4. As shown in Figure 4, given that the equilateral $\triangle ABC$ is inscribed in $\odot O, AB$ $=86$. If point $E$ is on side $AB$, and through $E$ a line $DG \parallel BC$ intersects $\odot O$ at points $D, G$, and intersects $AC$ at point $F$, and let $AE=x, DE$ $=y$. If $x, y$ are both positive integers, then $y=$ $\qquad$
4.12.

From the problem, we know that $E F=A E=x$.
By the symmetry of the circle, $F G=D E=y$.
By the intersecting chords theorem, we have $A E \cdot E B=D E \cdot E G$, which means $x(86-x)=y(x+y)$.
If $x$ is odd, then $x(86-x)$ is also odd, and in this case, $y(x+y)$ is even. Therefore, $x$ must be even, and $y$ must also be even.
From $x(86-x)=y(x+y)$, we get
$x^{2}+(y-86) x+y^{2}=0$.
Thus, $x=\frac{86-y \pm \sqrt{(y-86)^{2}-4 y^{2}}}{2}$.
By $\Delta=(y-86)^{2}-4 y^{2} \geqslant 0$, we get
$0<y \leqslant 28$.
Since $x$ is a positive integer, $\Delta$ must be a perfect square.

Since $\Delta=(86-3 y)(86+y)$, and $y=2,4$, $\cdots, 28$, it is verified that only when $y=12$, $\Delta$ is a perfect square. At this time, $x=2$ or 72.