3. A positive integer $M$, when added to 36, 300, and 596 respectively, results in the squares of three different positive integers. The sum of the smallest and the largest of these three different positive integers is twice the middle one. Then this positive integer $M=$
3.925 .

According to the problem, let $M+36=(k-m)^{2}$,
$M+300=k^{2}$,
$M+596=(k+m)^{2}(m>0)$.
Then $(M+36)+(M+596)=2\left(k^{2}+m^{2}\right)$.
Therefore, $M+316=M+300+16=k^{2}+m^{2}$.
Solving this, we get $m=4$.
$$
\begin{array}{l}
\text { Also, }(M+596)-(M+36) \\
=(k+m)^{2}-(k-m)^{2}=4 k m,
\end{array}
$$

Thus, $k=35$.
Therefore, $M=925$.