# On primitive Pythagorean triplets with same greatest term

For the purpose of this question, I'll define a primitive Pythagorean triplet as a triplet of positive integers $(a, b, c)$ such that

1. $a$ is odd;
2. $b$ is even;
3. $\gcd(a, b, c) = 1$;
4. $a^2 + b^2 = c^2$.

The set of all primitive Pythagorean triplets (PPTs) can be partitioned according to the value of $c$, the greatest term in the triplet. Let $[c]$ denote the class in this partition corresponding to $c$.

Empirically, I find that

1. the sizes of the classes are always powers of 2;
2. the lowest $c$ for a given class size is always divisible by 5.

Are these statements true in general? Where can I find proofs or counterexamples?

Below are a few examples of the findings described above:

$$\begin{array}{|c|l|r|} \hline \left|[c]\right| & \begin{array}{l} (a, b) \\ \end{array} & c \\ \hline 2^0 & \begin{array}{l} (3, 4) \\ \end{array} & 5 \\ \hline 2^1 & \begin{array}{l} (63,16) \\ (33,56) \end{array} & 65 \\ \hline 2^2 & \begin{array}{l} (1073, 264) \\ (943, 576) \\ (817, 744) \\ (47, 1104) \end{array} & 1105 \\ \hline 2^3 & \begin{array}{l} (32037,716) \\ (31323,6764) \\ (27813,15916) \\ (23067,22244) \\ (21093,24124) \\ (17253,27004) \\ (8283,30956) \\ (2277,31964) \end{array} & 32045 \\ \hline 2^4 & \begin{array}{l} (1177473,139136) \\ (1164447,223304) \\ (1129887,359384) \\ (1112703,409504) \\ (1074273,501736) \\ (1027743,591224) \\ (1015137,612616) \\ (927903,738104) \\ (782463,890816) \\ (661377,984064) \\ (540417,1055344) \\ (463263,1091416) \\ (448767,1097456) \\ (303873,1146064) \\ (279807,1152176) \\ (81567,1182856) \end{array} & 1185665 \\ \hline \end{array}$$

For each $i\in\{0,...,4\}$, the table gives the triplets in $[c]$ for the lowest value of $c$ such that $\left|[c]\right| = 2^i$.

For example, $1105$ is the lowest $c$ such that $\left|[c]\right| = 2^2 = 4$, and the PPTs in $[1105]$ are

$$\begin{array}{l} (1073, 264, 1105) \\ (943, 576, 1105) \\ (817, 744, 1105) \\ (47, 1104, 1105) \end{array}$$

• These results follow from understanding factorization in the ring of Gaussian integers $\Bbb Z[i]$. – Rolf Hoyer Jun 5 '16 at 3:30
• Great question! Really interesting topic – Zubin Mukerjee Jun 5 '16 at 4:27

This comes from Fermat's theorem on the sum of two squares. $65$ is the first number that is the product of two primes equivalent to $1 \bmod 4$. $32045$ is the first number that is the product of four of those. All of these numbers will have $5$ as a factor, and all after $5$ will have $13$ as a factor, etc. If the prime factors of form $1 \bmod 4$ are distinct and the factors of form $3 \bmod 4$ are only squares you can then use the Brahmagupta-Fibonacci identity to generate $2^n$ solutions where $n$ is the number of primes of form $1 \bmod 4$ Your requirement that the triple be primitive forces that all prime factors to be one one side or the other, which maintains the $2^n$ behavior. You can contrast the situation with $c^2=5^6=15625$ with solutions $(35,120,125), (44,11,125), (75,100,125)$ but only one is primitive. Thanks to kjo for pointing out that two of these are not primitive.
• @kjo: good point. I think one can turn this into a proof in favor of $2^n$, which takes chasing primes through the identity. Thinking on it. – Ross Millikan Jun 5 '16 at 3:55