How do you show that, for any integer, there is a triangle with side rational lengths and that integer area? 
In Cohen's book, Cohen's Number Theory Volume 1, the first exercise is to show that, for any integer, there is a triangle with side rational lengths such that the triangle has that integer as an area.

For example,
What are the side rational lengths for an area 2 triangle?
Given Heron's formula for a triangle of area $2$,
$$\sqrt{\frac{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}{16}}=2$$
How do we find the sides for a side rational triangle $(a,b,c)$ that satisfies this equation?
Another example, (9,10,17)/6 has area 1, and so on for each integer.

Looking for the method for solving the exercise, not necessarily a compendium of known triples with integer areas.

 A: 
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.
A: Well, the formula itself Geronova triangle.
$$S_g=\sqrt{(a+b+c)(a+b-c)(a-b+c)(b+c-a)}$$
If: $p,s,k,t$  -integers asked us. Then the solutions are.
$$a=(pt+ks)(k^2+t^2)ps$$
$$b=(pt-ks)((k^2+t^2)ps+(p^2+s^2)kt)$$
$$c=(pt+ks)(p^2+s^2)kt$$
$$S_g=4pskt(p^2t^2-k^2s^2)((k^2+t^2)ps+(p^2+s^2)kt)$$
A: For (p,s,k,t)=(2,1,1,1), the formula, 
given by "Individ" gives us the 
triangle $(a,b,c)=(12,9,15)$ and area $(A) =54$
Where $S_g=4A$
The area $A=54$ is an integer. So integer $(54)$ is represented as a triangle by the formula given by "Individ"
There are numerous formula's to represent the area's of  different triangle's. 
But if there is a general solution (regarding triangle's) representing all the integer's is anybody' guess.
A: 
The Complete Solution

I asked Henri Cohen for an answer and he was kind enough to email the following answer:
"By simple algebraic manipulations one can find a solution as a rational
function of the area $n$. For instance
$a=(2n-1)/2$
$b=n(4n^2+4n+5)/(4n^2-1)$
$c=(20n^2+4n+1)/(2(4n^2-1))$
"
So, for area $1$, we have rational triangle lengths $(1/2,13/3,25/6)$
For area $2$, we have rational triangle lengths $(3/2,58/15,89/30)$
For area $3$, we have rational triangle lengths $(5/2,159/35,193/70)$  and so on.
The easier question remains.  What "simple algebraic manipulations" is he talking about?
The first equation gives us the rational side $a$ for every integer area $n$.  
The sequence for side $a$ is ${1/2, 3/2, 5/2....}$ for the integer area sequence
$1, 2, 3....$
Solving the first equation for $n$ and plugging the answer into the other two equations gives the equations for $b$ or $c$ in terms of $a$ alone.  It is easy to see that rational side $a$ generates rational sides $b$ and $c$ by these equations.
$b=(2a+1)(a^2+2a+2)/(2a(a+1))$
$c=(5a^2+6a+2)/(2a(a+1))$
One can easily plug these equations into Heron's formula to check for consistency. So, by this, it has been shown that for every integer area there is a side rational triangle.
Another interesting paper that attempts to parameterize the angles of these triangles can be found here.
A: Solution given by Henri Cohen is equivalent to solution of triangle (a,b,c) with sides shown below:
$a=(2n-1)(4n^2-1)$
$b=2n(4n^2+4n+5)$
$c=(20n^2+4n+1)$
And Area $A= 4n(4n^2-1)^2$
Since "OP" is interested in integer $'n'$ as area he needs to divide the triangle sides by $[2(4n^2-1)]$
and so the area gets divided by square of $[(2)(4n^2-1)]$ which is equal to 
$4(4n^2-1)^2$ and he will be left with Area equal to $'n'$
