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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.

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.

$$a=(pt+ks)(k^2+t^2)ps$$

$$b=(pt-ks)((k^2+t^2)ps+(p^2+s^2)kt)$$

$$c=(pt+ks)(p^2+s^2)kt$$

$$S_g=4pskt(p^2t^2-k^2s^2)((k^2+t^2)ps+(p^2+s^2)kt)$$

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.

Partial Solution which may help .

Given $Min mathbb{N}$ 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 $Zin mathbb{Q}^+$ such that $$16M^2=(X+Y+Z)cdot X cdot Ycdot Z.$$

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

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.