I have not done an olympaid problem since last summer. Here's a geometric approach.
Solution 440Let's look at the curve
y−x2−nx−m=0. We want to intersect the line
y=0 twice in
Z2.
To this end, we consider its discriminant -
Δ=n2−4m. Naturally, we want to complete the square, so we choose
m=kn−k2, and thus we arrive at
Δ=n2−4kn+4k2=(n−2k)2.
We now need a definition. A path consists of consecutive horizontal and vertical moves in a point lattice. In our case the points are at a unit distance.
We next turn to the concrete problem. Our initial curve is
y−x2−x−2014. The person who is supposed to have a winning strategy wants to get an equation with coefficients satisfying the above relation. In other words, we need a path starting from
(1,2014) and eventually, in particular after finitely many steps, intersecting a line of the form
y−kx−k2=0 in a lattice point.
It is clear that for
k=1 there is no winning strategy (by symmetry the same holds for
k=−1). For
∣k∣=1 every path
(1,2014)↦(2,2014)↦⋯↦(kn,ln−1) will eventually intersect this line since the gradient is not
±1 (this sequence of moves works when
k is positive).
Hobbes moves always away from the line since if after his move the path intersects the line we are done. So suppose that our path intersects the line after Calvin's move and the point of intersection is not a lattice point.
Our goal is to choose
k so that after Hobbes' move, Calvin wins or needs to move in order to win. Obviously, this is possible only when
∣k∣=2. Indeed, let for example
k=2; trivial algebraic computation shows that if Hobbes moves upwards, then either we are done, or we need a move to the right; the dual statement holds if he moves downwards.
Hence, if we set
k=±2, it is always possible to get a path intersecting the line in a lattice point.