Question

a^5=b^6,c^3=d^4, and d-a=61,find the value of c-b

Answer 1

all are natural integers ? is it given ?

Answer 2

natural nos *

Answer 3

positive intergers

Answer 4

yes

Answer 5

using your formula for a and d (jonas0

if we let x = (b/c and y = (a.b)
we get (x - y^2)(x^2 + xy^2 + y^4) -61 = 0
solving this might help

Answer 6

is it wrong to say
\[64-3=61\]
d=64,a=3

Answer 7

is that just a guess?

Answer 8

yes

Answer 9

I mean positive integer.

Answer 10

so can we guess until b is positive int

Answer 11

@jonas - you deleted those values for a and b - have i got them right

Answer 12

I guess yes.... But guessing is not a nice method.

Answer 13

\[a=(\frac{ a }{ b })^6,d=(\frac{ c }{ d })^3\]

Answer 14

solution of that equation in x and y is

x = 5 , y = -2, or 2

Answer 15

a/b = 5, c/d = 2 or -2

Answer 16

a = 5b, c = 2d or c = -2d

- not sure how to go forward from here....

Answer 17

Did u get 593 @cwrw238

Answer 18

trial and error would be tedious i guess

Answer 19

593?

Answer 20

oh for c - b

no i've only got as far as the last relations i posted