Let the common real root of $x^2-5ax+1=0 \quad \text{and} \quad x^2-ax-5=0$ be $r$.

Since $r$ satisfies both equations, $r^2-5ar+1=0 \qquad ...(1)$ $r^2-ar-5=0 \qquad ...(2)$

Subtract (2) from (1): $\left(r^2-5ar+1\right)-\left(r^2-ar-5\right)=0$ $-4ar+6=0$ $4ar=6$ $ar=\frac{3}{2}$

Now use this in one of the equations, say (2): $r^2-ar-5=0$ $r^2-\frac{3}{2}-5=0$ $r^2=\frac{13}{2}$

Since $a>0$ and $ar=\frac{3}{2}>0$, we must have $r>0$. Hence $r=\sqrt{\frac{13}{2}}$

Now $a=\frac{3}{2r}=\frac{3}{2\sqrt{13/2}}$ Simplify: $a=\frac{3}{\sqrt{26}}$

Given $a=\frac{3}{\sqrt{2\beta}}$ compare with $a=\frac{3}{\sqrt{26}}$ so, $2\beta=26$ $\beta=13$

Therefore, the required integer is $\boxed{13}$