## Proof that level 9 is unreachable

Assign to a peg on lattice point (*x,y*) the value
*V*(*x,y*) = θ^{min{y,c-|x|}},

where θ is the positive root of θ^{2}=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+2*c*)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} [ θ^{y} + 2 ∑_{1≤x≤c-y} θ^{y} + 2 ∑_{x≥c-y+1} θ^{c-x} ]_{ }
= ∑_{y≤0} θ^{y} [1 + 2(*c*-*y*)] + 2 ∑_{y≤0} θ^{y}/(θ-1)_{ }
= [1 + 2*c* + 2/(θ-1)]∑_{y≤0} θ^{y} - 2 ∑_{y≤0} *y*θ^{y}_{ }
Note that ∑_{y≤0}[-*y*θ^{y}] = ∑_{i≥0} *i*θ^{-i} = ∑_{0≤j≤i} θ^{-i} = ∑_{j≥0} ∑_{i≥j} θ^{-i} = ∑_{j≥0} θ^{-j} θ/(θ-1)_{ }
So, *S* = [1 + 2*c* + 2/(θ-1)] θ/(θ-1) + 2 θ/(θ-1)^{2}_{ }
= (1 + 2*c*)θ/(θ-1) + 4 θ/(θ-1)^{2}

The net gain to *S* when jumping with a peg of value *v* is, at most,
*v*θ^{2} - *v*θ - *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 θ^{9} > 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*) = θ^{y-|x|}.
In this case, the sum of *V*(*x,y*) over the lower half plane
is *S* = (θ^{2}+θ)/(θ-1)^{2},
and the value of a single peg at (0,5) is θ^{5} which
equals *S*.
This allows for no other pegs to co-exist, an impossibility to achieve
within a finite number of steps.