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Module 4 – Chapter 6: Diophantine Equations
Pages 65 – 77
This chapter is slightly mis-named. In reality we’ll be looking at lots and lots of equation, and
what we’ll be searching for are Diophantine SOLUTIONS to them.
Which is to say, integers that make the equation true. We’re not interested in fractions or
irrational solutions, only integers. Integers that make the equation true are called Diophantine
Solutions. And equations that have some integer solutions are called Diophantine Equations.
For example:
3x = 5 is an equation and the solution is x 
5
3
So this is NOT an acceptable answer on our quest. The equation is not Diophantine, then.
Here’s another example:
2x  3y  6
Does this equation have Diophantine solutions?
Yes!
The x-intercept (3, 0) is a Diophantine solution. So is the y-intercept (0, 2).
It has non-Diophantine solutions, too. If x = 1 then y = 4/3. Not one we want.
The solutions go neatly into a set format.
Since our equations will be polynomials with rational coefficients, so I know that ALL solutions
are algebraic numbers. I’ll start with that set:
1
Let’s look a little bit harder at the solutions.
Did you know that the line drawn in the Cartesian plane is the set of all solutions?
2
y   x  2 is the equation re-written and ready to
3
graph. I used “Math GV” a nice free download grapher for this picture.
Let’s look more closely at that line:
The line is ALL of the solutions. Let’s pick out the Diophantine solutions:
(3, 0) (0, 2) (6, −2) (− 3, 4)…these are just a few of the ones we want to look at.
2
4
Solutions like (1, ) are NOT solutions we want. If we want to graph these, we’d graph points
3
on a dotted line.
Or we could say we want all numbers in [ 0 ] mod 3 as our x-values. Any multiple of 3 for x
will generate a Diophantine solution:
2
y  x2
3
Another, set-style representation is
Now the last half of this Module, we’ll focus exclusively on lines.
3
For right now, let’s expand into a very famous equation and look at it’s Diophantine solutions.
x2  y 2  z 2
Do you recognize this very famous equation?
When applied to right triangles, we call it The Pythagorean Theorem.
If x and y are the legs and z is the hypotenuse, then the formula always “works”.
y
z
x
So now if we have an isosceles right triangle with side length 1, the hypotenuse is 2 . This
situation is definitely NOT what we’re after.
2 is an algebraic number that is A solution but
it’s not a Diophantine solution.
There is a nice scalene right triangle that is Diophantine: 3 – 4 – 5. (9 + 16 = 25).
All of the Diophantine solutions for this application are called “Pythagorean Triples”. They’re a
very famous set in mathematics.
So, we want to find more of these!
Luckily there’s a handy theorem that we can use to create as many of these as we want.
Here’s what the theorem says:
First, pick two generators: m and n that are natural numbers.
The generators are NOT the leg lengths, they are used with formulas to create the leg lengths.
Here are the Rules for Generators of Pythagorean Triples.

They’re both natural numbers and m > n.

One is even and the other is odd.

They are relatively prime (i.e. (m, n) = 1)
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Once we’ve got the generators, we’re ready for the formulas:

x = leg 1 = m 2  n 2

y = leg 2 = 2mn

z = hypotenuse = m 2  n 2
Now let’s apply these rules:
Let’s pick m = 2 and n = 1.

They’re both natural numbers and m > n.

One is even and the other is odd.

