# Origins of Euclid's formula for Pythagorean triples

Hi, because of a question on this week's on Brilliant problem set, I've come across the Euclid's formula for generating Pythagorean triples for the first time.

The formula says that (in my own words) for a right angled triangle with sides a,b and hypotenuse c,

$a^2+b^2 = c^2 \iff (m^2-n^2)^2 + (2mn)^2 = (m^2 + n^2)^2$

for any 2 integers $m > n$. After staring at this formula for a while, I realised it's actually a simple and straightforward case of expansion on LHS and RHS, followed by a check on whether the terms matched on both sides.

This made me wonder if Euclid came up with this theorem because of some eureka moment (you know, like maybe one day he's just sitting alone at his desk and this idea that $(m^2-n^2) + (2mn)^2 = (m^2 + n^2)^2$ can be applied to the Pythagoras theorem just pops into his head) or was he driven by any other sort of motivation to derive this formula?

And this brings me to another question, suppose Euclid didn't leave any proof of the derivation of this formula because he thought that it was such a straight forward application of the idea of $(m^2-n^2)^2 + (2mn)^2 = (m^2 + n^2)^2$ to $a^2 + b^2 = c^2$. Would there had been any practical need to come up with a derivation from $a^2 + b^2 = c^2$?

I apologize if I sound like an idiot asking such questions, but well...I just couldn't get these questions off my mind :p

Note by Ghim Siang
7 years ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

I've added a note/comment that the expansion doesn't prove that that every pythagorean triple must be of the form $(m^2 - n^2, 2mn, m^2 + n^2)$ In fact, this statement is not true! What are the values of $m$ and $n$ that generate the triple $(9, 12, 15)$? We require $m^2 + n^2 = 15$, which you can verify has no solutions.

All that the expansion shows, is that $(m^2-n^2, 2mn, m^2 + n^2)$ is a Pythagorean triple. In fact, the classification is valid for primitive pythagorean triple, which are triples that have no common divisor. In the above example, the terms have a common divisor of 3.

Staff - 7 years ago

This is correct. A more general formula that applies to all pythagorean triples (as far as I know) is $a = k \cdot (m^2 - n^2), b = k \cdot 2mn, c = k \cdot (m^2 + n^2)$, where k is a positive integer.

- 7 years ago

Ghim, are you asking for the derivation or the 'need' of the derivation? Sorry, but I couldn't understand your point.

- 7 years ago

well, i was asking about the 'need' for the derivation from a^2 + b^2 = c^2 to the euclid's formula, and i guess alyosha's explaination pretty much makes sense!

- 7 years ago

http://www.popflock.com/learn?s=Euclid%27s_formula Hope this could help

- 2 years ago