$ \triangle ABC$ is a right triangle, $AX$ bisects $\angle A$ and ... Assume $ \triangle ABC$ is a right triangle where $\angle C= 90°$. Suppose that
$ \overline {AX} $ bisects $\angle A$ and point $X$ lies on $BC$. Assume that the circumcircle of triangle $AXB$ intersects $AC$ on $Y$.

a) Show that if lengths $BC$ and $CY$ are intergers divisible by a prime $p$, then $AY$ is also an intenger divisible by $p$.

b) Show that if $CY= k$ and $BC= 3k$ ($k$ is an interger), then the lengths of sides of the triangle $ABC$ must be intergers.

This is what I have done:

Aplying power of point of $C$, we get:

$$CY (CA) = (CX)(CB)$$ 

$$CY(CY+AY)= CB(CB -XB)$$

$$ (CY)^2+ CY(AY) = (CB)^2 - CB(XB)$$ 

$$CY(AY) = (CB)^2 - (CY)^2 - CB(XB)$$ 

$$ AY = \frac{(CB)^2 - (CY)^2 - CB(XB)}{CY}$$ 

But this doesn't prove that $AY$ is divisible by $p$ or that it is an interger. 



I did that considering this figure. I think I may be wrong because that might not be what a circumcircle is.
Since $\angle BAX = \angle XAY$ and $BXYA$ are concyclic, it means that $BX=XY$. Thus

$$
CX = BC - BX \land CX^2 = BX^2 - CY^2
$$
$$
(BC - BX)^2 = BX^2 - CY^2
$$
$$
BC^2 + BX^2 - 2BC\cdot BX = BX^2 - CY^2
$$
$$
BX = \frac{BC^2+CY^2}{2BC}
$$
From the power of the point:
$$
AC = \frac{BC\cdot BX}{CY} = \frac{BC^2 +CY^2}{2CY}
$$
Why is it divisible by $p$?
You might find it useful:
The power of point equation you wrote is correct.
$CY⋅CA=CX⋅CB$
Also, according to the internal bisector theorem:
$\frac{CX}{CA}=\frac{BX}{AB}$