# 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.

• Do you mean a single rational triangle or all of them? The simplest instance is sides $\,(5,29,30)/6.$ – Somos Dec 24 '18 at 0:13
• 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. – Pythagorus Dec 24 '18 at 1:53

## 5 Answers

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. •$P^2\ge4Q$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. – Pythagorus Dec 26 '18 at 11:46 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. $$a=(pt+ks)(k^2+t^2)ps$$ $$b=(pt-ks)((k^2+t^2)ps+(p^2+s^2)kt)$$ $$c=(pt+ks)(p^2+s^2)kt$$ $$S_g=4pskt(p^2t^2-k^2s^2)((k^2+t^2)ps+(p^2+s^2)kt)$$ • 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. – Pythagorus Dec 24 '18 at 7:57 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. 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 $$'n'$$ • 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. – Pythagorus Dec 30 '18 at 1:27

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.