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

Jack M
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Username Unknown
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4 Answers4

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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 $2$.)

2'5 9'2
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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 $5\cdot 13\cdot 17\cdot 29$.

André Nicolas
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    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
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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.

Tyrone
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JustMe
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The answer by alex jordan gave an example of four triples where $C=1105$ and all are primitives. More can be generated from the divisors of $1105$ and then multiplied by the respective co-factors. E.g. there are $9$ additional non-primitive triples where $C=1105$.

$$221\times(3,4,5)\quad 85\times (5,12,13)\quad 65\times(15,8,17)\quad 17\times(33,56,65)\quad 17\times(63,16,65)\quad 13\times(13,84,85)\quad 13\times(77,36,85)\quad 5\times(21,220,221)\quad 5\times(171,140,221)$$

To find multiple triples for $some$ $C$ we can try different values of $C=4t+1, t\in\mathbb{N}$ as shown below.

Given Euclid's formula $A=m^2-n^2, B=2mn, C=m^2+n^2$

$C= m^2+n^2\implies\space n=\sqrt{C-m^2}$ where $\biggl\lceil\sqrt{\frac{C}{2}}\space\space\biggr\rceil \le m\le\bigl\lfloor\sqrt{C}\bigr\rfloor\quad m,n\in\mathbb{N}$

For example, if $C=65,\biggl\lceil\sqrt{\frac{65}{2}}\space\space\biggr\rceil=6 \le m\le\bigl\lfloor\sqrt{65}\bigr\rfloor=8$ and we find

$$\sqrt{65-6^2}\notin \mathbb{N}\qquad\qquad \sqrt{65-7^2}=4\qquad\qquad \sqrt{65-8^2}=1$$ Our two triples are: $$f(7,4)=(33,56,65)\quad and \quad f(8,1)=(63,16,65)$$

The larger the value of $C$, the more likely it is to be part of multiple primitives but the next one up from the $1105$ sample has only $1.\quad$ Note: $(1105=4\times276+1)\quad (1109=4\times277+1)$

$$(141,1100,1109)$$

poetasis
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