, such as by using the Euclidean

Algorithm. So now we have a particular solution (x0, y0, z0). To

find all solutions, we let

x' = x - x0

y' = y - y0

z' = z - z0

and find that (x', y', z') must satisfy

Ax' + By' + Cz' = 0.

Next, x' must be divisible by gcd(B, C), and y' must be divisible by

gcd(A, C), and z' must be divisible by gcd(A, B), so we let

x" = x'/gcd(B, C)

y" = x'/gcd(A, C)

z" = z'/gcd(A, B),

and we let

A' = A/(gcd(A, C)*gcd(A, B))

B' = B/(gcd(A, B)*gcd(B, C))

C' = C/(gcd(A, C)*gcd(B, C))

so that now we have

A'x" + B'y" + C'z" = 0


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where A', B', and C' are pairwise relatively prime (that is, no two

have a common factor).

Now we would like to solve for x", so we decide what b must be mod A'.

B'y" + C'z" = 0 (mod A').

So we find an integer k that satisfies

B'k + C' = 0 (mod A'),

which is -C'/B' (mod A') and can be computed using the Euclidean

algorithm, and then y" - kz" must be divisible by A', so we let

y" = kz" + A'r

and then we solve for x"

A'x" = -B'y" - C'z"

= -B'(kz" + A'r) - C'z"

= -(B'k + C')z" - A'B'r

and so

x" = -B'r - z"(B'k + C')/A'

where that last number m = (B'k + C')/A' is an integer. Finally, we

choose any number for z" and we get

x" = -B'r - ms

y" = ks + A'r

z" = s

x' = gcd(B, C) * (-B'r - ms)

y' = gcd(A, C) * (ks + A'r)

z' = gcd(A, B) * s

x = x0 + gcd(B, C) * (-B'r - ms)

y = y0 + gcd(A, C) * (ks + A'r)

z = z0 + gcd(A, B) * s

and this last formula gives us all integer solutions to the equation

Ax + By + Cz = N

in the form of linear combinations of the two arbitrary integers r ands. Of course, this isn't the *only* way to express all integer

solutions. For example, we can change r to -t or t-s or t-7s, and get

new ways to write out all integer solutions. Or we can change s to

t-6r, for example. (We can't change them to just anything, but we can

change them to many different things.)

I hope you find this useful and informative. If you have any

questions about this or need more help, please write back, and I will

try to explain further.


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Systems with More Variables than Equations

Date: 12/11/2002 at 11:51:14

From: Sridar

Subject: Solving systems

Dr. Math. How do you solve this system?:

187y + 98x + 45z = 48

2y + 9x + 3z = 198

I tried solving it by trying to multiply equation 2 by 15 to eliminate

Z, but do you multiply again by another number to eliminate another

variable? I am totally stuck here.

P.S. I am in 4th grade but I am learning high school math! Isn't

that cool?

Date: 12/11/2002 at 13:11:27

From: Doctor Greenie

Subject: Re: Solving systems

Hello, Sridar -

With three variables and only two equations, you can't do anything

more than eliminate one variable. You will end up with a single

equation with two variables; the graph of such an equation is a

straight line, indicating that there are infinitely many solutions.

Usually, with problems like this, it is either stated or implied that

the solutions should be integers; in that case, you can perform an

analysis to determine the family of distinct solutions.

In your particular example, eliminating z gives you the equation

37x - 157y = 2922

The standard approach to finding the integer solutions of an equation

like this looks something like this:

37x - 157y = 2922

37x = 157y + 2922

x = (157/37)y + (2922/37)

x = 4y + (9/37)y + 78 + 36/37

x = (4y + 78) + (9y + 36)/37

Now since x must be an integer, and since 4y + 78 is an integer,

(9y + 36)/37 must also be an integer.

9y + 36

------- = n for some integer n

37

9(y + 4)

-------- = n

37

Since 9 and 37 are relatively prime, y + 4 must be a multiple of 37

for this expression to have an integer value. The smallest positive


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value for which this is the case is y=33. So we have as our first

solution in which both x and y are integers:

y = 33

x = (4y + 78) + (9y + 36)/37 = (132 + 78) + (333/37) = 210 + 9 = 219

2y + 9x + 3z = 198

66 + 1971 + 3z = 198

3z = -1839

z = -613

These integer values of x and y yield an integer value for z, so we

have our first solution to the original pair of equations in integers.

For the family of solutions, we have

9(y + 4)

-------- = n

37

where n is an integer. The general solution, for this expression to

be an integer, is

y = 33 + 37k

where k is any integer. To find the general solution, we then have

y = 33 + 37k

x = (4y + 78) + (9y + 36)/37

x = (132 + 148k + 78) + (297 + 333k + 36)/37

x = (148k + 210) + (9k + 9)

x = 157k + 219

2y + 9x + 3z = 198

(66 + 74k) + (1413k + 1971) + 3z = 198

3z = -1487k - 1839

z = -(1487/3)k - 613

The coefficient 1487 is not divisible by 3; this tells us that only

values of k which are multiples of 3 will yield integer values for z.

So now we can form the expressions for the family of integer solutions

to your problem.

y = 33 + 37(3k) where k is any integer (not the same k as above)y = 111k + 33

x = (4y + 78) + (9y + 36)/37

x = (444k + 132 + 78) + (999k + 297 + 36)/37

x = (444k + 210) + (27k + 9)

x = 471k + 219

2y + 9x + 3z = 198

(222k + 66) + (4239k + 1971) + 3z = 198

4461k + 2037 + 3z = 198

3z = -4461k -1839

z = -1487k - 613

The family of integer solutions to your pair of equations is


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(x, y, z) = (471k + 219, 111k + 33, -1487k - 613)

where k is any integer.

Equations like these, where there are more variables than equations,

and where solutions are to be integers, are called Diophantineequations. You can find many pages in the Dr. Math archives where

similar problems are discussed by performing a search of the archives

using the keyword

diophantine

I hope all this helps. If you have any questions about any of this,

try looking at some more examples in the archives. Then write back

if you still have questions.

diophantine equations in three variables

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