The Pythagorean Equation

Try this

$m = 2$ , $n = 1$ gives $(3, 4, 5)$ . Now try $m = 3$ , $n = 2$ and $m = 4$ , $n = 1$ , and write down the triples you get.
Try $m = 3$ , $n = 1$ , then $m = 4$ , $n = 2$ . Both times the three numbers have a common factor. What is wrong with $m$ and $n$ in each case?
When the slope $\frac{n}{m}$ of the line is rational, both coordinates of the intersection point $P$ are rational. Put the coordinates of $P$ over a common denominator: which three numbers are the numerators and the denominator?
Make $n$ just below $m$ (for example $m = 9$ , $n = 8$ ): the line is very steep, and the triangle is tall and thin. How much longer is the hypotenuse than the longer leg?

Positive integers $(a, b, c)$ with $a^{2} + b^{2} = c^{2}$ form a Pythagorean triple. When the three numbers have no common factor greater than $1$ , it is a primitive Pythagorean triple. Divide both sides by $c^{2}$ :

a^{2} + b^{2} = c^{2} ⟺ (\frac{a}{c})^{2} + (\frac{b}{c})^{2} = 1.

So finding Pythagorean triples is the same as finding points on the unit circle whose coordinates are both rational.

Predict first

In a primitive Pythagorean triple $(a, b, c)$ , what are the parities of the two legs $a$ and $b$ ?

Both oddBoth evenOne odd, one evenAny of the three

Choose an answer first; then you can see the result.

Why

One line sweeps out every Pythagorean triple

Step 1 / 6

Start from a point you know
$A (- 1, 0)$ is on the unit circle. Draw a line through $A$ with slope $t$ .

$y = t (x + 1)$
Find the other intersection
The other root
Write t as n/m
None is missed
When is it primitive?

Remove one condition

Break it

Counterexample lab: remove one condition

Conclusion: if $m$ and $n$ are coprime, $(m^{2} - n^{2}, 2 mn, m^{2} + n^{2})$ is a primitive Pythagorean triple

$m = 4$ , $n = 1$ : coprime, one even and one odd. This gives $(15, 8, 17)$ , and the gcd of the three numbers is $1$ .

A different kind of equation: infinite descent

Change the Pythagorean equation slightly: $x^{2} + y^{2} = 3 z^{2}$ . Does it have positive integer solutions?

x² + y² = 3z² has no positive integer solutions

Step 1 / 6

Suppose there is a solution
Suppose $(x, y, z)$ is a positive integer solution. $z$ is some positive integer; to start with, say it is $54$ .

$x^{2} + y^{2} = 3 z^{2}$
Mod 3
z cannot escape either
A smaller solution
Can it go on forever?
Any z does the same

The recipe for descent: suppose there is a solution → build a strictly smaller positive integer solution → positive integers cannot keep getting smaller forever → contradiction. You can also say it another way: take the solution with the smallest $z$ , then build one with an even smaller $z$ .

Practice

Exercise 1Basic

Take $m = 5$ and $n = 2$ . Write down the three numbers the formula gives.

Exercise 2Basic

A primitive Pythagorean triple has hypotenuse $25$ . What is its shorter leg?

Exercise 3Advanced

How many right triangles are there whose sides are all positive integers and which have a leg equal to $12$ ?

Exercise 4Challenge

Use infinite descent to prove that $x^{3} = 2 y^{3}$ has no positive integer solutions. (In other words, $32$ is not rational.)

Try this

Why

One line sweeps out every Pythagorean triple

Remove one condition

Counterexample lab: remove one condition

A different kind of equation: infinite descent

x² + y² = 3z² has no positive integer solutions

Practice

Mistake Museum · Exhibits from this stop

Thinking the formula gives every Pythagorean triple directly

Forgetting “positive integers” in a descent