UNSW MathSoc Championship Q3

We are interested in positive integral pairs $(m,n)$ that satisfy $m^2 + mn + n^2 > 10$ . Solving this quadratic inequality for m produces:

$m = \frac{-n \pm \sqrt{n^2 - 4(1)(n^2 - 10)}}{2} = \frac{-n \pm \sqrt{40 -3n^2}}{2}$

and since m is a positive integer, we ignore the negative root which leaves:

$m > \frac{-n + \sqrt{40 - 3n^2}}{2}$ (i).

and we also require a positive-valued discriminant, or $40 - 3n^2 > 0 \Rightarrow n^2 < \frac{40}{3} \Rightarrow 0 < n < \sqrt{40/3} \Rightarrow \boxed{n = 1,2,3}.$

Substituting each of these positive integral values for n back into (i) now yields:

$n = 1, m > \frac{-1 + \sqrt{40 - 3(1^2)}}{2} \Rightarrow m > 2.541$ , or $m = 3, 4, 5, ...;$

$n = 2, m > \frac{-2 + \sqrt{40 - 3(2^2)}}{2} \Rightarrow m > 1.646$ , or $m = 2, 3, 4, ...;$

$n = 3, m > \frac{-3 + \sqrt{40 - 3(3^2)}}{2} \Rightarrow m > 0.303$ , or $m = 1, 2, 3, ...$

We ultimately wind up with the following pairs: $(m,n) = (1,3); (2,2); (3,1)$ . Of these pairs, $\boxed{m = n = 2}$ gives the smallest positive integer greater than 10:

$3 \cdot 2^2 = \boxed{12}.$

Since $m ^ { 2 } + mn + n ^ { 2 }$ is unfactorable, so we can only do this.

Then, $m = 2$ , and $n = 2$ , so $12$ .