diophantine equation?

Question

find integers y and x that satisfy y^3=x^2.

between 100 and 1000

Answer 1

y^3 = x^2

Take the square root of both sides:
x = y*sqrt(y)

For sqrt(y) to be an integer, y must be a square,
so let y = z^2.

By substitution, we then have the parametric equations:

x = z^3
y = z^2

Now we can substitute any integer for z
and get an infinity of answers:

z = 1 implies x = 1, y = 1
z = 2 implies x = 8, y = 4
z = 3 implies x = 27, y = 9
etc.

If x and y are to be between 100 and 1000 inclusive, then
z^3 (= x) <= 1000, so z <= 10,
and z^2 (= y) >= 100, so z >= 10.
In other words, z can only be 10.

Thus, y = 100 and x = 1000 giving :
y^3 = x^2 is equivalent to 100^3 = 1000^2.

Answer 2

This will have an integer solution for any integer that is an even power of 5.

1^5 = 1
(x,y) = (1^3,1^2) = (1,1)
1*1 = 1

2^5 = 32
(x,y) = (2^3,2^2) = (8,4)
8*4 = 32

3^5 = 243
(x,y) = (3^3,3^2) = (27,9)
27*9 = 243

etc.

Answer 3

1^3 = 1^2.

Answer 4

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