\(25-y^2=8\left(x-2021\right)^2\)
=>\(8\left(x-2021\right)^2\le25\)
=>\(\left(x-2021\right)^2\le\frac{25}{8}\)
=>\(0\le\left(x-2021\right)^2\le\frac{25}{8}\)
mà x-2021 nguyên
nên \(\left(x-2021\right)^2\in\left\lbrace0;1\right\rbrace\)
TH1: \(\left(x-2021\right)^2=0\)
=>x-2021=0
=>x=2021
\(25-y^2=8\left(x-2021\right)^2\)
=>\(25-y^2=8\cdot0=0\)
=>\(y^2=25\)
mà y>0
nên y=5(nhận)
TH2: \(\left(x-2021\right)^2=1\)
=>\(\left[\begin{array}{l}x-2021=1\\ x-2021=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2022\left(nhận\right)\\ x=2020\left(nhận\right)\end{array}\right.\)
Ta có: \(25-y^2=8\cdot\left(x-2021\right)^2\)
=>\(25-y^2=8\cdot1=8\)
=>\(y^2=25-8=17\)
mà y nguyên
nên y∈∅
vậy: x=2021; y=5
