Inverse Pythagorean Triple

Suppose that $x = ma$ and $y = mb$ where $m,a,b$ are natural numbers and $a,b$ are coprime. The equation becomes $(a^2 + b^2)z^2 \; = \; a^2b^2m^2 \;.$ Since $a,b$ are coprime, so are $a^2+b^2,a^2b^2$ , and hence $m^2 \; = \; k(a^2+b^2) \qquad \qquad z^2 \;= \; ka^2b^2$ for some integer $k$ ; the second equation tells us that $k = n^2$ for some integer $n$ , and so $a^2 + b^2 \; = \; \tfrac{m^2}{n^2} \qquad \qquad z \; = \; nab \;.$ From this we deduce that $m = nc$ where $a^2+b^2 = c^2$ . Thus we deduce that $x \; = \; nac \qquad y \;= \; nbc \qquad z \; = \; nab$ where $n,a,b,c$ are natural numbers with $a,b$ coprime and $a^2 + b^2 = c^2$ . Thus the possible solutions $(x,y,z)$ with $z < 100$ are $(15n,20n,12n)$ and $(20n,15n,12n)$ for $1 \le n \le 8$ (arising from the $(3,4,5)$ Pythagorean triad) and $(65,156,60)$ and $(156,65,60)$ (arising from the $(5,12,13)$ Pythagorean triad). The distinct possible values of $z$ are thus $12n$ for $1 \le n \le 8$ , and the sum of these is $\boxed{432}$ .

The discussion provides some guide, but the extended solution is provided here. In this question and solution, since the answer is not dependent on the order of $x$ or $y$ , the $x, y$ may be swapped freely.

From the original equation, we put it into the form $\frac{y^2+x^2}{x^2y^2} = \frac{1}{z^2} \rightarrow \sqrt{x^2+y^2} = \frac{xy}{z}$ The hardest part is to find the natural number $z$ , which, from the original equation, can be rearrange to form $z=\frac{xy}{\sqrt{x^2+y^2}}$ For all primitive pythagorean triple whose $z < 100$ , there will be $z = 5, 25$ which may be reduced to the form natural number (and, to the greater extent, its multiples). Using the primitive pythagorean triple, we can get the following triples $(x, y, z)$ for a natural number $n$ $(x, y, z) = (15n, 20n, 12n), (175n, 600n, 168n), (65n, 156n, 60n)$

It should be noted here that although the triples of $(175n, 600n, 168n)$ also makes the equation true, but for natural number $n$ , the value of $z$ will go beyond 100 and will not used here. Jennifer Feith also observed that the triples of $(65n, 156n, 60n)$ also form the inverse triple, and for $n = 1$ , the $z=60$ may also be used. For all triples, the proof will omitted here but can be verified by calculations

Here, if one noticed the value of $z = 12n$ , any multiple of 12 below 100 would fulfill the equation above. Although $60$ appears in two separate sets, the $65, 156, 60$ and $75, 100, 60$ , the question only requires a distinct $z$ , so we will count $60$ only once. As $12(8) < 100$ , we then require $n = 1, 2, 3, \dots, 8$ , hence the sum of $z < 100$ satisfying the equation is $\sum_{n=1}^8 12n = 12\times(36) = 432$