IMO Problems

(1) x^3 = y^2 + 1
==> y^2 = x^3 - 1 = (x - 1)(x^2 + x + 1)
As y is a positive integer, so
x^2 + x + 1 = (x - 1)n^2 ... (where n is a positive rational number)
==> x^2 + (1 - n^2)x + (1 + n^2) = 0
As x is also a positive integer, so the discriminant is a perfect square, k^2, say.
(1 - n^2)^2 - 4(1 + n^2) = k^2
==> n^4 - 6n^2 - 3 = k^2
==> (n^2 - 3)^2 - k^2 = 12
==> (n^2 - 3 + k, n^2 - 3 - k) = (12, 1) or (6, 2) or (4, 3)
==> (n^2, k) = (9.5, 5.5) or (7, 2) or (6.5, 0.5)
As the square root of 9.5, 7 and 6.5 are not a rational number,
so there is no positive integer solution (x, y) of the question.

(2) x^2 = y^3 + 1
==> x^2 = (y + 1)(y^2 - y + 1)
Same as the previous question, so
y^2 - y + 1 = (y + 1)n^2 ... (where n is a positive rational number)
==> y^2 - (1 + n^2)y + (1 - n^2) = 0
As y is also a positive integer, so the discriminant is a perfect square, k^2, say.
(1 + n^2)^2 - 4(1 - n^2) = k^2
==> n^4 + 6n^2 - 3 = k^2
==> (n^2 + 3)^2 - k^2 = 12
==>( n^2 + 3 + k, n^2 + 3 - k) = (12, 1) or (6, 2) or (4, 3)
==> (n^2, k) = (3.5, 5.5) or (1, 2) or (0.5, 0.5)
The square root of 3.5 and 0.5 are not a rational number,
and as n is a positive rational number, therefore n = 1 and k = 2, and,
y = (1 + n^2 + k)/2 or (1 + n^2 - k)/2
so y = 2 or y = 0 (rejected as y is a positive integer), and when y = 2, then x = 3.
Therefore, the positive integer solution (x, y) is (3, 2).

(1) x^3 = y^2 + 1

y^2 = x^3 - 1 = (x - 1)(x^2 + x + 1)

So, there are 3 cases

Case 1: x - 1 = 0, then x = 1, y = 0 is a solution

Case 2: x^2 + x + 1 = 0. No positive integers solutions x and y

Case 3: x - 1 = x^2 + x + 1

x^2 + 2 = 0

No positive integers solutions x and y

So, the only solution is (1,0)

(2) x^2 = y^2 + 1 = (y + 1)(y^2 - y + 1)

So, there are 3 cases

Case 1: y + 1 = 0, rejected as y = -1 is not positive

Case 2: y^2 - y + 1 = 0. No positive integers solutions x and y

Case 3: y + 1 = y^2 - y + 1

y^2 - 2y = 0 => y = 0 or 2

So, there are two solutions (1,0) and (3,2)

(1) x=1 ,y=0
(2) x=1 ,y=0