7.4. Let $[a, b]$ denote the least common multiple of the positive integers $a, b$. Try to find the positive integer solutions of the equation $\left[x^{2}, y\right]+\left[x, y^{2}\right]=1996$.
7.4. The equation has no positive integer solutions.

Because $\left[x^{2}, y\right]$ and $\left[x, y^{2}\right]$ are both divisible by $x$ and $y$, their sum 1996 should also be divisible by $x$ and $y$. Since $1996=2^{2} \times 499$, and 499 is a prime number, $x$ and $y$ can only be $1, 2, 4, 499, 998$, and 1996. However, the last three numbers are impossible. Otherwise, one of $\left[x^{2}, y\right]$ and $\left[x, y^{2}\right]$ would be no less than $499^{2}>1996$, which is a contradiction. Conversely, if $x$ and $y$ are only 1, 2, or 4, they are too small, because in this case $\left[x^{2}, y\right] + \left[x, y^{2}\right]$ is much less than 1996.