Problem 1 of the Irish Mathematical Olympiad 1994

Let x, y be positive integers with y > 3 and x^2 + y^4 = 2[(x - 6)^2 +(y + 1)^2].

Prove that x^2 + y^4 = 1994.

Solution by Steve Dinh, a.k.a. Vo Duc Dien

Expand and combine the terms; we have

x^2 - 24x - y^4 + 2y^2 + 4y + 74 = 0 Solving for x gives x_{1,2} = 12 \sqrt{y{^4} - 2y^2 - 4y + 70} (*)

Therefore, y^4 - 2y^2 - 4y + 70 has to be a square of an integer and one possible scenario is that 2y^2 + 4y - 70 = 0 for the term inside the square root to equal y^4, and the positive value for y is y = 5. Now substitute y = 5 into (*), the positive value for x is x = 37.

Hence, 37^2 + 5^4 = 1994.

Note: If y is not bounded by the condition y > 3, we have y = 3 which makes x = 1 or x = 23. However, one has yet to show that the expression y^4 - 2y^2 - 4y + 70 is not a perfect square for other values of y.