Junior Mathematical Olympiad

First, we have $$(p+1{)}^{p-1}+(p-1{)}^{p+1}\ge (p+1{)}^{p-1}\ge (p-1{)}^{p-1}(1)$$ $$(p+1{)}^{p-1}+(p-1{)}^{p+1}<(p+1{)}^{p+1}+(p+1{)}^{p-1}=2(p+1{)}^{p+1}<(p+2{)}^{p+2}(2)$$ From (1) and (2) follows $$(p-1{)}^{p-1}\le q^q<(p+2{)}^{p+2}.$$ 1) Let $$q=p-1,(p+1{)}^{p-1}+(p-1{)}^{p+1}=(p-1{)}^{p-1}.$$ From $$(p+1{)}^{p-1}+(p-1{)}^{p+1}\ge (p+1{)}^{p-1}\ge (p-1{)}^{p-1}$$ we have $(p-1{)}^{p+1}=0$ and $(p+1{)}^{p-1}=(p-1{)}^{p-1}\Rightarrow p=1,q=0$. But $0$ is not a natural number, so $(p+1{)}^{p-1}+(p-1{)}^{p+1}=(p-1{)}^{p-1}$ has no solution in the set of natural numbers $\mathbb{N}$.

2) Let $q=p$, $(p+1{)}^{p-1}+(p-1{)}^{p+1}=p^p$.

If $p=1$, then $$(p+1{)}^{p-1}+(p-1{)}^{p+1}=1\text{ и }{p}^p=1.$$ So $(p,q)=(1,1)$ is a solution of $(p+1{)}^{p-1}+(p-1{)}^{p+1}=q^q$. If $p=2$, then $$(p+1{)}^{p-1}+(p-1{)}^{p+1}=4\text{ и }{p}^p=4.$$ Hence $(p,q)=(2,2)$ is a solution of $(p+1{)}^{p-1}+(p-1{)}^{p+1}=q^q$. If $p=3$, then $$(p+1{)}^{p-1}+(p-1{)}^{p+1}=32\text{ а }{p}^p=27.$$ Hence $(p,q)=(3,3)$ is not a solution of $(p+1{)}^{p-1}+(p-1{)}^{p+1}=q^q$. If $p\ge 4$, it holds $(p-1{)}^p>p^{p-1}$. We obtain $$(p+1{)}^{p-1}+(p-1{)}^{p+1}>(p+1{)}^{p-1}+p^{p-1}(p-1)>p^{p-1}+p^{p-1}(p-1)=p^p.$$ Hence, when $p\ge 4$, $(p+1{)}^{p-1}+(p-1{)}^{p+1}=p^p$ doesn't have a solution in $\mathbb{N}$.

3) Let $q=p+1$ $$(p+1{)}^{p-1}+(p-1{)}^{p+1}=(p+1{)}^{p+1}\iff$$ $$(p-1{)}^{p+1}=(p+1{)}^{p-1}((p+1{)}^2-1)\iff$$ $$(p-1{)}^{p+1}=(p+1{)}^{p-1}p(p+2)$$ Since $p$ and $p-1$ are coprime, the equation $(p+1{)}^{p-1}+(p-1{)}^{p+1}=(p+1{)}^{p+1}$ doesn't have a solution in $\mathbb{N}$.

Finally, the solutions of $(p+1{)}^{p-1}+(p-1{)}^{p+1}=q^q$ are $(p,q)=(1,1)$ and $(p,q)=(2,2)$.