Integer lengths in a triangle

Question

If $a,b,c$ are lengths of the sides of a right triangle, where $a$ is the hypotenuse, then is it possible that $c$, $b$, $\sqrt{a^2-ac}$, $\sqrt{a^2-ab}$ be all integers? I derived this in another geometry problem, but I don't know how to go about it.

Answer 1

As $\sqrt{a^2-ac}$, $\sqrt{a^2-ab}$ are integers, we get $a(b-c)\in\mathbb Z\implies b=c$ or $a\in\mathbb Z$. Clearly, $b\neq c$ and thus, $a\in\mathbb Z$. As $\triangle ABC$ is right angled triangle with integer sides, we get, $$a=m^2+n^2\qquad b=2mn\qquad c=m^2-n^2$$for some $m,n\in\mathbb N$. Let, $$C:=a(a-c)=(m^2+n^2)(2n^2)\qquad B:=a(a-b)=(m^2+n^2)(m-n)^2$$Now as $a(a-c)$ and $a(a-b)$ are perfect squares, we must have $2(m^2+n^2)$ and $m^2+n^2$ as perfect squaress which is absurd. Thus, the original hypothesis is false.

Answer 2

COMMENT.-If $a$ is irrational then $a=a_1\sqrt n$ so we have $\sqrt{a^2-ac}=d\in\mathbb N\Rightarrow a^2-ac=d^2$ which is impossible for $c$ positive integer (only valid for $c=0$ so $c$ cannot be a side of a triangle). Consequently you must have $(a,b,c)$ is a Pythagorean triangle.

Try now the problem with $a,b$ and $c$ integers.