They are relatively prime (i.e. (2, 1) = 1)
x = 4 – 1 =3
y = 2(2)(1) = 4
z=4+1=5
Our generators {2, 1} give us the 3 – 4 – 5 right triangle.
Let’s pick 2 different generators {5, 4}
x = 25 – 16 = 9
y = 2(5)(4) = 40
z = 25 + 16 = 41
Do these work? Let’s check:
92  402  1681  412
Note, too, that this new Pythagorean Triple is NOT similar to the earlier example. These side
lengths are not multiples of the {3, 4, 5} triangle.
5
So DIFFERENT generators give us totally DIFFERENT Pythagorean Triples each time we use
them.
Popper 07, Question 1
Pythagorean Triples are Diophantine solutions to x 2  y 2  z 2
A
True
B
False
On another note, here’s a geometric aside to Pythagorean Triples!
Associated with EVERY triangle is the incircle. This is a circle that is on the interior of a
triangle and it is tangent (shares one point) to each side of the triangle. The interesting thing
about incircles is that they stay inside the triangle no matter how you deform the triangle.
Here are two triangles with their incircles. I built one in Sketchpad and then dragged the vertices
to get the others.
6
Now, Pythagorean triples are special triangles, of course, but they are still in the set of all
triangles so each has an incircle. The amazing thing about this fact is that you can calculate the
radius of the incircle from the generators of the triple.
r = n(m  n)
p. 68
Who would have guessed at that!
Here’s a summary of what we’ve got on Pythagorean triples so far:

Pythagorean triples are a proper subset of scalene right triangles.

EACH triple is a Diophantine solution to the equation x 2  y 2  z 2 .

Each triple that is not similar to another triple is generated by a pair of generators that
follow rules given by a theorem
o
o
o
o
They’re both natural numbers and m > n.
One is even and the other is odd.
They are relatively prime (i.e. (m, n) = 1)
The triangle sides come from these formulas:



