[tex]n=2k+1\\n^2+2023=(2k+1)^2+2023=4k^2+4k+1+2023=4k^2+4k+2024=8(\frac12k^2+\frac12k+253)\\\\\frac{n^2+2023}8=\frac{(2k+1)^2+2023}8=\frac{4k^2+4k+2024}8=\frac{8(\frac12k^2+\frac12k+253)}8=\frac12k^2+\frac12k+253[/tex]
