32. For which natural numbers $y$ is the number $y^{2}+3^{y}$ a perfect square
79.32. Suppose that $y^{2}+3^{y}=x^{2}$, i.e., $x^{2}-y^{2}=3^{y}$. From this, we obtain that $x-y$ and $x+y$ are powers of three, specifically $3^{k}$ and $3^{y-k}$. Then $y=\left(3^{k}-3^{y-k}\right) / 2$. Since $\quad\left(3^{k}-3^{y-k}\right) / 2 \geqslant 3^{y-k-1} \geqslant 3^{(y-3) / 2}, \quad$ and for $y \geqslant 7$ this number is greater than $y$ (this is

![](https://cdn.mathpix.com/cropped/2024_05_21_55ce716f238acb5de3adg-226.jpg?height=420&width=417&top_left_y=647&top_left_x=1296)

Fig. 53 is not difficult to prove by induction), it remains to check the values of $y$ less than seven. Answer: $y=1$ or $y=3$.