Baltic Way 2019

At first we will prove that there are infinitely many triangles whose side lengths are consecutive integers and whose area also is an integer. Let the side lengths of the triangle be $2m-1$, $2m$ and $2m+1$, then by Heron's formula its area is $$S=\sqrt{\frac{6m(2m+2)2m(2m-2)}{16}}=m\sqrt{3(m^2-1)}(\text{Eq-7})$$ Now we show that the equation $3(m^2-1)=q^2$ has infinitely many integer solutions. We use the standard method for solving Pell's equations. As $q$ should be divisible by $3$ then we denote $q=3l$ and our equation turns into $m^2-1=3l^2$. It has a solution $m=2$ and $l=1$ and if $(m,l)$ is a solution then $(2m+3l,m+2l)$ also is a solution. Indeed, we can check, that $$(2m+3l{)}^2-1=3(m+2l{)}^2$$ is equivalent to $$4m+12ml+9l^2-1=3m^2+12ml+12l^2$$ what is equivalent to $m^2-1=3l^2$. We have proved that there are infinitely many triangles whose side lengths are $2m-1$, $2m$ and $2m+1$ and whose area is $S=m\sqrt{3(m^2-1)}$ where $\sqrt{3(m^2-1)}$ is an integer. Let $ABC$ be such a triangle with $AB=2m$, $BC=2m-1$ and $AC=2m+1$ and let $CH$ be the altitude drawn from $C$ to the side $AB$. As $S=\frac{CH\cdot AB}{2}$ we get that $CH=\sqrt{3(m^2-1)}$ what is an integer. And $$AH=\sqrt{A{C}^2-C{H}^2}=\sqrt{(2m+1{)}^2-(3m^2-3)}=m+2$$ is an integer, too. Now we can put our triangle $ABC$ into a coordinate plain. Let $A$ be the origin point $(0,0)$ and let $B$ be the point with coordinates $(2m,0)$. Then the point $C$ also has integer coordinates $(m+2,\sqrt{3m^2-3})$.