5. Find all solutions to the equation $m^{2}-3 m+1=n^{2}+n-1$, where $m$ and $n$ are natural numbers.

Solve the problems independently. You have 150 minutes for solving.

The use of notes, literature, or a pocket calculator is not allowed.

Mathematical Competition for High School Students in Slovenia

Optional Competition

March 30, 2005

## Problems for 3rd Year
II/5. The given equation can be rearranged to $m^{2}-3 m-\left(n^{2}+n-2\right)=0$. We solve the resulting quadratic equation: $m_{1,2}=\frac{3 \pm \sqrt{9+4 n^{2}+4 n-8}}{2}=\frac{3 \pm \sqrt{(2 n+1)^{2}}}{2}=\frac{3 \pm(2 n+1)}{2}$. Thus, $m_{1}=n+2$ and $m_{2}=-n+1$. Since $m_{2}$ is not a natural number for any natural value of the parameter $n$, this option is discarded. The given equation is solved by the pair $(m, m-2)$, where $m$ is any natural number.

Rearranging the equation into a quadratic equation

.1 point

Solving the quadratic equation and simplifying (eliminating the root, fraction) .............. 2 points

Justification that one of the solutions does not meet the requirements of the problem ......................... 2 points

Writing (infinitely many) solutions, e.g., as a pair $(m, m-2), m \in \mathbb{N}$ or $(n+2, n), n \in \mathbb{N}$ 2 points