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\author{}
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\begin{center}
    \fontsize{14pt}{18pt}\selectfont
    Minimum Prime Bound in Square Intervals
\end{center}

Min Min Oo(B.Sc(Maths),Independent researcher,Myanmar)

\textbf{Abstract}

This paper presents a proof of Legendre’s Conjecture by analyzing the distribution of odd composite and their elimination through odd composite sequences. The odd integer sequence is expressed as \(O_{\geq 3}=3+2r_O\), while odd composite numbers are represented as \(OC_{p}=p^2+2pr_p\), where \(p\) is prime and \(r_{p}\geq 0\). Each prime-square sequence introduces a systematic elimination of odd composite numbers, reducing the density of primes in the sequence. By sequentially applying elimination factors \((q-1)/q\) for primes \(q\geq 3\), we derive a lower bound for prime density:
\[
\frac{dN_p}{dr_O}\geq\frac{2}{p}.
\]

Considering the quadratic interval \([n^2,(n+1)^2]\), the total change in \(r_O\) is \(\Delta r_O=\frac{(n+1)^2-n^2}{2}=n+\frac{1}{2}\).
Combining this with the density bound yields a lowest bound of prime count per interval:
\[
\pi_{\left(n^2,(n+1)^2\right)}^{LB}
\geq
\left\lceil\frac{2}{p}\left(n+\frac{1}{2}\right)\right\rceil
=2
\qquad (p~n+1).
\]
Thus, every quadratic interval \([n^2,(n+1)^2]\) contains at least two primes, establishing Legendre’s Conjecture. This elimination-based framework provides a systematic proof based on the distribution of odd integers and prime-square sequences.

\begin{center}
Fig(1)
\end{center}

\includegraphics[width=420pt,height=400pt]{Fig1.png}


We can rewrite the odd integer sequence as follows:

\[O_{\geq 3} = 3 + 2r_{O}(r_{O} \geq 0)\]

Odd composite sequence:

\[OC_{p} = p^{2} + 2pr_{p}\ \left( p = \text{prime,}r_{p} \geq 0 \right)\]

Before the odd composite sequence corresponding to \(p=3\) begins at \(3^2=9\), all odd integers in \(O_{\geq 3}\) are prime. For example, \(3\), \(5\), and \(7\) are prime numbers. The odd composite sequence corresponding to \(p=3\) is \(9+6r_3\), where \(r_3\geq 0\).(See in Fig.1)

When the odd integers exceed \(3^2=9\), the gap in the odd composite sequence corresponding to \(p=3\) is \(2p=6\), while the gap in the odd integer sequence is \(2\). Therefore, every three consecutive odd integers contain one term from the odd composite sequence corresponding to \(p=3\), since \(\frac{6}{2}=3\). Thus, one out of every three odd integers is eliminated. If we assume that no other odd composite sequences exist, the remaining proportion is \(1-\frac{1}{3}=\frac{2}{3}\).

\[\frac{d}{dr_{O}}N_{p = 3} = \frac{2}{3}\ \left( N_{p = 3} = \text{lowest~prime~number~due~to~elimination~of~}OC_{= 3}\text{~sequence} \right)\]

When the odd integers exceed \(5^2=25\), the gap in the odd composite sequence corresponding to \(p=5\) is \(2p=10\), while the gap in the odd integer sequence is \(2\). Therefore, one out of every five odd integers is eliminated, since \(\frac{10}{2}=5\), leaving a proportion of \(\frac{4}{5}\). Applying this elimination rate to the proportion \(\frac{2}{3}\) remaining after considering the odd composite sequence \(9+6r_3\) for \(p=3\), we obtain \(\frac{2}{3}\times\frac{4}{5}=\frac{8}{15}\). This approach is used to estimate a lower bound for the proportion of odd integers remaining after successive odd composite sequences are considered.

\[\frac{d}{dr_{O}}N_{p = 5} \geq \frac{d}{dr_{O}}N_{p = 3} \times \frac{4}{5} = \frac{2}{3} \times \frac{4}{5} = \frac{8}{15}\]

When the odd integers exceed \(7^2=49\), the gap in the odd composite sequence corresponding to \(p=7\) is \(2p=14\), while the gap in the odd integer sequence is \(2\). Therefore, one out of every seven odd integers is eliminated, since \(\frac{14}{2}=7\), leaving a proportion of \(\frac{6}{7}\). Applying this elimination rate to the proportion \(\frac{8}{15}\) remaining after considering the odd composite sequences \(9+6r_3\) for \(p=3\) and \(25+10r_5\) for \(p=5\), we obtain \(\frac{8}{15}\times\frac{6}{7}=\frac{16}{35}\). This approach is used to estimate a lower bound for the proportion of odd integers remaining after successive odd composite sequences are considered.

\[\frac{d}{dr_{O}}N_{p = 7} \geq \frac{d}{dr_{O}}N_{p = 5} \times \frac{6}{7} = \frac{8}{15} \times \frac{6}{7} = \frac{48}{105}\]

For the general case:

\[
\frac{d}{dr_O}N_p
\geq \frac{2}{3}\times\frac{4}{5}\times\frac{6}{7}\times\cdots\times\frac{p-1}{p}
= \prod_{\substack{3\leq q\leq p\\ q\text{ prime}}}\frac{q-1}{q}
\]

\[
\prod_{\substack{3\leq q\leq p\\ p,q\text{ prime}}}
\frac{q-1}{q}
\geq \frac{2}{p}
\]

Thus:

\[\frac{d}{dr_{O}}N_{p} \geq \frac{2}{p}\]

For a window \((n + 1)^{2} - n^{2}\):

\[
\Delta r_O^{\text{for a window}}
= \frac{(n+1)^2-n^2}{2}
= n+\frac{1}{2}
\]

So the lowest bound of prime count:

\[
\pi_{\left(n^2,(n+1)^2\right)}^{LB}
\geq
\left\lceil
\frac{dN_p}{dr_O}
\times
\Delta r_O^{\text{for a window}}
\right\rceil
=
\left\lceil
\frac{2}{p}\left(n+\frac{1}{2}\right)
\right\rceil.
\]

We can assume that \(p\approx n+1\). Therefore,
\[
\begin{aligned}
\pi_{\left(n^2,(n+1)^2\right)}^{LB}
&\geq
\left\lceil
\frac{dN_p}{dr_O}
\times
\Delta r_O^{\text{for a window}}
\right\rceil \\[6pt]
&=
\left\lceil
\frac{2}{p}\left(n+\frac{1}{2}\right)
\right\rceil \\[6pt]
&\approx
\left\lceil
\frac{2}{n+1}\left(n+\frac{1}{2}\right)
\right\rceil \\[6pt]
&=
\left\lceil
2-\frac{1}{n+1}
\right\rceil \\[6pt]
&=2.
\end{aligned}
\]

Conclusion

Therefore, in every window
\(\left\lbrack n^{2},(n + 1)^{2} \right\rbrack\), there are at least two
primes.

\end{document}