Case 3

Next we study the case that $z = n = 3$ $(mod \ 4)$ and $y$ is odd.

Figure 5.21  

In this subsection we use methods that are very similar to the one used in Subsection 5.4.2 and Subsection 5.4.3, and hence we omit the detail of the argument.
In Fig. 5.21 we suppose that Predictions 5.1, 5.2, 5.3,5.4 and 5.5 are valid for $z=0,1,2,...,4m+2$ and for $y = 0,1,2,...,z$ . We also suppose that these predictions are valid for $z=4m+3$ and $y=0,1,...,2k-4$ . Under these assumptions we prove that $X= G1R(\{2k-3,4m+3\}) = 4m+k+2$ .

For $z=n=4m+2 = 2 (mod \ 4)$ by Prediction 5.1 the first number from the right end of the row is $6m+3$ .
We suppose that

$$G1R(\{2m+1,4m+2\}) = 3m+1$$ (5.51)

and

$$G1R(\{2k-3,4k-6\}) = 3k-5.$$ (5.52)

We also assume that

$$4m+2 > 4k-6 2k-3 < 2m+1.$$ (5.53)

If we are to prove that $X= G1R(\{2k-3,4m+3\}) = 4m+k+2$ , by Lemma 5.1 we have only to prove that $4m+k+2$ is the smallest number that does not belong to $L1(\{2k-3,4m+3\})$ .
$L11(\{2k-3,4m+3\}) \cup L12(\{2k-3,4m+3\}) \supset \{0,1,2,3,...,3k-3\}$ .
We have $L12(\{2k-3,4m+3\}) \supset \{3k-1,3k+1,...,4m-k+3\} \cup \{3k-2,3k,...,4m+k\}$ $\supset \{3k-2,3k-1,3k,...,4m-k+4\}$ .
$L13(\{2k-3,4m+3\}) \supset \{4m-k+3,4m-k+4,4m-k+5,...,4m+k+1\}$ . Therefore we have $X=4m+k+2$ .
The case that $z = n = 0$ $(mod \ 4)$ can be treated in the similar way, so we omit it.