39. If the value of $a+b$ can be determined (considering $m$ as a constant), then by the inverse of Vieta's theorem, there is at most one pair of positive integers $(a, b)$ that satisfies the condition.

From the condition, we know that $(a+b)^{2}=(a-b)^{2}+4 a b$ satisfies
$$\begin{aligned} 1+4 m & \leqslant(a+b)^{2} \\ & <5+4 \sqrt{4 m+1}+4 m \\ & =(\sqrt{4 m+1}+2)^{2} \end{aligned}$$

Thus,
$$\sqrt{4 m+1} \leqslant a+b<\sqrt{4 m+1}+2$$

Therefore,
$$a+b=\left\{\begin{array}{l} \sqrt{4 m+1} \text { or } \sqrt{4 m+1}+1, \text { if } \sqrt{4 m+1} \text { is an integer; } \\ {[\sqrt{4 m+1}]+1 \text { or }[\sqrt{4 m+1}]+2, \text { if } \sqrt{4 m+1} \text { is not an integer. }} \end{array}\right.$$

In summary, $a+b$ can only take values from two consecutive positive integers. Since $a b=m \equiv 2(\bmod 4)$, it follows that $a$ and $b$ are one odd and one even, meaning $a+b$ is odd. This way, we know that the value of $a+b$ is uniquely determined, and the proposition is proved.