# Pythagorean primitive triples

If $a,b,c$ are primitive Pythagorean triples such that $c^2=a^2+b^2$, prove that this is unique primitive Pythagorean triple using $c$ as hypotenuse.

I'm not sure if I phrased this correctly. I want to ask, if $c^2=a^2+b^2$ where $a,b,c$ are pairwise coprime, prove that there are no $a_1$ and $b_1$ such that $a_1,b_1,c$ that satisfy $c^2=a_1^2+b_1^2$, and they are also Pythagorean triple.

This is obviously true based on the short list I found on Wiki but I can't seem to prove it... Help?

EDIT Since this obviously seems false, my question now is, if the process described above false only for $c=5k$? Since it seems to be false for $c=65, 145, 185, 205, 265...$ Can one prove this?

• Not sure how to comment- but are how asking how to prove that given $c$ and $a$, $b$ is unique? – Ali Mar 1 '12 at 18:33
• Unless $C$ is prime, there can be one or more triples where $C_1$ is a factor of $C$ so the triple yields $C$ when multiplied by the co-factor. – poetasis Feb 22 '20 at 19:20

## 3 Answers

$63^2 + 16^2 = 33^2 + 56^2 = 65^2$

• Does this happen only for $c=5k$? – Lazar Ljubenović Mar 1 '12 at 18:36
• $21^2+220^2=140^2+171^2=221^2$ – Robert Israel Mar 1 '12 at 19:21
• @LazarLjubenović: No. See my answer and Bill's answer. – Aryabhata Mar 1 '12 at 19:36

Hint $\rm\ \ \ c^2 = N(a+b {\it i}) = a^2 + b^2,\ \ \ C^2 = N(A+B{\it i}) = A^2 + B^2\$ so by multiplicativity of norms

$\rm\quad\Rightarrow\ (cC)^2 = N(a\pm b{\it i})\:N(A + B{\it i}) = N(aA\mp bB + (aB\pm bA){\it i}) = (aA\mp bB)^2 + (aB\pm bA)^2$

e.g. $\ \ \ 5^2 = 4^2 + 3^2,\ \ 13^2 = 12^2 + 5^2\ \Rightarrow\ 65^2 =\: (48\mp 15)^2 + (20\pm 36)^2 =\: \{63^2+16^2,\: 33^2+56^2\}$

If the $c$ of the primitive pythagorean triple satisfies,

$c = m^2 + n^2$, then $b = 2mn$ and $a = m^2 - n^2$.

Looking in the guassian integers we see that $c = (m+in)(m-in)$. There are infinitely many $c$ which have multiple factorizations of the form $(a+ib)(a-ib)$.

So what you say is not true.

As to your edit. Any $c$ which has at least two primes factors (in $\mathbb{Z}$) of the form $4k+1$ will do and I believe any $c$ with that propertly must have at least two such prime factors. Note, if you want $m,n$ to be co-prime $c$ should not have any factor of the form $4k+3$.

Here is how you can use the above to generate Robert's example:

$65 = 5*13 = (2+i)(2-i)(3+2i)(3-2i)$

Writing it as

$[(3+2i)(2-i)] [(3-2i)(2+i)]$ gives us $(8-i)(8+i)$

which gives $m=8$, $n=1$ and so $65^2 = 16^2 + 63^2$.

Writing it as

$[(3+2i)(2+i)][ (3-2i)(2-i)]$ gives us $(4+5i)(4-5i)$

which gives $m=7$, $n=4$ and so $65^2 = 56^2 + 33^2$.