求整數解的數對

2010-03-02 7:10 pm
對所有實數x,y,恒使x^2+y^2-2m(x-2)-2n(y-3)≧0成立,則(m,n)的整數解有幾組

回答 (3)

2010-03-02 10:53 pm
✔ 最佳答案




對所有實數x,y,恒使x^2+y^2-2m(x-2)-2n(y-3)≧0成立,則(m,n)

的整數解有幾組

Sol

x^2+y^2-2m(x-2)-2n(y-3)≧0

x^2+x(-2m)+[y^2+ 4m-2ny+6n]>=0

D1/4=m^2-(y^2+ 4m-2ny+6n)<= 0

m ^2-y^2-4m+2ny-6n<=0

y^2-2ny+(-m^2+ 4m +6n)>=0

D2/4=n^2-(-m^2+ 4m +6n)<=0

n^2+m^2-4m-6n<= 0

m ^2-4m+4+n^2-6n+9<=13

(m-2)^2+(n-3)^2<=13

Set p=m-2,q=n-3

p^2+q^2<=13

(1) p=0 =>q^2<=13 =>7組

(2) p=+/-1 =>q^2<=12 =>2*7=14組

(3) p=+/-2 =>q^2<=9 =>2*7=14組

(3) p=+/-3 =>q^2<=4 =>2*5=10組

7+14+14+10=45組





2010-03-02 10:56 pm
x^2+y^2-2m(x-2)-2n(y-3)≧0
x^2+y^2-2mx+4m-2ny+6n≧0
(x-m)^2+(y-n)^2+4m+6n-m^2-n^2≧0
(x-m)^2+(y-n)^2-(m-2)^2-(n-3)^2+13≧0

So,13≧(m-2)^2+(n-3)^2
if n=0,m=0,1,2,3,4
if n=1,m=-1,0,1,2,3,4,5
if n=2,m=-1,0,1,2,3,4,5
if n=3,m=-1,0,1,2,3,4,5
if n=4,m=-1,0,1,2,3,4,5
if n=5,m=-1,0,1,2,3,4,5
if n=6,m=0,1,2,3,4

(m,n)的整數解有45組
2010-03-02 7:56 pm
(m,n)的整數解有45組


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