1. Find all prime numbers $p$ for which the number $7^{p-2}+9 p^{4}$ is a perfect square.
III/1. Let $7^{p-2}+9 p^{4}=m^{2}$. Then $7^{p-2}=m^{2}-9 p^{4}=\left(m-3 p^{2}\right)\left(m+3 p^{2}\right)$. Therefore, $m-3 p^{2}=7^{r}$ and $m+3 p^{2}=7^{p-r-2}$ for some $r \geq 0$ and $p-r-2>r$. Subtracting the equations, we get $6 p^{2}=7^{r}\left(7^{p-2 r-2}-1\right)$. Thus, $p=7$ or $r=0$. The first case is indeed a solution, as we get $m=196$.

In the second case, $6 p^{2}+1=7^{p-2}$. Clearly, $p=2, p=3$, and $p=5$ are not solutions, so let $p>7$. Since $6 p^{2}=7^{p-2}-1=(7-1)\left(7^{p-3}+7^{p-4}+\ldots+7^{1}+1\right)$, it must be that $p^{2}=7^{p-3}+7^{p-4}+\ldots+7^{1}+1$. We know that the line with equation $y=x$ and the exponential function $7^{x-1}$ intersect at $x=1$. For $x>1$, the exponential function grows faster. Therefore, $7^{x-1} \geq x$ or $7^{x} \geq 7 x$. Thus, for $p>7$, we can estimate

$$
\begin{aligned}
p^{2} & =7^{p-3}+7^{p-4}+\ldots+7^{1}+1 \geq \\
& \geq 7(p-3)+7(p-4)+\ldots+7 \cdot 1+1= \\
& =7((p-3)+(p-4)+\ldots+1)+1= \\
& =7 \frac{(p-3)(p-2)}{2}+1>p^{2}+\frac{5\left(p^{2}-7 p\right)}{2}= \\
& =p^{2}+\frac{5 p(p-7)}{2}>p^{2}
\end{aligned}
$$

In this case, there are no solutions, so $p=7$ is the only prime number for which the number $7^{p-2}+9 p^{4}$ is a perfect square.
Writing $7^{p-2}=\left(m-3 p^{2}\right)\left(m+3 p^{2}\right)$ ..... 1 point
Identifying $m-3 p^{2}=7^{r}$ and $m+3 p^{2}=7^{p-r-2}$ and concluding $6 p^{2}=7^{r}\left(7^{p-2 r-2}-1\right)$ ..... 2 points
Concluding $p=7$ or $r=0$ ..... 1 point
Concluding $m=196$ ..... 1 point
Estimating $7^{x} \geq 7 x$ ..... 1 point