$a^2 + b^2 = c^2$

There are, Primitive Pythagorean Triples, that share the same c value. For example, $63^2 + 16^2 = 65^2$ and $33 ^2 + 56^2 = 65^2$.

I have been trying to figure out why the following theorem for finding such triples works.

Take any set of primes. Ex: $5,13,17$.

Now take their product, $1052$, this is the new $c^2$ value. You can express that $c^2$ value as a triple by factoring the product of primes as Gaussian Integers:


Now if you take three of those Gaussian Integers and solve for their product:

ex, $(2+i)(3+2i)(4+i) = 9+32i$ and you have found an $a$ value $9$ and a $b$ value $32$ that works in the theorem. $9^2 + 32^2 = 1105$

You can continue with this:
$(2-i)(3+2i)(4+i) = a + bi$
$(2+i)(3-2i)(4+i) = a + bi$
$(2+i)(3+2i)(4-i) = a + bi$

In fact, the number of triples with this method is $2^{||p|| - 1}$ where $||p||$ is the number of primes used.

Can someone please explain why this method of finding these "c stuck triples" works the way it does?

EDIT: It appears that you cannot take any prime, $p$ but rather must use primes congruent to $1mod4$ according to Fermat's theorem on sums.


3 Answers 3


The system of equations:


Formulas you can write a lot, but will be limited to this. Will make a replacement.





The solution then is.






$p,s,k$ - integers.


1) $(a + bi)(a - bi) = a^2 + b^2$ always.

2) $(a + bi)(c + di)(e + fi) = g + hi \ne (a + bi)(c + di)(e - fi) = j + ki$

yet $(a - bi)(c - di)(e - fi) = g - hi \ne (a - bi)(c - di)(e + fi) = j - ki$


3) $(a + bi)(c + di)(e + fi)(a - bi)(c - di)(e - fi) = (g+hi)(g - hi) = g^2 + h^2 =K$


4) $(a + bi)(c + di)(e - fi)(a - bi)(c - di)(e + fi) = (j+ki)(j - ki) = j^2 + k^2 = pqr$

So $pqr = g^2 + h^2 = j^2 + k^2$.

You do have to choose primes that a gaussian factorable which seems to me like begging the question.


Write the Pythagorean equation $a^2 + b^2 = c^2$ under the form $N(z) =c^2$ , where $z = a + ib$ in $Z[i]$ and N denotes the norm map of $Q(i)/Q$ . Since the norm is multiplicative, it is sufficient to solve the same problem where c is replaced by a prime p factor of c. In your example(s), you take such a p to be congruent to 1 mod 4, so that it is totally decomposed in $Z[i]$, and the usual proof shows that $p$ (hence $p^2$) is indeed a norm. Process as you did, and we are through.

Let me propose another way to find all the "$c$ stuck triples", as you call them, by exploiting more heartily the arithmetic of the Gaussian integers. The parametrization of the integer triples verifying $a^2 + b^2 = c^2$ in $Z$ is classically known, but here is a « Galois solution ». Let us begin with $N(z) = 1$, with $z$ in $Q(i)$ . An element $X + iY$ in $Q(i)$ has norm 1 iff it has the form $(x +iy)/(x-iy)$ (Hilbert 90 !), equivalently iff $X = (x^2 - y^2)/(x^2 + y^2) , Y = 2xy/(x^2 + y^2)$. Since these expressions are homogeneous in x, y, we can take x, y in $Z$. Coming back to the original Pythagorean equation, we get the usual parametrization of the Pythagorean triples : $a = m^2 – n^2 , b = 2mn , c = m^2 + n^2 $. If we fix c, all the « c stuck triples » (a,b,c) will be given as above, starting from (m,n) such that $c = m^2 + n^2$ . More precisely, the quotient $(x+iy)/(x-iy)$ as above, with $x^2 + y^2 = c$, will yield all the "c stuck triples". These are parametrized as follows : $x + iy = t. (m + in)$, with $N(t) = 1$.

EDIT Particular case : if $t$ is in $Z [i]$, then $t$ is a unit (invertible), hence a power of $i$, and we recover the « manipulations » in your example(s). General case : as before, $t = \bar s /s$, with $s$ in $Z[i]$, hence $s. (x + iy) = \bar s . (m + in) (*)$, which means that $s$ divides $\bar s . (m + in)$ in $Z[i]$. Let us consider the prime factors $\pi$ of $s$. Denoting by $p$ the prime number under $\pi$, ramification theory in quadratic fields distinguishes three cases : (1) $p$ is inert ; (2) $p$ is totally ramified ; (3) $p$ is totally split (this corresponds to $p\equiv 1\pmod 4$). In the first two cases, there is only one prime ideal over $p$, hence $\pi$ and $\bar \pi$ differ by a unit, and we can simplify in $s$ and $\bar s$ . In the third case, $\pi $ and $\bar \pi$ are coprime, hence $\pi$ divides $(m + in)$ (and $p$ divides $c$). Conversely, taking for $s$ such a $\pi$, we get (x + iy) as in (*) . Conclusion : if no prime factor of $c$ is congruent to 1 mod 4, we are in the particular case ; if there is at least one prime factor of $c$ congruent to 1 mod 4, the solutions are more complicated.

  • $\begingroup$ Oups ! Sorry, I made a mistake at the very end and cannot conclude that N(u) = 1 . $\endgroup$ Apr 12, 2016 at 21:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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