# Proof explanation: if a triangle is Heronian then a reduced version is Heronian

I have a question about the proof of Lemma 1 in "Determination of Heronian triangles" by Carlson. Part of this lemma claims: if the triangle with sides $$na,nb,$$ and $$nc$$ (where $$n,a,b,c\in\mathbb Z$$) is Heronian (i.e. its area is an integer), then the triangle with sides $$a,b,$$ and $$c$$ is Heronian (we call this the reduced triangle).

Heron's formula implies that the area of the first triangle is $$A'=n^2A$$ where $$A$$ is the area of the second (reduced) triangle. We suppose that $$A'$$ is an integer. Then the proof of the lemma claims that, because $$A=\sqrt{s(s-a)(s-b)(s-c)}$$ is the square root of an integer (where $$s:=\frac{a+b+c}{2}$$ is the semiperimeter of the reduced triangle), it must be the case that $$A$$ is an integer because it is rational by $$A=\frac{A'}{n^2}$$.

Most of this makes sense to me, except

how do we know that $$s(s-a)(s-b)(s-c)$$ is an integer?

We know that $$(A')^2=n^4(s(s-a)(s-b)(s-c))$$ is an integer and $$a,b,c$$ are integers, but I cannot see how this implies that $$s(s-a)(s-b)(s-c)$$ is an integer because $$s=\frac{a+b+c}{2}$$ need not be an integer in general.

Question in full: if $$n,a,b,c\in\mathbb Z$$ and $$n^2\sqrt{s(s-a)(s-b)(s-c)}\in\mathbb Z$$ where $$s:=\frac{a+b+c}{2}$$, then how do we show that $$s(s-a)(s-b)(s-c)\in\mathbb Z$$?

Edit: in the accepted answer here it is claimed that, if $$s\notin\mathbb Z$$, then $$\sqrt{s(s-a)(s-b)(s-c)}\notin\mathbb Q.$$ I don't see how this is true either, but an explanation of this would clear up my confusion with my original question.

Notice first of all that $$a+b+c$$ cannot be odd for a Heronian triangle, for in that case $$(a+b+c)/2$$ would not be integer. Let's show that if $$na$$, $$nb$$, $$nc$$ form a Heronian triangle, then $$a+b+c$$ is even and consequently $$s(s−a)(s−b)(s−c)$$ is integer. We can consider $$n$$ prime without loss of generality.

If $$n>2$$ then $$a+b+c$$ is even, because it has the same parity as $$n(a+b+c)$$. We are then left with $$n=2$$: in this case we will show by contradiction that if $$a+b+c$$ is odd then triangle $$2a$$, $$2b$$, $$2c$$ is not Heronian.

If $$a+b+c$$ is odd, quantities $$s=a+b+c$$, $$\alpha=s-2a$$, $$\beta=s-2b$$, $$\gamma=s-2c$$ are all odd, hence they are congruent to $$\pm1$$ modulo $$4$$. But $$\alpha+\beta+\gamma=s$$, hence these four quantities cannot all be congruent modulo $$4$$: if $$s\equiv+1$$ then two among $$\alpha$$ $$\beta$$ $$\gamma$$ are congruent to $$+1$$ and the other is congruent to $$-1$$, while if $$s\equiv-1$$ then two among $$\alpha$$ $$\beta$$ $$\gamma$$ are congruent to $$-1$$ and the other is congruent to $$+1$$. In all cases we have then $$s\cdot\alpha\cdot\beta\cdot\gamma\equiv -1 \pmod 4$$, thus this number cannot be a perfect square (perfect squares are congruent to $$0$$ or $$1$$ modulo $$4$$) and triangle $$2a$$, $$2b$$, $$2c$$ is not Heronian. This completes the proof.

The square root of an integer $$n$$ cannot be a rational number, unless $$n$$ is a perfect square. Suppose, by contradiction, that $$\sqrt{n}=p/q$$ where $$p$$ and $$q$$ are integers without prime cofactors (and $$q\ne1$$). Then $$n=p^2/q^2$$. But $$p^2$$ and $$q^2$$ don't have prime cofactors either, because $$p^2$$ has the same prime factors as $$p$$ and $$q^2$$ has the same prime factors as $$q$$. Hence $$n$$ is not integer, which is absurd.
• Yes I know that the square root of an integer is either irrational or integral, but how do we know that $s(s-a)(s-b)(s-c)$ is an integer? – Dave Feb 23 at 20:01
• Thanks for the updated answer. I just found a proof that $s$ is integral here: citeseerx.ist.psu.edu/viewdoc/… as well. – Dave Feb 25 at 19:13