Problem 917

## Solution.

By rearranging the given expression, we have

$$\begin{gathered} m^{2}-4 m n=n-4 \\ m^{2}+4=n(4 m+1) \end{gathered}$$

or $n=\frac{m^{2}+4}{4 m+1}$.

By extending the fraction by a factor of 16 and completing the numerator to a difference of squares, we get

$$n=\frac{16 m^{2}+64}{16 \cdot(4 m+1)}=\frac{16 m^{2}-1+65}{16 \cdot(4 m+1)}=\frac{(4 m+1)(4 m-1)+65}{16 \cdot(4 m+1)}$$

or

$$16 n=4 m-1+\frac{65}{4 m+1}$$

Since $\frac{65}{4 m+1}$ must be an integer, the number $4 m+1$ must be a divisor of 65. Since $65=5 \cdot 13$, the only possibilities for $4 m+1$ are $1, 5, 13$, and 65.

From $4 m+1=1$ we get $m=0$, which is not possible.

From $4 m+1=5$ we get $m=1$ and $n=1$, and from $4 m+1=13$ we get $m=3$ and $n=\frac{3^{2}+4}{4 \cdot 3+1}=1$.

From $4 m+1=65$ we get $m=16$ and $n=\frac{16^{2}+4}{4 \cdot 16+1}=4$.

1 point

Therefore, all such pairs of numbers $(m, n)$ are $(1,1), (3,1)$, and $(16,4)$.

Note: Points can also be awarded for correct solutions without proving that these are all the solutions. For both pairs of solutions $(1,1)$ and $(3,1)$, 1 point is awarded (which corresponds to the second-to-last point in the scoring scheme), and for the pair $(16,4)$, 1 point is awarded (which corresponds to the last point in the scoring scheme).

## SCHOOL/CITY COMPETITION IN MATHEMATICS

## 2nd grade - high school - A variant

## February 15, 2022.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.