\(x.y^2+2xy+x=128y\)
\(x(y^{2}+2y+1)=128y\)
\(x(y+1)^{2}=128y\)
\(x=\frac{128y}{(y+1)^{2}}\)
Để

là số nguyên,

phải chia hết cho

.
Vì

và

là hai số nguyên liên tiếp, chúng luôn nguyên tố cùng nhau, nên

và

cũng nguyên tố cùng nhau.
Do đó, để

thì

phải là ước của

.
Các ước chính phương của

(

) là:

.

• Nếu (y+1)2=1open paren y plus 1 close paren squared equals 1(𝑦+1)2=1:
• • y+1=1⇒y=0⇒x=128⋅01=0y plus 1 equals 1 implies y equals 0 implies x equals the fraction with numerator 128 center dot 0 and denominator 1 end-fraction equals 0𝑦+1=1⇒𝑦=0⇒𝑥=128⋅01=𝟎
  • y+1=-1⇒y=-2⇒x=128⋅(-2)1=-256y plus 1 equals negative 1 implies y equals negative 2 implies x equals the fraction with numerator 128 center dot open paren negative 2 close paren and denominator 1 end-fraction equals negative 256𝑦+1=−1⇒𝑦=−2⇒𝑥=128⋅(−2)1=−𝟐𝟓𝟔
• Nếu (y+1)2=4open paren y plus 1 close paren squared equals 4(𝑦+1)2=4:
• • y+1=2⇒y=1⇒x=128⋅14=32y plus 1 equals 2 implies y equals 1 implies x equals the fraction with numerator 128 center dot 1 and denominator 4 end-fraction equals 32𝑦+1=2⇒𝑦=1⇒𝑥=128⋅14=𝟑𝟐
  • y+1=-2⇒y=-3⇒x=128⋅(-3)4=-96y plus 1 equals negative 2 implies y equals negative 3 implies x equals the fraction with numerator 128 center dot open paren negative 3 close paren and denominator 4 end-fraction equals negative 96𝑦+1=−2⇒𝑦=−3⇒𝑥=128⋅(−3)4=−𝟗𝟔
• Nếu (y+1)2=16open paren y plus 1 close paren squared equals 16(𝑦+1)2=16:
• • y+1=4⇒y=3⇒x=128⋅316=24y plus 1 equals 4 implies y equals 3 implies x equals the fraction with numerator 128 center dot 3 and denominator 16 end-fraction equals 24𝑦+1=4⇒𝑦=3⇒𝑥=128⋅316=𝟐𝟒
  • y+1=-4⇒y=-5⇒x=128⋅(-5)16=-40y plus 1 equals negative 4 implies y equals negative 5 implies x equals the fraction with numerator 128 center dot open paren negative 5 close paren and denominator 16 end-fraction equals negative 40𝑦+1=−4⇒𝑦=−5⇒𝑥=128⋅(−5)16=−𝟒𝟎
• Nếu (y+1)2=64open paren y plus 1 close paren squared equals 64(𝑦+1)2=64:
• • y+1=8⇒y=7⇒x=128⋅764=14y plus 1 equals 8 implies y equals 7 implies x equals the fraction with numerator 128 center dot 7 and denominator 64 end-fraction equals 14𝑦+1=8⇒𝑦=7⇒𝑥=128⋅764=𝟏𝟒
  • y+1=-8⇒y=-9⇒x=128⋅(-9)64=-18y plus 1 equals negative 8 implies y equals negative 9 implies x equals the fraction with numerator 128 center dot open paren negative 9 close paren and denominator 64 end-fraction equals negative 18𝑦+1=−8⇒𝑦=−9⇒𝑥=128⋅(−9)64=−𝟏𝟖 

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