Triangles,Line Up!

$\begin{aligned} &\left|\{(a,b,c) \in \mathbb{N}^3 : a\leq b\leq c \text{ and there is a triangle with side lengths } a,b,c \text{ and perimeter } 2018\}\right| \\ &= \left|\{(a,b,c) \in \mathbb{N}^3 : a\leq b\leq c < a+b \text{ and } a+b+c=2018\}\right| \\ &= \left|\{(a,b,c) \in \mathbb{N}^3 : 2018-b-c \leq b \leq c < \frac{a+b+c}{2} \text{ and } a+b+c = 2018\}\right| \\ &= \left|\{(b,c) \in \mathbb{N}^2 : 1009-\frac{c}{2} \leq b \leq c < 1009\}\right| \\ &= \left|\{(b,c) \in \mathbb{N}^2 : 1009-\lfloor\frac{c}{2}\rfloor\leq b\leq c \text{ and } 673\leq c\leq 1008\}\right| \\ &= \sum_{c=673}^{1008} \left|\{b \in \mathbb{N} : 1009-\lfloor\frac{c}{2}\rfloor \leq b \leq c\}\right| \\ &= \sum_{c=673}^{1008} c - \left(1009-\lfloor\frac{c}{2}\rfloor\right) + 1 \\ &= \sum_{c=673}^{1008} \frac{3}{2}c + \frac{1}{4}((-1)^c - 1) - 1008 \\ &= \sum_{c=673}^{1008} \frac{6c - 4033}{4} + \sum_{c=673}^{1008}\frac{1}{4}(-1)^c \\ &= \frac{1}{2}\left(\frac{6(673)-4033}{4} + \frac{6(1008)-4033}{4}\right)(1008-673+1) + 0 \\ &= \boxed{84840} \end{aligned}$

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For the bonus question: Let $p$ be a prime number and consider the triangle with integral sidelengths $a,b,c$ and perimeter $2p$ . Then by Heron's formula, the area $A$ of this triangle satisfies $A^2 = p(p-a)(p-b)(p-c).$ Suppose that $A$ is an integer. Then since $p$ divides $A^2$ and $p$ is a prime, it follows that $p^2$ divides $A^2$ and therefore $p$ divides $A^2 / p = (p-a)(p-b)(p-c)$ As once again $p$ is a prime, $p$ must divide at least one of $p-a, p-b, p-c$ . However, since all of these are positive integers less than $p$ , none of them is divisible by $p$ . This is a contradiction, so it follows that $A$ cannot be an integer.

We have proven: $\text{If a triangle has integral sidelengths and prime semiperimeter, then its area is not an integer.}$

In our case, the semiperimeter is $1009$ , a prime number, so there are $\boxed{\text{zero}}$ such triangles with integral area.