How do you show that, for any integer, there is a triangle with side rational lengths and that integer area?

Partial Solution which may help .

Given $M\in \mathbb{N}$ we are supposed to find $a,b$ and $c \in \mathbb{Q}$ such that $M^2=s(s-a)(s-b)(s-c) $ where $s=\frac{a+b+c}{2}$

This is equivalent to finding $a,b$ and $c$ such that $$16M^2=(a+b+c)(a-b+c)(a+b-c)(-a+b+c)$$

Let $(a-b+c)=X,(a+b-c)=Y$ and $(-a+b+c)=Z$

Then we are supposed determine $X,Y$ and $Z\in \mathbb{Q}^+$ such that $$16M^2=(X+Y+Z)\cdot X \cdot Y\cdot Z.$$

Let $(X+Y+Z)=2^k\cdot M$ and $X \cdot Y\cdot Z=\frac{16M}{2^k}$

The solution to this system exists when $P^2 \geq 4Q$ where $P=2^k M-X$ and $Q=\frac{16M}{2^k}$

That is solution will exists for some bigger $k$ as LHS of $P^2 \geq 4Q$(After making signs of all terms positive by transposing) includes $2^{2k}$ but RHS have highest exponent $2^{k+1}$

We are left to find $Y,Z \in \mathbb{Q}^+$ such that $(2^kM-Y-Z)(YZ)=16M{2^k}$ and $2^k M-Y-Z >0$. Note that this is a curve in $\mathbb{R}^2. Which is connected

When are rational then $Y,Z$ then $X=2^k M-Y-Z$ which is rational and hence the system of equations $(a-b+c)=X,(a+b-c)=Y$ and $(-a+b+c)=Z$ admits rational solution because they are linear equations.

Note-Other properties of Triangle are automatically satisfied because if $X>0$ then $a+c>b$ and so on.