Chứng minh rằng không thể có 2 số nguyên m,n để đẳng thức sau thỏa mãn 2m2+n2=2023

$$\eqalign{ & 2{m^2} + {n^2} = 2023 \cr & {A^2}\,\,co\,\,chu\,\,so\,\,\tan \,\,cung\,\,la\,\,\,0,{\rm{ }}1,{\rm{ }}4,{\rm{ }}5,{\rm{ }}6,{\rm{ }}9. \cr & \Rightarrow 2{A^2}\,\,co\,\,chu\,\,so\,\,\tan \,\,cung\,\,la\,\,0;\,\,2;\,\,8 \cr & \Rightarrow {m^2};\,\,{n^2}\,\,deu\,\,phai\,\,co\,\,chu\,\,so\,\,\tan \,\,cung\,\,la\,\,1 \cr & \Rightarrow m,\,\,n\,\,\tan \,\,cung\,\,la\,\,1\,\,hoac\,\,9 \cr & 2{m^2} + {n^2} = 2023 \cr & \Rightarrow 2{m^2} < 2023 \cr & \Leftrightarrow {m^2} < {{2023} \over 2} \Leftrightarrow m \le 31 \cr & m\,\,\tan \,\,cung\,\,la\,\,1\,\,hoac\,\,9 \Rightarrow m \in \left\{ {1;9;11;19;21;29;31} \right\} \cr & m = 1 \Rightarrow {n^2} = 2021\,\,\left( {loai} \right) \cr & m = 9 \Rightarrow {n^2} = 1861\,\,\left( {loai} \right) \cr & m = 11 \Rightarrow {n^2} = 1781\,\,\left( {loai} \right) \cr & m = 19 \Rightarrow {n^2} = 1301\,\,\left( {loai} \right) \cr & m = 21 \Rightarrow {n^2} = 1141\,\,\left( {loai} \right) \cr & m = 29 \Rightarrow {n^2} = 341\,\left( {loai} \right) \cr & m = 31 \Rightarrow {n^2} = 101\,\,\left( {loai} \right) \cr & \Rightarrow DPCM \cr} $$