6. Given the system of equations $\left\{\begin{array}{l}\frac{x}{a}+\frac{y}{b}=1, \\ x^{2}+y^{2}=50\end{array}\right.$ has only integer solutions. Then the number of real pairs $(a, b)$ that satisfy the condition is $\qquad$ .
6.60.

It is easy to know that the equation $x^{2}+y^{2}=50$ has 12 sets of integer solutions: $(\pm 1, \pm 7), (\pm 7, \pm 1), (\pm 5, \pm 5)$, which correspond to 12 integer points on the circle. For each pair of real numbers that satisfy the condition, there corresponds a line in the coordinate system. Therefore, the problem can be transformed into finding the number of lines passing through these 12 integer points. It is easy to know that the lines satisfying the condition fall into two categories:
(1) Lines passing through one of the integer points and tangent to the circle, totaling 12 lines;
(2) Lines passing through two of the integer points, totaling $\mathrm{C}_{12}^{2}=66$ lines.
Since the given line equation is in intercept form, lines passing through the origin or parallel to the coordinate axes should be excluded. Therefore, the number of lines satisfying the condition is $12+66-3 \times 6=60$. That is, there are 60 pairs of real numbers $(a, b)$ that satisfy the condition.