Question

Tỉ̀m các cặp số nguyên x , y thỏa mãn : x² + x +3 bằng y² .

giúp mik luôn nha .

Answer 1

\(\left\{{}\begin{matrix}x^2+x+3=\dfrac{1}{4}\left(2x+1\right)^2+\dfrac{11}{4}\\\left(2y\right)^2-\left(2x+1\right)^2=11\end{matrix}\right.\)

\(\left\{{}\begin{matrix}y=\pm3\\\left\{{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\end{matrix}\right.\) thử lại ok !

Answer 2

different way

ta có: \(x^2< x^2+x+3< x^2+4x+4\)

\(\Rightarrow x^2< y^2< \left(x+2\right)^2\)

x,y nguyên => y²=(x+1)²

<=> x²+x+3=x²+2x+1

<=> x=2

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