tìm số nguyên`x;y` thỏa mãn : `x^5+y^2=xy^2+1`

Đáp án:

$ \left[ \begin{array}{l}x=1, y \in \mathbb{Z} \\ (x;y) = \{(0;\pm1),(-1;\pm1),(3;\pm11)\end{array} \right.$

Giải thích các bước giải:

$x^5+y^2=xy^2+1$

$\Leftrightarrow x^5-1= y^2.(x-1)$

+ Nếu $x=1 \Rightarrow y \in \mathbb{Z} (\text{TM})$

+ Nếu $x \ne 1$

$\to y^2= x^5+ x^3 + x^2 + x+1$

$\Leftrightarrow (2y)^2 = (2x^2+x)^2 + 2x^2 + (x+2)^2 > (2x^2 + x)^2$

Mặt khác : 

$(2x^2 +x+2)^2 – 4.(x^4 +x^3+x^2+x+1)$

$= 4x^4 + 4x^3 +9x^2 + 4x +4-4(x^4 +x^3+x^2+x+1)$

$= 5x^2 \ge 0$

$\Rightarrow (2x^2 +x)^2 < (2y)^2 \le (2x^2+x+2)^2$

$° (2y)^2 =(2x^2 +x+2)^2$

$\Rightarrow x =0 \to y=x$ hoặc $y=-x$

$° (2y^2)= (2x^2 + x+1)^2$

$\Leftrightarrow 4.(x^4 +x^3 +x^2+x+1)=4x^4 +4x^3-5x^2 +2x+1$

$\to x^2 – 2x-3=0$

$\Leftrightarrow  \left[ \begin{array}{l}x=-1 \to y=\pm 1\\x=3 \to y= \pm 11 \end{array} \right.$

Vậy $ \left[ \begin{array}{l}x=1, y \in \mathbb{Z} \\ (x;y) = \{(0;\pm1),(-1;\pm1),(3;\pm11)\end{array} \right.$