Happy Right Triangle New Year!

2017 is a prime number, so all Pythagorean triples containing 2017 will be primitive Pythagorean triples. Primitive Pythagorean triples have the form $(2st,s^2-t^2,s^2+t^2),$ where $s$ and $t$ are coprime positive integers, one of which is even. $2st$ and $s^2-t^2$ are the lengths of the legs, but they are not necessarily in that order.

2017 is odd, so it cannot equal $2st.$

Case 1 : $2017=s^2-t^2$

Factoring gives $2017=(s+t)(s-t).$ Given that 2017 is prime and $s$ and $t$ are positive integers,

$\begin{aligned} s+t &=2017 \\ s-t &=1 \\ \end{aligned}$

This gives $s=1009$ and $t=1008.$ There is exactly 1 Pythagorean triple for this case. Incidentally, it is $(2017,2034144,2034145).$

Case 2 : $2017=s^2+t^2$

Given that 2017 is prime, it can be expressed as the sum of two squares in exactly one way only if it has the form $4k+1.$ It does have this form, so there is exactly 1 Pythagorean triple for this case. Incidentally, $s=44,$ $t=9,$ and the Pythagorean triple is $(792,1855,2017).$

Thus, there are exactly $\boxed{2}$ Pythagorean triples that contain 2017.