Pythagorean Monstrosities

It's really difficult to find any efficient method of solving this problem without knowing Euler's formula for Pythagorean triples or simply just hacking away with modular arithmetic.

Solution 1: Notice that $a \equiv 1 \mod{4}$ and $b \equiv 1 \mod{4}$ , which means that $c \equiv 2 \mod{4}$ , which is impossible.

Solution 2:

$a = k(m^2 - n^2)$

$b = 2kmn$

$c = k(m^2 + n^2)$

where $a$ , $b$ , and $c$ is your triple, and $m$ , $n$ , and $k$ are positive integers such that $m>n$ .

There is no way that any of the listed numbers can be $b$ because they are odd. Notice that $a \equiv 1 \mod{3}$ and $c \equiv 1 \mod{3}$ , so we can find the difference between these two numbers to get that $c-a = 2kn^2 = 47592121476$ is therefore divisible by 3. Since k cannot have a factor of 3, this means it must be part of the $n^2$ , meaning it must be divisible by 9, and it is not. Therefore, these numbers are not part of the same Pythagorean triple.

It should also be noted that there isn't really an elementary method of checking whether or not a set of positive integers are in fact part of Pythagorean triples without direct computation or solving for $m$ and $n$ here. Almost any question of this type that is even remotely reasonable to do on pencil/paper will have to result in the answer being "not part of the same triple".