We begin by noting that if one were able to reach a particular lattice point P on level five, one could, by doing the same thing from a starting position farther to the left or to the right, arrive at any specified lattice point on level five. It's all or nothing for level five. We settle the matter by showing that an arbitrary but definite lattice point P on level five is inaccessible.

We introduce mathematics to the situation by assigning a placevalue to every lattice point in the plane. Every value is a power of a number x, a number which shall be specified shortly. The exponent of the power is simply the number of unit steps in directions parallel to the axes in a shortest path from the distinguished point P to the lattice point in question. Thus P, itself, bears the value x⁰, or 1; the four lattice points adjacent to P are valued at x, the eight lattice points which are two steps from P have value x², and so on. As a result, the rows and columns of lattice points are assigned sequences of consecutive powers of x. (See Figure 3)

The value of a set of men is taken to be the sum of the place-values of the positions occupied by the men. We investigate now the change in value of a set of men that is incurred by a checker-jump.

There are three kinds of jumps - those which carry a man (i) closer to P, (ii) farther from P, (iii) the same distance from P. In every jump we begin with two adjacent points occupied and we end up with these places empty and a third point occupied. In every move of type (i) (see Figure 4), we gain a value xⁿ while losing two higher powers xⁿ⁺¹ and xⁿ⁺². Accordingly, the change is given by

For a move of type (ii), we obtain a change in value xⁿ⁺² - (xⁿ⁺¹+ xⁿ) = xⁿ( x² - x - 1). Now it is time to specify the number x. We choose x so that there is no change in value for a jump of type (i). For this we need

Thus x = (-1 ± √5) / 2. Taking x to be the positive root, (√5 - 1) / 2, we obtain a value between 0 and 1. In fact, x is the reciprocal of the famous Golden Mean, which crops up in so many different contexts. Note that x² = 1 - x. For a move of type (ii), then, we see that the change in value is

After a move of type (ii), then, the value of the men remaining is less than the value of the set of men before the move was made. Because a move of type (iii) is merely a jump across one of the two lattice lines which pass through P, we see that a move of type (iii) also reduces the value of a set of men: xⁿ - (x^n-1+ xⁿ) = -x^n-1. In summary, every jump either leaves the value of a set of men unchanged or it makes it less; under no circumstances does the value increase.

A man at the point P would have, itself, the value 1. Thus a set of men capable of sending a man to the point P would have to begin with a value of at least 1. A set with value less than 1 would require its value to be increased in order to place a man at P, and this cannot be achieved through checker moves. Now the value of the entire half-plane which constitutes the starting zone is easily determined by considering the points in columns. The column directly below P contributes the values x⁵+ x⁶+ x⁷... , a geometric progression with sum x⁵/(1-x) (recall that 0 <x < 1). The two columns, one on each side of this "central" one, each gives

Similarly there are two columns x⁷+ x⁸+ x⁹..., for a total of of 2x⁷/(1 - x). Altogether, the grand total for the infinity of columns is

the latter bracket being a geometric progression. Evaluating the series, and using 1 - x = x², we obtain

Thus even a single unoccupied place in the starting zone means that the starting set has a value < 1, and this is too small to send a man to level five.