x = leg 1 = m 2  n 2
y = leg 2 = 2mn
z = hypotenuse = m 2  n 2
o The radius of the triple’s incircle is r = n(m  n).
Putting it together:
x2  y 2  z 2
Let’s use m = 4 and n = 3
x = 16 – 9 = 7
y = 2(4)(3) = 24
z = 16 + 9 = 25
r = 3 (4 – 3) = 3
72  242  252  625
note that the generators do not appear in the equation!
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Popper 07, Question 2
(6, 5, 21) is a Pythagorean Triple with generators 3 and 2
A
True
B
False
One common error is to assume that if the hypotenuse is a natural number, then the leg lengths
are, too. Let’s look at an example of this.
B
A
C
6 cm
Here’s a right triangle with a hypotenuse of 6 cm. It is quite common for students to get all
excited and think that the leg lengths are natural numbers and that they’ve got a Pythagorean
triple in the problem. But they’re wrong with this case!
x 2  y 2  36
We’re working with:
Suppose x = 1, then y =
35
Suppose x = 2, then y =
32
Suppose x = 3, then y =
27
Suppose x = 4, then y =
20
Suppose x = 5, then y =
11
8
And if we suppose x = 6, then y = 0.
Do NOT make assumptions about things mathematical without doing some checking ok?
The perimeter and area of this triangle are irrational! Not nice natural numbers at all.
Hypotenuse is a natural number
All right triangles
Popper 07, Question 3
3x  4 y  96
Given
The Diophantine solutions to this equation will be a proper subset of all the solutions.
A.
True
B.
False
9
Here’s another surprising tie-in to Pythagorean triples on page 69.
The question is simple enough
How many non-collinear points in the plane can be spaced at integer distances from one another?
(and have NICE, nice point coordinates…they meant to say this!)
The answer is startling.
Now “non-collinear” means not all on the same line so you’re going to have to work with
triangles in answering the question.
You might be tempted to say that an isosceles right triangle with leg length one will work.
Unfortunately, that hypotenuse is 2 .
It’s not a Diophantine solution to the question.
You might try a right triangle with legs, 1 and 2…but that hypotenuse is 5 .
You might try a triangle that is equilateral with all side lengths 1. That sounds good until you
look at the point coordinates of the apex angle.
(
(0, 0)
3
1/2,
2
)
This is one answer. And all similar triangles to this will work, too.
But look at those point coordinates!
Ugly as homemade sin, as my grandmother used to say.
(1, 0)
So that’s 3 of them. To get 4 of them just add a point to the left one away from the apex, though
6 of them is easy.
10
1
1
1
1
1
1
1
1
1
You can continue to work with this and get numbers of points that are integer distances apart but
with really ugly point coordinates.
Or for a second answer you might remember Pythagorean triple right triangles:
5
3
These are all integer distances apart. All nice point coordinates if
we put the origin at the right angle.
Perfect! Here’s 3 points, integer distances.
4
Now the authors come up with a clever way to re-use this triangle and create similar right
triangles that maintain the nice coordinates and natural number distances for ANY natural
number of points.
Let’s look at how this would work for 4 points.
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B
A
3(12)
C
4(5)
3(5)
5(5)
3(13)
D
I know that (3, 4, 5) and (5, 12, 13) are 2 Pythagorean triples.
By careful multiplication I can come up with a pair of nested similar triangles that have natural
number distances among 4 points.
The first multiplication is 3(5) so AD is now 15 units long. Then I multiply the rest of the
3 – 4 – 5 triangle by 5. Now it’s (15, 20, 25). Then, and this is the clever part, I multiply the
other (5, 12, 13) triangle by 3 … I’ve already multiplied 5(3) so AD will be a side for BOTH
triangles. Then 3(12) is AC and 3(13) is DC (a new hypotenuse).
To get 5 points that are integer numbers apart, I’ll get the second round of nested triangles inside
a triangle similar to a (7, 24, 25) triangle. I’ll change point D to point F by multiplying the old
(3, 4, 5) triangle by both 5 and, now, 7. I’ll multiply the old (5, 12, 13) by both 3 and 7, and I’ll
multiply the new enclosing triangle (7, 24, 25) by (3)(5).
origin
3(5)(7)
B
4(5)(7)
5(5)(7)
24(3)(5)
C
3(12)(7)
E
3(13)(7)
25(5)(3)
F
Now in the homework, show the changes to get 6 points at integer distances from one
another…do it so clearly that the GRADER gets it, ok!
12
