IRMO 1994

[geda]

Siano $ x $ e $ y $ due interi positivi con $ y>3 $ e

$ x^2+y^4=2[(x-6)^2 + (y+1)^2] $.

Dimostrare che $ x^2+y^4=1994 $.

[jordan]

Da qui.

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. $

A ugly exercise..
It is a quadratic equation in x with discriminant $ \Delta: = y^4 - 2(y + 1)^2 + 72 $, but for all y > 5 we have $ (y^2 - 2)^2 < \Delta < y^4 $; now $ \Delta = (y^2 - 1)^2 $ has no integral solution. So $ y \in \{4,5\} $, and it is only computational to verify that the unique solution is (x,y) = (37,5).