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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.

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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.

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  • $\begingroup$ 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? $\endgroup$ – Dave Feb 23 at 20:01
  • $\begingroup$ @Dave See my edited answer. $\endgroup$ – Aretino Feb 24 at 9:28
  • $\begingroup$ Thanks for the updated answer. I just found a proof that $s$ is integral here: citeseerx.ist.psu.edu/viewdoc/… as well. $\endgroup$ – Dave Feb 25 at 19:13
  • $\begingroup$ I see. That's substantially the same argument I used. $\endgroup$ – Aretino Feb 25 at 20:57

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