In another shift, the authors move to discuss a new equation on page 72.
x4  y4  z2
It turns out that this nice equation has NO Diophantine solutions whatsoever.
And looking at another interesting equation on page 75:
x3  y3  1 .
This is a polynomial with rational coefficients so all solutions will be algebraic numbers. There
will be NO transcendental numbers in the solution set.
And the thing that is interesting with this equation is that there are EXACTLY TWO
Diophantine solutions: (0, 1) and (1, 0). All other solutions have 1 or 2 irrational numbers in the
point pair that is the solution. Imagine that.
In fact, solutions to the general equation:
xn  y n  zn
were given by the British mathematician Dr. Wiley in the 1990’s. His solution was predicted but
not proved by our old friend Fermat.
Now
On to Linear Diophantine Equations.
We’ll abandon all equations with powers of x and y higher than 1. And we’ll call them LDE’s.
All of our equations will look like
Ax  By  C
Theorem 1
An LDE has Diophantine solutions whever the GCD of (A, B) divides C.
Example:
10 x  15 y  6
(10, 15) = 5
5 does NOT divide 6
this equation has NO Diophantine Solutions
13
Example:
x  5 y  15
(1, 5) = 1
1 divides 15
This equation HAS Diophantine solutions.
Example
3x  4 y  11
(3, 4) = 1
1 divides 11. This equation HAS Diophantine solutions.
Popper 7, Question 4
3x  2 y  7
Given:
A
This equation has NO Diophantine solutions.
B
This equations does too have Diophantine solutions.
Theorem 2
by Dr. Lamé
If an LED has Diophantine solutions, then the number of steps to the solution is less than or
equal to 5 times the smaller of A and B, the coefficients in the equation.
Example:
3x  2 y  7
The number of steps will be  5  min(3, 2)  10
Note the “manglish” on “min”
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Theorem 3
If you have one Diophantine solution to Ax  By  C , then all the others can be calculated by
using the following formula.
Let ( x0 , y0 ) be the solution that you have and d  ( A, B ) , the GCD.
x  x0 
B
t
d
y  y0 
A
t
d
t is any integer
An example:
30 x  12 y  18
Theorem 1
(30, 12) = 6, 6 divides 18. There are Diophantine solutions
Theorem 2
the number of steps is less than or equal to 5(12) = 60
Theorem 3
(3, 6) is a Diophantine solution
(how to get this, later!)
All other solutions can be found by varying t in the formulas:
x  3
12
t  3  2t
6
y  6 
30
t  6  5t
6
If t = 0 we get our original point.
If t = 1, we get (1, 1)
15
If t = 3, we get (9, 21).
16
Now, naturally, there is the question of how we got that first Diophantine solution!
There are five steps to finding it:
Step 1
Check Theorem 1
Step 2
Rewrite the equation totally to
where d is the GCD of A and B
and p and q are dummy variables.
Step 3
Use the Euclidean Algorithm to find p and q.
Step 4
Multiply both sides of the changed equation to retrieve
the original equation.
Step 5
Use Theorem 3 to generate all desired solutions.
Let’s use the same equation that we started with:
Step 1
we did this.
Step 2
work with:
Step 3
Ap  Bq  d
30 x  12 y  18
30 p  12q  6
30  2(12)  6
2
12 30
12 = 2(6) + 0
2
6 12
with a remainder of 6
with a remainder of 0
DONE!
Go to the next-to-the-last step and use it with a change: subtract 2(12) from both
sides
30  2(12) = 6
Pattern match it to
30 p  12q  6
30 (1) + 12 ( 2) = 6
17
Step 4
I need for the right hand side to be 18, so I’ll multiply both sides by 3…and I’ll
carefully preserve my original coefficients by multiplying INSIDE the
parentheses.
30 (1 x 3) + 12 (  2 x 3) = 18
So my ( x0 , y0 ) is (3,  6).
Step 5
All my Diophantine solutions look like:
x  3
12
t  3  2t
6
y  6 
30
t  6  5t
6
Where t is any integer.
There will be lots of practice on this in the second video.
Right now:
Brilliant numbers.
A Brilliant number is a number with n prime factors each which is the same length. We name
the number for the number of factors it has:
Example:
9 = 3(3)
9 is a 2-Brilliant number
121 = 11(11)
121 is a 2-Brilliant number
143 = 11 (13)
143 is a 2-Brilliant number
989 = 23 (43)
989 is a 2-Brilliant number
18
Now sometimes we actually describe the number of digits in the number along with telling how
Brilliant it is:
3, 589
is a 4 digit 2-Brilliant number
3589 = 37(97)
875
is a 3 digit 4-Brilliant number
875 = 5(5)(5)(7)
55, 913
is a 5 digit 4-Brilliant number
55913 = 11(13)(17)(23)
Now, on the other hand
26 is not Brilliant. 26 = 2(13) and the factors are differing lengths
