AUT_ABooklet_2023

Let $N=2023$. We assume that the equation $x^3-N{x}^2-Nx+r=0$ has three rational solutions $\frac{a}{k}$, $\frac{b}{k}$, $\frac{c}{k}$, where $a,b,c$ are integers and $k$ is a positive integer with $\gcd (a,b,c,k)=1$. According to Vieta we have $\frac{a}{k}+\frac{b}{k}+\frac{c}{k}=N$ and $\frac{b}{k}\cdot \frac{c}{k}+\frac{a}{k}\cdot \frac{c}{k}+\frac{a}{k}\cdot \frac{b}{k}=-N$. This is equivalent to $$\begin{array}{rlrl}a+b+c& =kN& \Rightarrow a^2+b^2+c^2+2(bc+ac+ab)& =k^2{N}^2\\ bc+ac+ab& =-k^{2}N& \Rightarrow a^2+b^2+c^2& =k^2{N}^2+2k^{2}N=k^{2}N(N+2).\end{array}$$

In a next step, we recognize that $k$ cannot be even. If it were, we would have $a^2+b^2+c^2\equiv 0(\mathrm{mod}4)$, from which we obtain that $a,b,c$ are all even, as 0 and 1 are the only quadratic residues modulo 4. This contradicts the assumption that $\gcd (a,b,c,k)=1$. For odd values of $k$, we have $k^2\equiv 1(\mathrm{mod}8)$. Furthermore, we have $N=2023\equiv 7(\mathrm{mod}8)$. From this, we obtain $k^{2}N(N+2)\equiv 1\cdot 7\cdot 1\equiv 7(\mathrm{mod}8)$. The sum of three perfect squares can never be congruent to 7 modulo 8, which can easily be verified by adding all possible combinations (the only quadratic residues modulo 8 are 0, 1 and 4). It follows that the above equation can never have three rational solutions.

(Josef Greilhuber) $\square$