Toán nâng cao

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chaudoublelift

Giải

Tìm giá trị x,y nguyên dương sao cho : x2−y2+2x−4y−10=0x^2 - y^2+ 2x - 4y-10=0(1)
Giải:
Ta có:
(1)⇔x2−y2+2x−4y−10=0⇔(x2+2x+1)−(y2+4y+4)=7(1)⇔x^2 - y^2+ 2x - 4y-10=0⇔(x^2+2x+1)-(y^2+4y+4)=7
⇔(x+1)2−(y+2)2=7⇔(x+1−y−2)(x+1+y+2)=7⇔(x−y−1)(x+y+3)=7⇔(x+1)^2-(y+2)^2=7⇔(x+1-y-2)(x+1+y+2)=7⇔(x-y-1)(x+y+3)=7
Do x,y∈Z⇒x−y−1,x+y+3∈Zx,y\in Z⇒x-y-1,x+y+3\in Z nên (x−y−1)(x+y+3)=7=(−1)(−7)=1.7(x-y-1)(x+y+3)=7=(-1)(-7)=1.7
Ta có bảng sau:
x−y−1∣−7−117x+y+3∣−1−771x∣−5−533y∣1−54−5 \begin{matrix} x-y-1&|&-7&-1&1&7\\ x+y+3&|&-1&-7&7&1\\x&|&-5&-5&3&3\\ y&|&1&-5&4&-5\end{matrix}
Vậy (x,y)=(−5,1);(−5,−5);(3,4);(3,−5)(x,y)=(-5,1);(-5,-5);(3,4);(3,-5) thỏa mãn PT(1)
 
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