This is just an aside on yet another type of number. Note that Brilliant numbers are a proper
subset of the natural numbers. In fact, the two subsets Brilliant and Not-brilliant are a partition
on the set of natural numbers. 1, 2, and 3 are not 2-Brilliant because, for example: 2 has 2
factors: 2 and 1, but one is not a prime factor. You could argue that it’s a 1-Brilliant number,
though but you’d still have to deal with 1 in your argument! This is the type of thing that keeps
the coffee and beer flowing at international math conferences!
Popper 07, Question 5
16, 900 = 130(130)
A
True
B
False
is a 5 digit 2-Brilliant number
End of video 1
19
Diophantine equations are those with some integer solutions – not every solution has to be
diophantine, just some of them, to qualify. We’ve seen an example of an equation that has NO
diophantine solutions:
x4  y 4  z 4
This means that the distinction has meaning and that both kinds of subsets of the set of all
equations with rational coefficients have set elements.
Let’s look at 10 x  25 y  7
(10, 25) = 5 and 5 does not divide 7
Here’s another equation in the not-diophantine subset!
However:
30 x  19 y  15
(30, 19) = 1 and 1 divides 15
So this one is Diophantine.
Popper 8, Question 1
The two subsets: Diophantine and Not-diophantine are a partition on the set of all polynomials
with rational number coefficients.
A
True
B
False
20
Let’s look at another example of solving an LDE
210 x  45 y  30
(Linear Diophantine Equation)
(210, 45) = 15 and 15 divides 30
Set up the alternate equation in p and q:
210p + 45q = 15
(so the last step will be MBS by 2!)
Using the Euclidean Algorithm:
210 = 45(4) + 30
45 = 30(1) + 15
30 = 15(2) + 0
End EA. Go to the next to the last equation and rewrite it by subtracting 30(1) from both sides:
45 – 30(1) = 15
Note that the 30(1) does not match the p/q equation…
210 is needed. Solve the first equation for 30:
30 = 210 – 45(4)
SUBSTITUTE
45 – [210 – 45(4)] = 15
45 – 210 + 45 (4) = 15
distribute
45(5) – 210 = 15
combine like terms 45(1) + 45 (4) = 45 (5)
210 (−1) + 45 (5) = 15
matches the p/q. Now multiply to match original (by 2)
210 ( −2) + 45 (10) = 15
( x0 , y0 ) is (−2, 10)
21
All Diophantine solutions are:
x  2 
45
t  2  3t
15
y  10 
210
t  10  14t
15
t is any integer
If t = 1 (1, − 4) is a Diophantine solution, for example.
Popper 08, Question 2
Given an original equation:
5x + 7y = 6
Suppose we do all the work and get down to 5(2) + 7(−1) = 3.
What is ( x0 , y0 ) ?
A
(2, −1)
B
(4, −2)
C
It cannot be determined at this point in the work.
22
Another example:
7 x  17 y  185
Yes! There are Diophantine solutions.
(7, 17) = 1
Set up the p/q equation:
7p + 17q = 1
Euclidean Algorithm work:
NOTE THE SWITCH ON THE COEFFICIENTS!
17 = 7(2) + 3
7 = 3(2) + 1
3 = 1(3) + 0
grab the next to the last equation! Reformat!
7 – 3(2) = 1
oops! Hate that 3! Go up one further: 3 = 17 – 7(2)
Substitute
7 – [17 – 7(2)](2) = 1
note the second times 2 there!
7 – [17(2) – 7(4)] = 1
Distribute the second times 2 first
7 – 17(2) + 7(4) = 1
Distribute the times – 1
7 (5) – 17 (2) = 1
Check above 17 is positive…move it inside
7(5) + 17(−2) = 1
NOW retrieve the original equation. Multiply “inside”
7(185(5)) + 17(185(−2)) = 185
7(925) + 17 (−370) = 185
( x0 , y0 ) is (925, −370)
23
All Diophantine solutions are:
x  925  17t
y  370  7t
t is any integer
Popper 08, Question 3
In the above equation (942, −377) is one of the solutions.
A
True
B
False
And now, one last example:
Find all Diophantine solutions to
6 x  64 y  2
(6, 64) = 2
2 divides 2!
Set up the p/q equation:
6p + 64q = 2
This one is nice! We’ll multiply BS by 1 at the end/
Euclidean Algorithm work:
64 = 6(10) + 4
6 = 4(1) + 2
4 = 2(2) + 0
24
We’ll work with
6 – 4(1) = 2
and we’re missing that coefficient of 64 so we’ll substitute
64 – 6(10) = 4
6 – [64 – 6(10)] = 2
6 – 64 + 6(10) = 2
6(11) – 64 = 2
not quite a match to 6p + 64q = 2
6(11) + 64 (−1) = 2
( x0 , y0 ) is (11, −1).
All the other Diophantine solutions come from varying the integer t in
x  11 
64
t  11  32t
2
6
y  1  t  1  3t
2
t is any integer
Popper 08, Question 4
If you use t = 0 in the above equation you get the ( x0 , y0 ) from the work with the steps to a
solution.
A
True
B
False
25
Popper 08, Question 5
x  x0 
B
t
d
y  y0 
A
t
d
t is any integer
Using this formula with the A and B coefficients from the original equation gives you:
A
all solutions to the equation
B
a proper subset of all solutions
26