Question

find two natural numbers, such that the difference of their squares is a cube and the difference of their cubes is a square. What is the answer in the smallest possible numbers?

Answer 1

i wud start somthing like below :-

x^2-y^2=z^3
(x+y)(x-y) = z^3

Answer 2

Since you want the smallest possible number, lets look for powers of 2

Answer 3

x+y = 2^n
x-y = 2^n-3

Answer 4

what do i do after this?

Answer 5

it should be like this actually
x+y = 2^n
x-y = 2^k
n+k = 3m

Answer 6

so thats the equation? what about after that?

Answer 7

all are natural numbers, next let me think

Answer 8

n+k = 3m

when m=1,
n=2, k=1 will give , x=3 and y = 1.

Answer 9

see if they satisfy the second constraint, which is, x^3-y^3 must be a square
but its not a square, so discard m=1

Answer 10

move to m=2

Answer 11

n+k = 6
since n > k, lets start wid n = 4, k = 2
that gives x = 10, y = 6

Answer 12

see if they satisfy the second constraint, which is, x^3-y^3 must be a square
10^3-6^3 is a perfect square !! so your required minimum natural numbers satisfying both constraints are 10 and 6

Answer 13

see if that makes some sense,

Answer 14

yes it does thanks :D

Answer 15

cool :)