Question

已知a^5-b^4=0,c^3-d^2=0,c-a=19,其中a,b,c,d均为正整数,则b-d=?

Answer 1

a^5 = b^4

所以a = x^4,b = x^5 x 是整数

c^3 = d^2

所以c = y^2,d = y^3 y 是整数

c-a = y^2-x^4 = 19

(y+x^2)(y-x^2) = 19

由于19是质数

所以y+x^2 = 19,y-x^2 = 1

解得 y = 10,x = 3

b-d = (x^5)-(y^3) = 243-1000 = -757

我猜你打错了 应该是问d-b吧 ..