IRL_ABooklet_2023

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 $$\begin{array}{rl}N(a,b)N(x,y)& =(x^2-xy+y^2)(a^2-ab+b^2)\\ & =(x-\xi y)(x-\bar{\xi }y)(a-\xi b)(a-\bar{\xi }b)\\ & =(xa-\xi ya-\xi xb+{\xi }^{2}yb)(xa-\bar{\xi }ya-\bar{\xi }xb-{\bar{\xi }}^{2}yb)\end{array}$$ 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 $N(x,y)=x^2-xy+y^2=(x-\frac{1}{2}y{)}^2+\frac{3}{4}{y}^2\ge 0$, hence, for any solution triple $(x,y,z)$ we have $490z^2\le 2023$ and so $z\le 2$. 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 $$7=N(u,v),\text{or equivalently}28=(2u-v{)}^2+3v^2.$$ 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, $$\begin{array}{rl}3\cdot 7& =N(2,1)N(2,3)=N(1,5),\\ 3\cdot 7& =N(2,1)N(3,1)=N(5,4),\\ 3\cdot 7& =N(2,1)N(3,2)=N(4,5).\end{array}$$ 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 $$\begin{array}{rl}3\cdot 7\cdot 73& =N(1,5)N(9,1)=N(4,41)\\ & =N(5,1)N(8,9)=N(31,44)\\ & =N(5,1)N(9,1)=N(44,13)\\ & =N(5,1)N(9,8)=N(37,41).\end{array}$$ 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.