Shortest primitive Pythagorean triangle hypotenuse with 32 distinct such triangles sharing it?

N.B., the list of such triangles was not requested as part of the solution.

The answer is the product of the first 6 $4n+1$ primes: $5, 13, 17, 29, 37, 41$ . The primitive Pythagorean triangle hypotenuse always is a product of such primes and the multiplicity of a given hypotenuse is $2^{n-1}$ where is the number of distinct factors in the factorization in the hypotenuse's length.

This integer sequence supplies the answer and states the size of the solution set of a primitive Pythagorean triangle hypotenuse value.

Here is the list of such triangles, even though that list was not required for the problem's solution: $\begin{array}{rrr} 822696 & 48605303 & 48612265 \\ 1547863 & 48587616 & 48612265 \\ 1840344 & 48577417 & 48612265 \\ 4206377 & 48429936 & 48612265 \\ 5031817 & 48351144 & 48612265 \\ 6365833 & 48193656 & 48612265 \\ 7383024 & 48048343 & 48612265 \\ 8073576 & 47937143 & 48612265 \\ 8707776 & 47826007 & 48612265 \\ 10400983 & 47486544 & 48612265 \\ 12118584 & 47077513 & 48612265 \\ 13908384 & 46580137 & 48612265 \\ 14399273 & 46430736 & 48612265 \\ 15368407 & 46119024 & 48612265 \\ 16162697 & 45846696 & 48612265 \\ 17598504 & 45314953 & 48612265 \\ 19412183 & 44568144 & 48612265 \\ 21173033 & 43759056 & 48612265 \\ 21561864 & 43568777 & 48612265 \\ 22469496 & 43107703 & 48612265 \\ 23281176 & 42674807 & 48612265 \\ 24544343 & 41961024 & 48612265 \\ 26394487 & 40822584 & 48612265 \\ 27827784 & 39859337 & 48612265 \\ 28353264 & 39487273 & 48612265 \\ 28590473 & 39315864 & 48612265 \\ 29737897 & 38455296 & 48612265 \\ 30473184 & 37875287 & 48612265 \\ 31114776 & 37350007 & 48612265 \\ 32898647 & 35788704 & 48612265 \\ 33410167 & 35311656 & 48612265 \\ 33640873 & 35091936 & 48612265 \\ \end{array}$

I missed the condition that the Pythagorean triple be primitive. I'll leave my solution below, anyway -- there are two shorter hypotenuses if you remove the primitive condition: 29,641,625 and 32,846,125.

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There is a known method to compute the number of ways a number can be expressed as the sum of two squares.

Let $n$ be the number of ways to write our number $c$ as the sum of two squares of positive integers (ignoring order ). Then we compute $n$ using the following steps:

• Consider the prime factorization of $c$ . Split up your prime factors into $2$ , primes of the form $4k+1$ , and primes of the form $4k+3$ .
• If any of the primes of the form $4k+3$ is raised to an odd power, then your number cannot be written as the sum of two squares.
• Say there are $r$ prime factors of the form $4k+1$ . Set $B=\prod_{k=1}^r (b_k+1)$ , where the $b_k$ s are the exponents on the primes of these factors.
• The number of ways to write $c$ as the sum of two squares is $n=\frac{1}{2}B$ if $B$ is even and $n=\frac{1}{2}(B-(-1)^a)$ if $B$ is odd, where $a$ is the exponent of $2$ in the prime factorization.

Since we want $n=32$ , either $B=63,64,65$ , depending on $a$ . Since we wish to minimize $c$ , assume that there are no prime factors of the form $4k+3$ .

• If $B=63=7\cdot 3\cdot 3$ then $a$ (the exponent of $2$ in the prime factorization) must be odd. Thus $c$ is minimized when $a=1$ . We have the following possibilities for $c$ : $\begin{aligned} c &= 2\cdot 5^6\cdot 13^2\cdot 17^2 \\ c &= 2\cdot 5^8\cdot 13^6 \\ c &= 2\cdot 5^{20}\cdot 13^2 \\ c &= 2\cdot 5^{62} \\ \end{aligned}$ The smallest of these is the first, which is on the order of $10^8$ .
• If $B=65 = 13\cdot 5$ then $a$ must be even. Thus $c$ is minimized when $a=0$ . We have the following possibilities for $c$ : $\begin{aligned} c &= 5^12 \cdot 13^4 \\ c &= 5^64 \end{aligned}$ The smallest of these is the first, which is on the order of $10^{12}$ .
• If $B=64 = 2^6$ then $a$ can be either even or odd. Thus $c$ is minimized when $a=0$ . Consider $c = 5^1\cdot 13^1\cdot 17^1\cdot 29^1\cdot 37^1\cdot 41^1$ any other possibility for $c$ will involve raising one of those factors to an exponent of at least $3$ . For any factor other than 5, this would mean multiplying by a factor greater than $41$ , thus increasing $c$ . However, $c=5^3\cdot 13^1\cdot 17^1\cdot 29^1\cdot 37^1$ is indeed smaller. This is on the order of $10^7$ and thus smaller than either of our previous two cases as well. Thus the correct answer is $c=5^3\cdot 13^1\cdot 17^1\cdot 29^1\cdot 37^1=\boxed{29641625}$