1. Find all prime numbers $p$ and $q$ for which $p^{4}-q^{6}$ is a power of a prime number. (The numbers 7 and 8 are powers of a prime number, but 6 is not.)
III/1. Let's write $p^{4}-q^{6}=r^{n}$ for some prime $r$ and natural number $n$. The expression $p^{4}-q^{6}$ can be factored as $p^{4}-q^{6}=\left(p^{2}-q^{3}\right)\left(p^{2}+q^{3}\right)$. Since $p^{4}-q^{6}>0$, the primes $p$ and $q$ cannot be equal, so they are coprime. Let $d$ be the greatest common divisor of $p^{2}-q^{3}$ and $p^{2}+q^{3}$. Then $d$ also divides $2 p^{2}$ and $2 q^{3}$, so due to the coprimality of $p$ and $q$, $d$ can be at most 2.

If $d=2$, then $r=2$ and $p^{2}-q^{3}=2$ and $p^{2}+q^{3}=2^{n-1}$. Adding these equations and dividing by 2, we get $p^{2}=1+2^{n-2}$ or $2^{n-2}=p^{2}-1=(p-1)(p+1)$. The numbers $p-1$ and $p+1$ must both be powers of 2, which is only possible if $p=3$. But then we get a contradiction $q^{3}=7$.

Therefore, $d=1$. Since the numbers $p^{2}-q^{3}$ and $p^{2}+q^{3}$ are coprime and their product is a power of a prime, we must have $p^{2}-q^{3}=1$ or $p^{2}=q^{3}+1=(q+1)\left(q^{2}-q+1\right)$. Since $q$ is a prime, $q+1>2$ and $q^{2}-q+1>2$, so neither of the factors $q+1$ and $q^{2}-q+1$ is equal to 1. If we want their product to be the square of a prime, the factors must be equal. From the equation $q+1=q^{2}-q+1$ we get $q=2$ and from $p^{2}=1+q^{3}$ we get $p=3$. Finally, we check that $3^{4}-2^{6}=17$ is indeed a power of a prime. The only solution is thus $p=3, q=2$.
Calculation of possible gcd of $p^{2}+q^{3}$ and $p^{2}-q^{3}$ ..... 2 points
Conclusion $p^{2}-q^{3}=1$ in the case $\operatorname{gcd}\left(p^{2}+q^{3}, p^{2}-q^{3}\right)=\mathbf{1}$ ..... 1 point
Writing the solution $p=3, q=2$ in the case $\operatorname{gcd}\left(p^{2}+q^{3}, p^{2}-q^{3}\right)=\mathbf{1}$ ..... 1 point
Conclusion $p^{2}+q^{3}=2^{n-1}$ and $p^{2}-q^{3}=2$ in the case $\operatorname{gcd}\left(p^{2}+q^{3}, p^{2}-q^{3}\right)=\mathbf{2}$ ..... 1 point
Writing $p=3$ and concluding no solution in the case $\operatorname{gcd}\left(p^{2}+q^{3}, p^{2}-q^{3}\right)=\mathbf{2}$ ..... 1 point

2. method. Again, we write $p^{4}-q^{6}=r^{n}$. By considering the parity, we conclude that all three of $p, q, r$ cannot be odd at the same time. Since $r>0$, we also have $p>q$. Due to the positivity of all terms, we also conclude that $p, q, r$ are pairwise distinct. Since at least one is even, we can consider 3 cases. If $p=2$, then due to $p>q$ we have $q=1$. In this case, there is no solution. If $q=2$, we can factor the expression: $\left(p^{2}-8\right)\left(p^{2}+8\right)=r^{n}$, or $p^{2}-8=r^{a}$ and $p^{2}+8=r^{b}$, where $a, b \in \mathbb{N}_{0}$, and $b>a$. We again consider two cases: if $a=0$, we get the solution $p=3, q=2, r=17$. If $a>0$, we can subtract the equations and factor out $r$: $16=r\left(r^{b-1}-r^{a-1}\right)$. This gives us $r \mid 16$, from which we get $r=2=q$, which again does not give a solution. The only remaining case is when $r=2$. We again write $p^{2}-q^{3}=2^{a}$ and $p^{2}+q^{3}=2^{b}$, where $a, b \in \mathbb{N}_{0}$ and $a1$. It follows that both $p-1$ and $p+1$ are powers of 2, which is only possible if $p=3$, from which we get $q^{3}=7$. Again, there is no solution. If $a>1$, we add the equations and factor out 4: $2 p^{2}=4\left(2^{b-2}-2^{a-2}\right)$. From this, we get $p=2$, which again does not give a solution.

Observations that $p, q$ and $r$ cannot all be odd, that they are pairwise distinct, and that $p>q \ldots 1$ point

Handling the case $p=2$

1 point

Handling the case $q=2$ (cases $a=0$ and $a>0$, writing the solution)