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

Question

已知a^5-b^3=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 ..