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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.

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    $\begingroup$ Do you mean a single rational triangle or all of them? The simplest instance is sides $\,(5,29,30)/6.$ $\endgroup$ – Somos Dec 24 '18 at 0:13
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    $\begingroup$ I just mean a single rational triangle with area $2$. Since $(9,10,17)/6$ has area $1$, and you just gave us one for area $2$, then the next question is exactly "how" did you get your simplest instance? Then, we need to find area $3$, I guess. In Cohen's Number Theory book in chapter 1, the first exercise is to show that for all integer areas, one can prove there is a rational triangle. So, "how" do we do this? We don't need all of them for the proof, just one per integer area, for each integer. $\endgroup$ – Pythagorus Dec 24 '18 at 1:53
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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.

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    $\begingroup$ $P^2\ge4Q$ is not necessary to conclude that if $Y$ and $Z$ are rational, then $X$ must be rational to produce integer $M$. And concluding this much is trivial toward trying to establish that integer $4M=\sqrt{(X + Y + Z) \cdot X \cdot Y \cdot Z}$ where $X$, $Y$, and $Z$ (all three) are rational, which is the question. $\endgroup$ – Pythagorus Dec 26 '18 at 11:46
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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.

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

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

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

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

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    $\begingroup$ If $p,s,k,t$ have to be integers, then the parameterization you gave does not even recover the integer areas $1$ or $2$. So, it does not help. $\endgroup$ – Pythagorus Dec 24 '18 at 7:57
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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.

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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 $'n'$

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  • $\begingroup$ You don't need to do any division to get the Area equal to $n$. Just simply take Henri Cohen's three equations and plug them into Heron's formula that I gave when I posted the original question and you get $n$ as the area. $\endgroup$ – Pythagorus Dec 30 '18 at 1:27
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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.

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