IRL ABooklet

a. The formula in part (a) can be checked by direct calculation
$$(x^2 - xy + y^2)(a^2 - ab + b^2) = (xa - by)^2 - (xa - by)(ay + bx - by) + (ay + bx - by)^2.$$
Alternatively, we may let $\xi, \bar{\xi}$ be the complex roots of the quadratic $x^2 - x + 1$, so that $\xi^2 = \xi - 1$, as well as $\bar{\xi}^2 = \bar{\xi} - 1$, and $N(x, y) = (x - \xi y)(x - \bar{\xi} y)$. Then
$$\begin{aligned} N(a, b)N(x, y) &= (x^2 - xy + y^2)(a^2 - ab + b^2) \\ &= (x - \xi y)(x - \bar{\xi} y)(a - \xi b)(a - \bar{\xi} b) \\ &= (xa - \xi ya - \xi xb + \xi^2 yb)(xa - \bar{\xi} ya - \bar{\xi} xb - \bar{\xi}^2 yb) \end{aligned}$$
obtained by multiplying the first and third brackets, and second with fourth brackets. Now using $\xi^2 = \xi - 1$ and $\bar{\xi}^2 = \bar{\xi} - 1$ we get
$$N(a, b)N(x, y) = N(ax - by, ay + bx - by).$$

b. To solve (b) we note that $N(x, y) = x^2 - xy + y^2 = (x - \frac{1}{2}y)^2 + \frac{3}{4}y^2 \geq 0$, hence, for any solution triple $(x, y, z)$ we have $490z^2 \leq 2023$ and so $z \leq 2$. If $z = 2$ then $2023 - 490 \cdot 4 = 63 = 3^2 \cdot 7$. Thus $3^2|N(x, y)$. Writing $x = 3u$ and $y = 3v$, we would like to solve
$$7 = N(u, v), \quad \text{or equivalently} \quad 28 = (2u - v)^2 + 3v^2.$$
It is easy to find the following four non-negative solutions, see Problem 21:
(u, v) = (1, 3), (2, 3), (3, 1), (3, 2), giving rise to
*(x, y) = (3, 9), (6, 9), (9, 3), (9, 6).

If $z = 1$ then $2023 - 490 = 1533 = 3 \cdot 7 \cdot 73$. We first solve $N(a, b) = 3$, $N(u, v) = 7$, $N(s, t) = 73$, and then use the formula from part (a) to construct solutions to $N(x, y) = 3 \cdot 7 \cdot 73$. Because the numbers are small, we easily find positive solutions:
$3 = N(a, b)$ is equivalent to $12 = (2a - b)^2 + 3b^2$
which has solutions $(a, b) = (1, 2), (2, 1)$;
$7 = N(u, v)$ is equivalent to $28 = (2u - v)^2 + 3v^2$
which has solutions $(u, v) = (1, 3), (2, 3), (3, 1), (3, 2)$;
$73 = N(s, t)$ is equivalent to $292 = (2s - t)^2 + 3t^2$
which has solutions $(s, t) = (8, 9), (1, 9), (9, 1), (9, 8)$.
Applying the formula from part (a) we obtain, for example,
$$\begin{align*} 3 \cdot 7 &= N(2,1)N(2,3) = N(1,5), \\ 3 \cdot 7 &= N(2,1)N(3,1) = N(5,4), \\ 3 \cdot 7 &= N(2,1)N(3,2) = N(4,5). \end{align*}$$
By symmetry we have $N(x, y) = N(y, x)$, hence $N(5, 1) = N(1, 5) = 3 \cdot 7$.
Applying the formula from part (a) again, we find
$$\begin{align*} 3 \cdot 7 \cdot 73 &= N(1,5)N(9,1) = N(4,41) \\ &= N(5,1)N(8,9) = N(31,44) \\ &= N(5,1)N(9,1) = N(44,13) \\ &= N(5,1)N(9,8) = N(37,41). \end{align*}$$
Using symmetry, we get eight solutions with $z = 1$:
$$(x, y) = (4, 41), (41, 4), (31, 44), (44, 31), (44, 13), (13, 44), (37, 41), (41, 37).$$
Finally, if $z = 0$ then $2023 = 17^2 \cdot 7$ and we can look for solutions $(x, y) = (17u, 17v)$. This leads us to solve $N(u, v) = 7$ for which we earlier found four solutions $(u, v) = (1, 3), (2, 3)$ and their symmetric counterparts. This leads to four solutions with $z = 0$:
$$(x, y) = (17, 51), (51, 17), (34, 51), (51, 34).$$
Altogether we found 16 different solution triples, as required.