AND BY DEFINITION, FOR I ∈ {0,1, . . . , `}, KI(N−I−KI) = M. SO, TO...

2, and by definition, for i ∈ {0,1, . . . , `}, k

i

(n−i−k

i

) = m. So, to complete the proofof the sufficiency we only need to show that each k

i

is an integer. For this one has bydirect computation that k

`

= (a+ 1)b and k

`−1

=ab. Fori∈ {0,1, . . . , `−2}, if one hadthatk

i

were a fraction with denominator 2 (in lowest terms) one would conclude from theequationk

i

(n−i−k

i

) = m that mis a fraction with denominator 4, which is impossiblesince m is an integer.Necessity: Given i ∈ {0,1, . . . , `}. Since we have an arrow from n−i to m, there is apositive integer k

i

≤(n−i)/2 such that k

i

(n−i−k

i

) =m. Solving this equation for k

i

and using the fact that k

i

≤(n−i)/2 we get(n−i)

²

−4mk

i

= (n−i)−p