anonymous
  • anonymous
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?
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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ganeshie8
  • ganeshie8
i wud start somthing like below :- x^2-y^2=z^3 (x+y)(x-y) = z^3
ganeshie8
  • ganeshie8
Since you want the smallest possible number, lets look for powers of 2
ganeshie8
  • ganeshie8
x+y = 2^n x-y = 2^n-3

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anonymous
  • anonymous
what do i do after this?
ganeshie8
  • ganeshie8
it should be like this actually x+y = 2^n x-y = 2^k n+k = 3m
anonymous
  • anonymous
so thats the equation? what about after that?
ganeshie8
  • ganeshie8
all are natural numbers, next let me think
ganeshie8
  • ganeshie8
n+k = 3m when m=1, n=2, k=1 will give , x=3 and y = 1.
ganeshie8
  • ganeshie8
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
ganeshie8
  • ganeshie8
move to m=2
ganeshie8
  • ganeshie8
n+k = 6 since n > k, lets start wid n = 4, k = 2 that gives x = 10, y = 6
ganeshie8
  • ganeshie8
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
ganeshie8
  • ganeshie8
see if that makes some sense,
anonymous
  • anonymous
yes it does thanks :D
ganeshie8
  • ganeshie8
cool :)

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