These Triangles Broke My Calculator

A primitive Pythagorean triple is one of the form $(u^2-v^2,2uv,u^2+v^2)$ where $u,v$ are coprime positive integers of opposite parities, with $u > v$ . Moreover this formula enumerates all primitive Pythagorean triples uniquely. Now $100!$ is even, and hence can only appear as a leg of a primitive Pythagorean triple if we can find coprime integers with opposite parities such that $100! = 2uv$ . Thus we want to know the number of ways in which $\tfrac12 100!$ can be written as a product of two coprime positive integers of opposite parities.

There are $25$ distinct prime factors in $\tfrac12 100!$ , and the index of $2$ in $100!$ is $97$ . Let us write $\tfrac12 100! \; = \; 2^{96} \prod_{j=1}^{24} p_j^{a_j}$ where $p_1,p_2,...,p_{24}$ are the $24$ odd primes less than $100$ , and $a_1,a_2,..,a_{24}$ are positive integers. For any subset $A$ of $\{1,2,3,...,24\}$ , we define two numbers $2^{96} \prod_{j \in A} p_j^{a_j} \hspace{2cm} \prod_{j \not\in A} p_j^{a_j}$ and let $u_A$ be the larger of the two, and $v_A$ the smaller. Then $u_A,v_A$ are coprime and of opposite parities, $u_A > v_A$ , and $2u_Av_A = 100!$ . Moreover, any decomposition of $100! = 2uv$ where $u,v$ are coprime and of opposite parities must be achieved in this manner. Thus there are as many primitive Pythagorean triples with $100!$ as one leg as there are subsets of $\{1,2,3,...,24\}$ , namely $2^{24} = \boxed{16777216}$ .

Copied from Mark Hennings solution: A primitive Pythagorean triple is one of the form where are coprime positive integers of opposite parities. The factorization of 100! is {{2,97},{3,48},{5,24},{7,16},{11,9},{13,7},{17,5},{19,5},{23,4},{29,3},{31,3},{37,2},{41,2},{43,2},{47,2},{53,1},{59,1},{61,1},{67,1},{71,1},{73,1},{79,1},{83,1},{89,1},{97,1}} where the left member of each pair is the prime and the right member is the power. The factorization list has 25 elements. The only even prime is 2. Each prime must be represented in only one variable; otherwise a non-primitive triple would be created. $2$ and any subset of the others goes into one variable; the remainder of the others gives into the other variable. Therefore, the problem is how many subsets of 24 things are there. That number is $1\,000\,000\,000\,000\,000\,000\,000\,000_2 \Rightarrow 100000000_8 \Rightarrow 1000000_{16} \Rightarrow 16777216_{10}$ .