Given that a,b,c,d are positive integers such that b^3/2=a and d^5/4 =c. If b-d= 9 then find the value of a-c

• From hit n trial method
  we get
  b=25 and d=16
  ((5)^(2*3/2))=a
  a=125
  ((2)^(4*5/4))=c
  c=32
  answer 125-32=93
• 12 years agoHelpfull: Yes(13) No(0)
• b=a^2/3
  d=c^4/5
  now b-d=9
  a^2/3-c^4/5=9
  (a^1/3+c^2/3)(a^1/3-c^2/3)=3*3 ---------by (x^2-y^2=(x+y)(x-y))
  so a^1/3+c^2/3=3
  and a^1/3-c^2/3=3
  add both expression
  then 2a^1/3=6
  a^1/3=3
  a=27
  putting this value in upper expression we get c=0
  so answer is a-c=27
  here u can see that value of a=27,b=9,c=0,d=0
  
• 12 years agoHelpfull: Yes(2) No(3)
• NO such value is possible with satisfying a,b,c,d to be INTEGERS.
  
  Because, given b-d = 9 which is an odd Integer.
  So one of b,d must be odd and one even Integer. [like (11,2),(13,4).,]
  [since difference between 2 odd or 2 even integers is an even Integer]
  Multiplication of two odd integers results an odd integer.so if 'a' is an odd
  Integer then a^b is an odd Integer.
  So one of b^3 , d^5 is an odd integer.
  
  But according to a,c must be integers,
  b^3 must be divisible by '2'and d^5 must be divisible by '4'.
  One of them is impossible BECAUSE 2 or multiple of 2 cant divide an odd number perfectly.
  
  SO THERE IS NO ANSWER FOR THIS QUESTION.
• 12 years agoHelpfull: Yes(1) No(3)