Right,Triangle!

By Euclid's formula, every primitive Pythagorean Triple can be written as $a = m^2 - n^2$ , $b = 2mn$ and $c = m^2 + n^2$ for some relatively prime $m$ and $n$ not both odd. In this case, $a$ and $c$ are odd and $b$ is even. Thus, if the triple is primitive, $2018 = b = 2mn$ and $mn = 1009$ . Since 1009 is prime, $m$ and $n$ must be 1009 and 1, a contradiction. Thus, there is no primitive Pythagorean triple containing 2018.

To look for non-primitive Pythagorean triples, we must find a primitive triple with at least one side in the set $\{1, 2, 1009\}$ . Since there are no primitive triples with side lengths 1 or 2, we need only consider 1009. By the formulae for $a$ , $b$ and $c$ above, we have two possibilities:

• $1009 = c = m^2 + n^2$ : the only way to represent 1009 as the sum of two squares is $1009 = 15^2 + 28^2$ . Thus $a=559,b=840,c=1009$ is a primitive Pythagorean triple and $a=1118, b=1680, c=2018$ is a Pythagorean triple containing 2018.
• $1009 = a = m^2 - n^2$ . Then $1009 = (m+n)(m-n)$ and, since 1009 is prime, $m+n = 1009$ and $m-n = 1$ . Thus $m=505$ and $m=504$ . Since these numbers are consecutive, they are relatively prime and not both odd. Thus $a=1009, b=509040,c=509041$ is a primitive Pythagorean triple and $a=2018, b=1018080,c=1018082$ is a Pythagorean triple containing 2018.

There are exactly two Pythagorean triples containing 2018.

The general formulae for Pythagorean triplets is a = 2kxy, b = (x^2 - y^2)k, c = k(x^2 + y^2), where a^2 + b^2 = c^2. If k = 1, the solution is called primitive. We have 2 choices; either 2018 is a leg, or 2018 is the hypotenuse. If the former, then 2018 = 2kxy, or kxy =1009, a prime number. Hence k = 1, x = 1009, and y =1. The sides are a = 2xy = 2018, b = x^2 - y^2 = 1018001 - 1 = 1018000, c = x^2 + y^2 = 1018001 + 1. Since k = 1, we go to the other alternative, Where c = 2018. Here, we determine that 2018 = k(x^2 + y^2), k = 2, and x^2 + y^2 = 1009 . A few trials give x=28, y = 15 as the only solution. Then a = 4xy =1680, b = 2(x^2 - y^2) = 450, c = 2(x^2 + y^2)= 2018. Ed Gray