Almost Equilateral

let the side lengths of the triangle be $s-1, s ,s+1$ , with $s\in \mathbb{N}$ . then by herons formula, the area is $A = \dfrac{1}{4} \sqrt{(3s)(s+2)(s)(s-2)} = \dfrac{s}{4} \sqrt{3 (s^2-4)} = \dfrac{3}{4} sk$ for A to be integer, we require there to be a $k \in \mathbb{N}$ such that $s^2-4= 3k^2 \implies \left(\dfrac{s}{2}\right)^2 -3 \left( \dfrac{k}{2}\right)^2 = 1$ notice that from the left eqn $(s,k)$ must both be evens, meaning the equation in the right is a pell's equation in $\left(\dfrac{s}{2} , \dfrac{k}{2}\right)$ . For equations of the form $x^2-Dy^2= 1 \to x_n +y_n \sqrt{D} = (x_1+y_1\sqrt{D})^n$ with $(x_1,y_1)$ being the fundemental solution to the system(i.e the smallest one). Its easy to see with $D=3$ that the fundemental solution is $(2,1)$ . so we have $\left(\dfrac{s}{2} , \dfrac{k}{2}\right) = \left(\dfrac{ (2+\sqrt{3})^n+ (2-\sqrt{3})^n}{2} , \dfrac{(2+\sqrt{3})^n- (2-\sqrt{3})^n}{2\sqrt{3}}\right)\\ \implies A = \dfrac{3}{4} sk = \dfrac{\sqrt{3}}{4} \left((2+\sqrt{3})^{2n}- (2-\sqrt{3})^{2n}\right) = \dfrac{\sqrt{3}}{4} \left((7+4\sqrt{3})^{n}- (7-4\sqrt{3})^{n}\right)$ The answer follows