2024 CGMO

Proof. Let $m^2+r=2^k$ and $n^2+r=2^l$ with $k,l\in {\mathbb{Z}}_{>0}$. First, we analyze the 2-adic valuations: Since $v_2(m^2)=v_2(2^k-r)=v_2(r)$, we conclude $v_2(r)$ must be even. Let $v_2(r)=2t$ where $t\in {\mathbb{Z}}_{\ge 0}$, and write $r=2^{2t}{r}_1$, $m=2^t{m}_1$. Similarly, $v_2(n^2)=v_2(r)$, so let $n=2^t{n}_1$. Define $k_1=k-2t$ and $l_1=l-2t$, which gives the reduced system: $$m_1^2+r_1=2^{k_1},$$ $$n_1^2+r_1=2^{l_1}.$$ Claim: $r_1\ge 7$. Since $r_1$ is odd, both $m_1$ and $n_1$ must be odd. As $n_1\ge 3$ (because $n>m\ge 1$), we have $l_1>3$. Thus $n_1^2+r_1\equiv 1+r_1\equiv 0(\mathrm{mod}8)$, implying $r_1\equiv 7(\mathrm{mod}8)$. * Therefore $r_1\ge 7$. This also shows $k_1\ge 3$. Now consider the difference: $$2^{l_1}-2^{k_1}=(n_1-m_1)(n_1+m_1).$$ Since $2^{k_1}$ divides the product $(n_1-m_1)(n_1+m_1)$, and both factors are even but cannot both be divisible by $4$ (which would imply $m_1$ and $n_1$ are both even), we have: $$2^{k_1-1}\mid (n_1-m_1)\text{or}{2}^{k_1-1}\mid (n_1+m_1).$$ This leads to the lower bound: $$\begin{array}{rl}{n}_1& \ge 2^{k_1-1}-m_1=\frac{1}{2}(m_1^2+r_1)-m_1\\ & =\frac{3m_1^2}{7}+\left(\frac{m_1^2}{14}+\frac{r_1}{2}-m_1\right)\\ & \ge \frac{3m_1^2}{r_1}+\left(\frac{m_1^2}{14}+\frac{7}{2}-m_1\right)(\text{since }{r}_1\ge 7)\\ & =\frac{3m_1^2}{r_1}+\frac{1}{14}(m_1-7{)}^2\\ & \ge \frac{3m_1^2}{r_1}=\frac{3m^2}{r}.\end{array}$$

$$n=2^t{n}_1\ge n_1\ge \frac{3m^2}{r}>\frac{2m^2}{r},$$