3. Parity

If we work with real numbers then the equation is equivalent to $y=(11-2x)/6$. Thus, for every real number $t$ we have a solution $(x,y)=(t,(11-2t)/6)$, showing that there are infinitely many solutions.

However, there are no solutions at all if we required that $x$ and $y$ are integers. Indeed, if $x$ and $y$ are integers then $2x+6y$ is an even integer, and so cannot be equal to $11$.

By comparison with the previous example, we might be tempted to say yes: both the left and right hand sides are even, so there is no problem with parity. However, just because there is no problem with parity, we cannot conclude that there are no other problems. In fact, the left hand side is always divisible by $3$, but the right hand side is not, so in fact there are no integer solutions.

Suppose that $n$ is even, say $n=2m$. Then we can cover each row of the board with $m$ non-overlapping horizontal dominos, and thus cover the whole board with $2m^2$ non-overlapping dominos. The case where $n=6$ and $m=3$ is like this:

Now suppose that $n$ is odd. We could try to cover the board like this:

This fails, because there is one extra square that we have not covered. However, this does not really prove anything. Our first unimaginative approach has failed, but perhaps that is just because we were not clever enough; perhaps there is a different approach of stupendous complexity and cunning that will cover the whole board? In fact that is not the case, but we need a proper proof to explain why. Fortunately, that is not very difficult. We are assuming that $n$ is odd. There are $n^2$ squares in total, and this number is also odd. On the other hand, each domino covers two squares, so any set of $m$ non-overlapping dominos covers $2m$ squares, and this number is even. Thus, it is impossible for a set of non-overlapping dominos to cover the whole board.

Now suppose we consider a nicked $n\times n$ board (where $n\ge 2$), with two opposite corners removed. The total number of squares is $n^2-2$. If $n$ is odd then $n^2-2$ is again odd, so the nicked board still cannot be covered by disjoint dominos. Suppose instead that $n$ is even, say $n=2m$. Then the nicked board has $4m^2-2=2(2m^2-1)$ squares, and this number is even. We might be tempted to deduce that the board can be covered by disjoint dominos, but that would be too hasty. Suppose that we colour the squares in the usual chessboard pattern. The full $(2m)\times (2m)$ board will then have $2m^2$ white squares and $2m^2$ black squares. The two removed corners can be joined by a diagonal line, and all the squares on that line will have the same colour, so in particular, the two removed corners will have the same colour. Suppose for the sake of example that they are both white. This nicked board will then have $2m^2-2$ white squares and $2m^2$ black squares. However, every domino will cover one white square and one black square. Thus, any set of disjoint dominos will cover the same number of black squares as white squares. As the nicked board has more black squares than white squares, it cannot be covered by a disjoint set of dominos.

We can imagine placing the tiles on a grid that is coloured like a chessboard. It is not hard to see that however you place the first six tiles, each of them will cover two white squares and two black squares. (One way to see this is to note that each of the first six tiles can be divided into two non-overlapping dominos.) That makes $12$ white squares and $12$ black squares altogether. However, the last T-shaped tile is different: however you place it, it will cover either three black squares and one white square, or three white squares and one black square. Thus, the full set of seven tiles will cover either $15$ black and $13$ white, or $13$ black and $15$ white. However, any $4\times 7$ rectangle will cover $14$ squares of each colour. This proves that the seven tiles cannot cover such a rectangle.

Euler used a parity argument to show that no such tour is possible. Indeed, there are four sections of land in the city: the north bank (N), the south bank (S), the western island (W) and the eastern island (E). We will call these nodes. Nodes N, S and E are each connected to three bridges, and W is connected to five bridges: in each case, the number is odd. Now suppose we have a tour of the city that crosses each bridge precisely once. Let $P$ be a node that is neither the beginning nor the end of the tour. Every time we reach $P$ on one bridge, we must leave on a different bridge. Thus, if we visit $P$ a total of $n$ times, then we will have crossed $2n$ of the bridges that touch $P$. However, $P$ has an odd number of bridges, so we cannot have crossed all of them, which is a contradiction.