\(\Leftrightarrow x^2-1=2y^2\)

\(\Leftrightarrow x^2=2y^2+1\)

Dễ nhận thấy: \(2y^2+1\) lẻ suy ra x lẻ suy ra x=2k+1

\(\Rightarrow\left(2k+1\right)^2=2y^2+1\)

\(\Rightarrow4k^2+4k+1=2y^2+1\)

\(\Leftrightarrow4k^2+4k=2y^2\Leftrightarrow2\left(k^2+k\right)=y^2\)

Suy ra y chan. Suy ra y=2

Thay vao suy ra x=3

Ta có: \(y^2 + 2(x^2 + 1) = 2y(x + 1)\)

=>\(y^2+2x^2+2=2xy+2y\)

=>\(y^2-2(x+1)y+2x^2+2=0\quad(1)\)

Δ=\(\left\lbrack2\left(x+1\right)\right\rbrack^2-4\cdot1\cdot\left(2x^2+2\right)\)

\(=4\left(x^2+2x+1\right)-4\left(2x^2+2\right)=4\left(x^2+2x+1-2x^2-2\right)=4\left(-x^2+2x-1\right)=-4\left(x-1\right)^2\)

Để (1) có nghiệm thì Δ>=0

=>\(-4\left(x-1\right)^2\ge0\)

=>\(\left(x-1\right)^2\le0\)

=>x-1=0

=>x=1(nhận)

(1) sẽ trở thành: \(y^2-2\left(1+1\right)y+2\cdot1^2+2=0\)

=>\(y^2-4y+4=0\)

=>\(\left(y-2\right)^2=0\)

=>y-2=0

=>y=2(nhận)

Vậy: (x;y)∈{(1;2)}

Ta có: \(y^2 + 2(x^2 + 1) = 2y(x + 1)\)

=>\(y^2+2x^2+2=2xy+2y\)

=>\(y^2-2(x+1)y+2x^2+2=0\quad(1)\)

Δ=\(\left\lbrack2\left(x+1\right)\right\rbrack^2-4\cdot1\cdot\left(2x^2+2\right)\)

\(=4\left(x^2+2x+1\right)-4\left(2x^2+2\right)=4\left(x^2+2x+1-2x^2-2\right)=4\left(-x^2+2x-1\right)=-4\left(x-1\right)^2\)

Để (1) có nghiệm thì Δ>=0

=>\(-4\left(x-1\right)^2\ge0\)

=>\(\left(x-1\right)^2\le0\)

=>x-1=0

=>x=1(nhận)

(1) sẽ trở thành: \(y^2-2\left(1+1\right)y+2\cdot1^2+2=0\)

=>\(y^2-4y+4=0\)

=>\(\left(y-2\right)^2=0\)

=>y-2=0

=>y=2(nhận)

Vậy: (x;y)∈{(1;2)}

Ta có bảng :

x-2             17             1

x                19             3

2y+1           1              17

y                1               8

Vậy (x;y) = (19;1) ; (x;y) =( 3 ;8 )