With \(m \geq 4\), we have

by Corollary 3.5 with \(x = \frac{n}{m}\). When \(m = 3\), we have

by Corollary 3.6 with \(x = \frac{n}{m}\).

Since \(c {\gt} 0\), we have \(\pm \frac{p}{2} + \sqrt{\left(\frac{p}{2}\right)^2 + c} {\gt} \pm \frac{p}{2} + \left\lvert \frac{p}{2}\right\rvert \geq 0.\) The statement holds trivially when \(x = 0\). Assume that \(x {\gt} 0\). Since \(x {\lt} \frac{p}{2} + \sqrt{\left(\frac{p}{2}\right)^2 + c}\), we have \(x - p {\lt} -\frac{p}{2} + \sqrt{\left(\frac{p}{2}\right)^2 + c}\). Thus,

Since \(x {\gt}\frac{p}{2} + \sqrt{\left(\frac{p}{2}\right)^2 + c} {\gt} 0\), we have \(x - p {\gt} -\frac{p}{2} + \sqrt{\left(\frac{p}{2}\right)^2 + c} {\gt}-\frac{p}{2} + \sqrt{\left(\frac{p}{2}\right)^2 + c} {\gt} 0\). Hence,

Note that

Setting \(p := 1 - \frac{6}{m}\) and \(c := 6 \left(\frac{n - r}{m}\right) - 4\), we have \(c {\gt} 0\) and so, by Lemma 6 part (b), we obtain that \(b^2 + 2b + 4 - 3a = b ^2 - \left(1 - \frac{6}{m}\right) b - \left(6\left(\frac{n - r}{m}\right) - 4\right) {\gt} 0\).

Note that

Setting \(p := 1 - \frac{6}{m}\) and \(c := 6 \left(\frac{n - r}{m}\right) - 4\), we have \(c {\gt} 0\) and so, by Lemma 6 part (b), we obtain that \(b^2 + 2b + 4 - 3a = b ^2 - \left(1 - \frac{6}{m}\right) b - \left(6\left(\frac{n - r}{m}\right) - 4\right) {\gt} 0\). We can also see from the above derivation that \(b {\gt} 0\). Now,

Setting \(p := 4 \left(1 - \frac{2}{m}\right)\) and \(c := 8 \left(\frac{n - r}{m}\right)\), we have \(c {\gt} 0\) (as \(n - r \geq 2m - (m-3) = m+3\)) and so, by Lemma 5 part (a), we obtain that \(b^2 - 4a = b^2 - 4 \left(1 - \frac{2}{m}\right) b - \frac{8n - r}{m} {\lt} 0.\)

Let \(\ell = \lceil p \rceil .\)

Note that \(p {\gt} \ell - 1\).

We can take \(m\) to be the least integer such that \(2m + 1 \geq \ell \). Indeed, for all \(i = 0,\ldots ,k-1\), we have that \(b_i \geq b_0 = 2m + 1 \geq p\) and \(b_i \leq b_{k-1} = 2(m+(k-1))+ 1 = 2m+1+2(k-1).\)

If \(\ell \) is even, then \(2m+1 = \ell + 1\).

Hence,

If \(\ell \) is odd, then \(2m+1 = \ell \).

Hence,

Proof by cases on \(n \mod b_1\)

Proof by cases on \(n \mod b_1\)

First, consider the case when \(m \geq 4\) and \(n \geq 53m\). By Lemma 4 part (a), we have \(u(n,m) - \ell (n,m) \geq 4.\) It follows from Lemma 10 that there exist odd integers \(b_0, b_1\) in the interval \([\ell (n,m), u(n,m)]\) such that \(b_1 = b_0+2\). Let \(r'\) be the remainder when \(n - b_0\) is divided by \(m\). Note that \(r' \leq m - 1\) and \(n - b_0 - r' \equiv 0 \pmod{m}.\) If \(r' \geq m - 2\), set \(r\) to \(r'-2\) and \(b\) to \(b_1\). Since \(r' \leq m-1\), we have that \(r = r' - 2 \leq m - 3\). Also, \(r = r'-2 \geq m-2-2 = m - 4 \geq 4 - 4 = 0\). Then setting \(b\) to \(b_1\), we have that \(n - b - r = n - b_1 - (r'-2) = n - b_0 - r' \equiv 0 \pmod{m}\). Hence, \(m\) divides \(n - b - r\). Otherwise, we have \(r' \leq m - 3\). Setting \(r\) to \(r'\) and \(b\) to \(b_0\), we have that \(n - b - r = n - b_0 - r' \equiv 0 \pmod{m}.\) Hence, \(m\) divides \(n - b - r\). Next, consider the case when \(m = 3\) and \(n \geq 159m\). We set \(r\) to 0. By Lemma 4 part (b), we have \(u(n,m) - \ell (n,m) \geq 6.\) It follows from Lemma 10 that there exist odd integers \(b_0, b_1, b_2\) in the interval \([\ell (n,m), u(n,m)]\) such that \(b_1 = b_0+2\) and \(b_2 = b_1 + 2\). Since \(b_1 \equiv b_0 + 2 \pmod{3}\) and \(b_2 \equiv b_1 + 2 \equiv b_0 + 4 \equiv b_0 + 1 \pmod{3}\), it follows that for some \(b \in \{ b_0, b_1,b_2\} \), we have \(n - b - r \equiv n - b \equiv 0 \pmod{3}\).

Omitted.