Three primitive Pythagorean triples with the same c

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

• In the sense that $a^2+b^2=c^2$? – Jack M May 28 '14 at 21:00
• – user2357112 May 28 '14 at 21:04

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 $5\cdot 13\cdot 17\cdot 29$.

• 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 $2$.)

André Nicolas found the solutions of $a^2+b^2=1105$. Question was solutions for $c = 1105$ not $c$ being square root of $1105$. To get the solution of the equations you will need to square each of André Nicolas' complex answers. For example, his $9+32i$ solution, if you square it you get $-943+576i$, which was the 3rd solution of alex.jordan's solution. The other $3$ are related in the same method.