x3次方-y3次方=XY+61

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xy均為正整數
x^3 – y^3 = xy + 61 > 61
x > y, x-y > 0

x^3 – y^3 = (x-y)^3 + 3xy(x-y)
(x-y)^3 = xy – 3xy(x-y) + 61
(x-y)^3 = xy(1-3(x-y)) + 61 > 0
1-3(x-y) < 0

So, (x-y)^3 < 61
(x-y)^3 = 1, 8, 27
x-y = 1, 2, 3

x-y=1, xy(1-3) + 61 = 1, xy = 30, x=6, y=5

x-y=2, xy(1-6) + 61 = 8, xy = 53/3 (X)

x-y=3, xy(1-9) + 61 = 27, xy=17/4 (X)

So, x=6, y=5

x^3-y^3=xy+61,x,y=正整數=?Ans:F(x,y)=y^3+xy+61如果y=1,則x^3=x+62 => x>5F(5,1)=58, F(5,3)=22, F(5,4)=-20F(6,1)=148, F(6,4)=67, F(6,5)=0=> (x,y)=(6,5)......ans


2014-06-05 15:23:36 補充：
補充:

令x=y+a, a>=1正整數

x^3=y^3+3ay^2+3ya^2+a^3

xy=y(y+a)=y^2+ay

代入原式裡面:

y^3+3ay^2+3ya^2+a^3=y^3+(y^2+ay)+61

0=(3a-1)y^2+a(3a-1)y-(61-a^3).....(1)

判別式為:

D(a)=[a(3a-1)]^2+4(3a-1)(61-a^3)

=(3a-1)[(3a-1)a^2+4(61-a^3)]

=(3a-1)(244-a^3-a^2)

=b^2

2014-06-05 15:24:14 補充：
又D(a)=(3a-1)(244-a^3-a^2)

=-(3a-1)(a-5.93)(a^2+6.93a+41.13)>0

=> (3a-1)(a-5.93)<0

=> 6>a>=1......(2)

D(1)=2*(244-2)

=2*242

=2*2*121

=(2*11)^2

=b^2

b=2*11

a=1代入(1):

2014-06-05 15:24:33 補充：
0=2y^2+2y-60

=y^2+y-30

=(y-5)(y+6)

y=5 => x=6.....ans

Eq.(2)顯示答答只有1ㄍ

2014-06-05 15:25:47 補充：
答答 改為 答案