Q12/118) Given 4 positive integers a,b,c and d such that a^5=b^4, c^3=d^2 and c−a=19, what is d−b?

as c^3 = d^2 so c will be a square let c = x^2

as a^5 = b^4 so a = y^4 for some y

now
c-a = 19
=> x^2 - y^4 = 19
=> (x-y^2)(x+y^2) = 19
hence x - y^2 =1 and x+y^2 = 19 as 19 is a prime
so x = 10 and y = 3

so a = y^4 or a^5 = y^20 = b^ 4 or b= y^5 = 243

c= x^2 => c= 100 and hence d^2 = 10^6 and so d = 1000

d-b = 1000 - 243 = 757