Given a triangle $ABC$, $\angle C =90 ^{\circ}$ and $M$ is an interior point
$MA, MB, MC$ are positive integers.
Find $AB=c, BC=a$ and $CA=b$ where $a,b,c \in \mathbb Z$ and $MA, MB, MC$.
I need to make an example of such a triangle and example of such a point $M$. There are an infinite number of solutions? (Without multiplying by $k$)
My work so far:
Let $C(0;0); A(0;b); B(a;0)$ and $M(x;y)$. Then $$MA=\sqrt{x^2+(y-b)^2} \in \mathbb Z$$ $$MB=\sqrt{(x-a)^2+y^2} \in \mathbb Z$$ $$MC=\sqrt{x^2+y^2} \in \mathbb Z$$ $c^2=a^2+b^2$
Addition
We have:
1) triangle $ABC$, $\angle C =90 ^{\circ}$, point $M \in \triangle ABC$.
2) $MA, MB, MC$ are positive integers.
Find:
$AB, BC, CA, MA, MB, MC \in \mathbb Z$
Let $MA=r,MC=q,MB=p,\angle CMB=\alpha,\angle CMA=\beta$ , applying cosine's law,
$a^2=p^2+q^2-2pq\cosα$
$b^2=q^2+r^2-2rq\cosβ$
$c^2=p^2+r^2-2pr\cos(360-α-β)$
$\cos{\alpha},\cos{\beta}$ must be rational. Also $\sinα\sinβ$ is rational(c is integer), let $\cos{\alpha}=\dfrac{n_1}{m_1},\cos{\beta}=\dfrac{n_2}{m_2}$
$\sqrt{\big(1-\dfrac{n_1^2}{m_1^2}\big)\big(1-\dfrac{n_2^2}{m_2^2}\big)}=\dfrac{l}{m1m2}$
$⇔(m_1^2-n_1^2)(m_2^2-n_2^2)=l^2$
$(i)$special case of $\angle AMB=\angle AMC\angle=BMC= 120°$, $(ii)$composite-number $l=(2d+1)(2e+1)$ are solutions of this equation.
$(i)$This leads possible case is $\angle AMB=\angle BMC=\angle CMA=120°$ by conditions of obtuse triangles. Thinking area of triangle,
$ab=\dfrac{\sqrt3}2(pq+pr+qr)$
This equation hold if ABC is regular triangle, or $\sin \angle ABC=\dfrac{n\sqrt3}m$.
$(ii)$ Since $m_1,n_1,m_2,n_2,d,e$ are pythagorean triples, organizing equations of cosine's laws on coprime $(m,n)(s,t)$, we get next parametric equation.
$2q^2+2pq(\dfrac{2mn}{m^2+n^2})+2qr(\dfrac{2st}{s^2+t^2})=2pr\dfrac{4mnst-(m^2-n^2)(s^2-t^2)}{(m^2+n^2)(s^2+t^2)}$