
Problem 8'.3. Find all natural numbers $n$ such that there exists an integer number $x$ for which $499\left(1997^{n}+1\right)=x^{2}+x$.


Solution: Let $n$ be a solution of the problem. Then $(2 x+1)^{2}=$ $1996 \cdot 1997^{n}+1997$. If $n=1$ we get $(2 x+1)^{2}=1997^{2}$ and so $2 x+1= \pm 1997$. Therefore $x=998$ and $x=-999$ satisfy the conditions of the problem. Let $n \geq 2$. Now $(2 x+1)^{2}$ is divisible by 1997 , which is a prime number, and so $(2 x+1)^{2}$ is divisible by $1997^{2}$. But this is impossible, since $1996 \cdot 1997^{n}+1997$ is not divisible by $1997^{2}$ when $n \geq 2$. The only solution is $n=1$.
