Maclaurin 2017 Q3
3. How many solutions are there to the equation
where \(m\) and \(n\) are integers?
Maclaurin Olympiad 2017, Question 3. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: 16
By rewriting the equation in the form
and then completing the square on each side, we obtain
Taking the square root of each side, we get
Hence there are two cases to consider.
\(\boldsymbol{m^2 - 21 = n^2 - 4}\)
In this case, we have
so that
Therefore, because 17 is prime, \(m - n\) and \(m + n\) are equal to 1 and 17, or \(-1\) and \(-17\), in some order.
Thus in this case there are four solutions for \((m, n)\), namely \((\pm 9, \pm 8)\).
\(\boldsymbol{m^2 - 21 = -(n^2 - 4)}\)
In this case, we have
Hence
Now a square is non-negative, so that \(-5 \le m, n \le 5\).
Thus in this case there are twelve solutions for \((m, n)\), namely \((0, \pm 5)\), \((\pm 5, 0)\), \((\pm 3, \pm 4)\) and \((\pm 4, \pm 3)\).
Therefore altogether there are sixteen solutions to the given equation.
Solution © UK Mathematics Trust, from the Maclaurin Olympiad 2017 solutions.