Annual hypotenuse

Key fact 1 : If $(a,b,c)$ is a primitive Pythagorean triple, i.e. coprime positive integers such that $a^2 + b^2 = c^2$ , then there exist $p > q$ such that $c = p^2+q^2$ . (And $\{a,b\} = \{p^2 - q^2, 2pq\}$ .)

Key fact 2 : An odd number $x$ can be written as the sum of two squares $x = a^2 + b^2$ , with coprime $a,b$ , if and only if $x$ is a product of prime numbers of the form $4n+1$ . This sum-of-squares representation is unique iff $x$ is prime.

(If $x$ is not prime but of the form $x = pr$ with $p$ prime, let $p = a_1^2 + b_1^2$ and $r = a_2^2 + b_2^2$ . Let $a_{3,4} = |a_1a_2 \pm b_1b_2|$ and $b_{3,4} = |a_1b_2 \mp b_1a_2|$ . Then $x = a_3^2 + b_3^2 = a_4^2 + b_4^2$ are two different representations of $x$ as a sum of squares.)

Solution : A Pythagorean triple $(a,b,2019)$ is either primitive or not. If it is primitive, then $2019 = p^2 + q^2$ must be the sum of two squares; however, since 2019 contains a prime factor 3, which is not of the form $4n+1$ , this cannot be the case.

If it is not primitive, $(a,b,2019) = 3 \cdot (a',b',673)$ , with coprime $a',b'$ . Since $673$ is a prime number of the form $4n+1$ , it can be written as the sum of two squares, $(p')^2 + (q')^2$ uniquely . This results in only one possible Pythagorean triple; there should be no others beside the one Marty found.

Specifically, we have $673 = 23^2 + 12^2$ , so that $a' = 23^2 - 12^2 = 385, b' = 2\cdot 23\cdot 12 = 552$ ; multiply by 3 to find $a = 3\cdot 385 = 1155$ and $b = 3\cdot 552 = 1656$ .