Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can someone give me an example of Three primitive Pythagorean triples with the same c?

share|cite|improve this question
In the sense that $a^2+b^2=c^2$? – Jack M May 28 '14 at 21:00
Helpful: – user2357112 May 28 '14 at 21:04
up vote 5 down vote accepted

We produce an answer with hypotenuse $c=5\cdot 13\cdot 17$. So we need to find three relatively prime pairs $\{m,n\}$ of opposite parity such that $m^2+n^2=5\cdot 13\cdot 17$.

It is simplest to calculate using complex numbers, although the same goal can be achieved by using the Brahmagupta Identity. Note that $c$ factors over the Gaussian integers as $(2+i)(2-i)(3+2i)(3-2i)(4+i)(4-i)$.

First one: Note that $(2+i)(3+2i)(4+i)=9+32i$. That gives $m=9$, $n=32$.

Second one: Do the same calculation using $(2-i)(3+2i)(4+i)$.

Third one: Same calculation, using $(2+i)(3-2i)(4+i)$.

Remark: There is a fourth, using $(2+i)(3+i)(4-i)$. For an example with $8$ instead of $4$, we can play the same game with $5\cdot 13\cdot 17\cdot 29$.

share|cite|improve this answer
To add some detail, you need to find $c=m^2+n^2$ in three or more ways - the primitive triples being $m^2-n^2, 2mn, m^2+n^2$. A prime of the form $p=4s+1$ has one such expression. Note that $(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2=(ac-bd)^2+(ad+bc)^2$ (the identity corresponds to the use of complex numbers in the answer) so $pq$ gives two possibilities and $pqr$ gives four (where $p,q,r$ are primes congruent to $1$ mod $4$). The only chance for exactly three is $p^2q$ but it seems that one of these has a factor $p$ and is therefore imprimitive (not checked in detail). – Mark Bennet May 28 '14 at 21:19

This is a list of the first Pythagorean triples. $1105$ actually has four primitive breakdowns as $a^2+b^2$, and my skimming did not find any earlier ones with three.

$$\begin{align} 1105^2 &=47^2+1104^2\\ &=264^2+1073^2\\ &=576^2+943^2\\ &=744^2+817^2 \end{align}$$

(Maybe @AndreNicolas's answer can be explored more to prove that the number of breakdowns for any $c$ that is the hypotenuse of a Pythagorean triple is a power of $2$.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.