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

Question

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 $s(s-a)(s-b)(s-c)\in\mathbb Z$?

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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.

Answer 1

EDITED ANSWER.

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.

ORIGINAL ANSWER

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.

Answer 2

Lemma: If the side lengths $(a, b, c)$ of a Heronian $\triangle{ABC}$ has a common divisor $n$, then the triangle with the sides reduced by $n$ times is also Heronian.

Proof: Notice first, if we will prove the Lemma for a common prime divisor $p$ of $a, b, c$, applying the same statement to the triangle reduced on the previous iteration, we will prove for any common divisor $n$ of $a, b, c$, since $n$ is a product of some finite set of prime numbers (Fundamental theorem of arithmetic).

Suppose $p$ is a common prime divisor of $a, b, c$, so there exist three integer numbers $x, y, z \in \mathbb{N}$ as the sides of some $\triangle{XYZ}$ such that $a = px, b = py, c = pz$

Since $\triangle{ABC}$ is a Heronian, then the area of the triangle $\triangle{ABC}$ is integer: $S_{\triangle{ABC}} \in \mathbb{N}$.

According to Heron's formula the areas of triangles $\triangle{XYZ}$ and $\triangle{ABC}$ are \begin{gather*} S_{\triangle{XYZ}}=\frac{1}{4}\sqrt{(x+y+z)(-x+y+z)(x-y+z)(x+y-z)}\\ S_{\triangle{ABC}}=\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)} \end{gather*}

Let \begin{gather*} S=(x+y+z)(-x+y+z)(x-y+z)(x+y-z) \in \mathbb{N} \end{gather*}

Then \begin{gather*} S_{\triangle{XYZ}}=\frac{1}{4}\sqrt{S}\\ S_{\triangle{ABC}}=\frac{p^2}{4}\sqrt{S} \end{gather*}

From there \begin{alignat*}{2} S \cdot p^4&=16 S_{\triangle{ABC}}^2 \end{alignat*}

Since $S$ is integer, $p^4, 16, S_{\triangle{ABC}}^2$ are all perfect squares. It means $S$ is a perfect square too, then one concludes that $\sqrt{S} \in \mathbb{N}$ and $S_{\triangle{XYZ}}=\frac{1}{4}\sqrt{S}$ is at least rational: $S_{\triangle{XYZ}} \in \mathbb{Q}$.

If $\delta=x+y+z$ is even then $\alpha=\delta-2a, \beta=\delta-2b, \gamma=\delta-2c$ are all even. $S=\delta \cdot \alpha \cdot \beta \cdot \gamma = 16 \cdot N$, where $N$ is some integer number. Since $S$ and $16$ are both perfect squares, $N$ is a perfect square too.

$S_{\triangle{XYZ}}=\frac{1}{4}\sqrt{S}=\sqrt{N} \in \mathbb{N}$

That would enough to conclude the triangle $\triangle{XYZ}$ is Heronian.

Now we will show that $x+y+z$ is even.

Notice first of all that $s=a+b+c$ cannot be odd for a Heronian triangle, for in that case $-a+b+c=s-2a$, $a-b+c=s-2b$, $a+b-c=s-2c$ are all odd too, $(a+b+c)(-a+b+c)(a-b+c)(a+b-c)$ the product of four odd numbers is odd, the square root of odd number cannot be even, thus the area of this triangle as a quarter of an odd number cannot be integer.

If $p>2$ then $x+y+z$ is even, because it has the same parity as $a+b+c=p(x+y+z)$.

When $p=2$, in this case we will show by contradiction that if $x+y+z$ is odd then triangle $2x, 2y, 2z$ is not Heronian.

First, if $a=2x, b=2y, c=2z$ then

$S_{\triangle{ABC}}=\sqrt{(x+y+z)(-x+y+z)(x-y+z)(x+y-z)}$

If $x+y+z$ is odd, quantities $\delta=x+y+z$, $\alpha=\delta-2a, \beta=\delta-2b, \gamma=\delta-2c$ are all odd, hence they are congruent to $±1$ modulo $4$. But $\alpha+\beta+\gamma=\delta$, hence these four quantities cannot all be congruent modulo $4$: if $\delta \equiv +1$ then two among $\alpha, \beta, \gamma$ are congruent to $+1$ and the other is congruent to $-1$, while if $\delta \equiv -1$ then two among $\alpha, \beta, \gamma$ are congruent to $-1$ and the other is congruent to $+1$. In all cases we have then $\delta \cdot \alpha \cdot \beta \cdot \gamma \equiv -1 \mod 4$, thus this number cannot be a perfect square (perfect squares are congruent to $0$ or $1$ modulo $4$), $S_{\triangle{ABC}} = \sqrt{\delta \cdot \alpha \cdot \beta \cdot \gamma}\notin \mathbb{N}$ and triangle $2x, 2y, 2z$ is not Heronian.

This completes the proof.