## Proof that level 9 is unreachable

Assign to a peg on lattice point (x,y) the value V(x,y) = θmin{y,c-|x|},
where θ is the positive root of θ2=θ+1 (≈ 1.618) and c is a positive integer.
Initially, the sum S(c) of the value of all pegs is the sum of V(x,y) over the lower half plane,
which is (1+2c)θ/(θ-1) + 4θ/(θ-1)2.  This is shown as follows.

```
S = ∑y≤0 ∑x V(x,y)
= ∑y≤0 [ V(0,y) + 2 ∑x≥1 V(x,y) ]
= ∑y≤0 [ θy + 2 ∑1≤x≤c-y θy + 2 ∑x≥c-y+1 θc-x ]
= ∑y≤0 θy [1 + 2(c-y)] + 2 ∑y≤0 θy/(θ-1)
= [1 + 2c + 2/(θ-1)]∑y≤0 θy - 2 ∑y≤0 yθy
Note that ∑y≤0[-yθy] = ∑i≥0 iθ-i =  ∑0≤j≤i θ-i = ∑j≥0 ∑i≥j θ-i = ∑j≥0 θ-j θ/(θ-1)
So, S = [1 + 2c + 2/(θ-1)] θ/(θ-1) + 2 θ/(θ-1)2
= (1 + 2c)θ/(θ-1) + 4 θ/(θ-1)2
```

The net gain to S when jumping with a peg of value v is, at most, vθ2 - vθ - v = 0.
Therefore, the total value of all pegs cannot exceed its initial value.

Consider c = 9.  S(9)=66.687+.  Since the value of a peg at (0,9) is θ9 > 76,
it follows that we cannot get a peg to the point (0,9).
If we could get a peg to any point on or above the line y = 9
then we could get one to (0,9) by stopping when some peg reaches a point (x,9),
and then retracing all the jumps performed shifted left by x, or shifted right by -x when x is negative.

Note that the same argument can be used to show that Conway's problem, in which diagonal jumps are not permitted cannot admit a finite sequence of moves leading to a peg at level 5, by using the potential function V(x,y) = θy-|x|.  In this case, the sum of V(x,y) over the lower half plane is S = (θ2+θ)/(θ-1)2, and the value of a single peg at (0,5) is θ5 which equals S.  This allows for no other pegs to co-exist, an impossibility to achieve within a finite number of steps.