Square-Product-Square Factors

Since $a$ and $b$ are coprime, both cannot be even. There are then two cases left: (1) $a$ and $b$ are both odd, and (2) $a$ and $b$ have opposite parities. Note that in either case, $z = a^2 + ab + b^2$ will be odd, so its factors are odd as well.

Case (1) $a$ and $b$ are both odd:

Let $p = \frac{b - a}{2}$ and $q = \frac{b + a}{2}$ . Since $a$ and $b$ are odd co-prime integers, $p$ and $q$ are also co-prime integers. Then $a = q + p$ and $b = q - p$ , so $z = a^2 + ab + b^2 = (q + p) + (q + p)(q - p) + (q - p)^2 = p^2 + 3q^2$ . Then by this theorem , the odd factors of $p^2 + 3q^2$ are also in the same form $r^2 + 3s^2$ where $r$ and $s$ are co-prime.

Case (2) $a$ and $b$ have opposite parities:

Let $p = b - a$ and $q = b + a$ . Since $a$ and $b$ are odd co-prime integers, $p$ and $q$ are also co-prime integers. Then $a = \frac{q + p}{2}$ and $b = \frac{q - p}{2}$ , so $z = a^2 + ab + b^2 = (\frac{q + p}{2}) + (\frac{q + p}{2})(\frac{q - p}{2}) + (\frac{q - p}{2})^2 = \frac{1}{4}(p^2 + 3q^2)$ . Then by this theorem , the odd factors of $p^2 + 3q^2$ are also in the same form $r^2 + 3s^2$ where $r$ and $s$ are co-prime, so the odd factors of $z = \frac{1}{4}(p^2 + 3q^2)$ are also in the same form $r^2 + 3s^2$ where $r$ and $s$ are co-prime.

In either case, the factors of $z$ are in the form of $r^2 + 3s^2$ where $r$ and $s$ are co-prime.

Let $c = s - r$ and $d = s + r$ . Since $r$ and $s$ are odd co-prime integers, $c$ and $d$ are also co-prime integers. Then $s = \frac{d + c}{2}$ and $r = \frac{d - c}{2}$ , so $r^2 + 3s^2 = (\frac{d - c}{2})^2 + 3(\frac{d + c}{2})^2 = c^2 + cd + d^2$ , which tells us that every positive integer factor of $z = a^2 + ab + b^2$ can be expressed as $c^2 + cd + d^2$ for some coprime integers $c$ and $d$ , so the statement is true .