國中數學,不定方程式求正整數解

1)設勾 = a , 股 = b , 則 a <= b

周長值 = 面積值

a + b + √(a^2 + b^2) = ab/2

2√(a^2 + b^2) = ab - 2(a+b)

4(a^2 + b^2) = (ab)^2 - 4ab(a+b) + 4(a+b)^2

4a^2 + 4b^2 = (ab)^2 - 4ab(a+b) + 4a^2 + 4b^2 + 8ab

0 = (ab)^2 - 4ab(a+b) + 8ab

ab - 4(a+b) + 8 = 0

ab - 4(a+b) + 16 = 8

(a - 4)(b - 4) = 8

(a - 4)(b - 4) = 1*8 或 2*4

由a < = b 得 :

(a - 4) = 1 , (b - 4) = 8 -----> a = 5 , b = 12

得直角△三邊長 ( 5 , 12 , √(5^2 + 12^2) = 13 )
或
(a - 4) = 2 , (b - 4) = 4 -----> a = 6 , b = 8

得直角△三邊長 ( 6 ,8 , √(6^2 + 8^2) = 10 )



2)不妨設x<=y<=z :

xyz+xy+yz+xz+x+y+z=1989

x(yz+y+z+1) + (yz+y+z) = 1989

x(yz+y+z+1) + (yz+y+z+1) = 1989+1

(x+1)(yz+y+z+1) = 1990

(x+1)(y+1)(z+1) = 1990 = 2*5*199

由 x<=y<=z 得:

x+1 = 2--------->x = 1
y+1 = 5--------->y = 4
z+1 = 199----->z = 198

若x,y,z的最小值不小於3,則方程無解。




3) x^2 + y^2 = 328

不妨先設x <= y ,

則 x^2 <= 328/2 = 164

x <= 12.8...

x <= 12 (由於x是正整數)

又平方數尾只能是 0, 1, 4, 5, 6, 9,

x^2 及 y^2 必同為4尾或9尾,

因此對 x 試驗 2 , 3 , 7 , 8 , 12 即可:

只有當 x = 2 時,得 2^2 + 18^2 = 328

因此 (x , y) 共兩組解 (2,18) , (18,2)