Problem 2179

a. Every $a$ of the form $a=4k-3$ with $k \geq 2$ is friendly. Indeed the numbers $m=2k-1>0$ and $n=k-1>0$ satisfy the given equation with $a=4k-3$:
$$((2k-1)^{2}+(k-1))((k-1)^{2}+(2k-1)) = (4k-3)k^{3} = a(m-n)^{3}.$$
Hence $5, 9, \ldots, 2009$ are friendly and so $\{1,2, \ldots, 2012\}$ contains at least 502 friendly numbers.

b. We show that $a=2$ is not friendly. Consider the equation with $a=2$ and rewrite its left-hand side as a difference of squares:
$$\frac{1}{4}\left((m^{2}+n+n^{2}+m)^{2}-(m^{2}+n-n^{2}-m)^{2}\right) = 2(m-n)^{3}$$
Since $m^{2}+n-n^{2}-m = (m-n)(m+n-1)$, we can further reformulate the equation as
$$(m^{2}+n+n^{2}+m)^{2} = (m-n)^{2}\left(8(m-n)+(m+n-1)^{2}\right)$$
It follows that $8(m-n)+(m+n-1)^{2}$ is a perfect square. Clearly $m>n$, hence there is an integer $s \geq 1$ such that
$$(m+n-1+2s)^{2} = 8(m-n)+(m+n-1)^{2}$$
Subtracting the squares gives $s(m+n-1+s) = 2(m-n)$. Since $m+n-1+s > m-n$, we conclude that $s<2$. Therefore the only possibility is $s=1$ and $m=3n$. However then the left-hand side of the given equation (with $a=2$) is greater than $m^{3}=27n^{3}$, whereas its right-hand side equals $16n^{3}$. The contradiction proves that $a=2$ is not friendly.