5. Given $0<a<b<c<d<300$,
$$
a+d=b+c, b c-a d=91 \text {. }
$$

Then the number of ordered quadruples of positive integers $(a, b, c, d)$ that satisfy the above conditions is $\qquad$.
5.486.

Let $y>x>0, x, y$ be positive integers, and let
$(a, b, c, d)=(a, a+x, a+y, a+x+y)$.
Then $b c-a d=(a+x)(a+y)-a(a+x+y)$
$=x y=91=7 \times 13$.
Thus, $(x, y)=(1,91),(7,13)$.
When $(x, y)=(1,91)$,
$(a, b, c, d)=(a, a+1, a+91, a+92)$,
then $a$ can take 207 values;
When $(x, y)=(7,13)$,
$(a, b, c, d)=(a, a+7, a+13, a+20)$,

then $a$ can take 279 values.
And there are no repeated values in these two sets, therefore, there are a total of $207+279=486$ quadruples of positive integers that satisfy the condition.