Integer as Sum of Polygonal Numbers/Lemma 2

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Theorem

Let $n, m \in \R_{>0}$ such that $\dfrac n m \ge 1$.

Define $I$ to be the open real interval:

$I = \openint {\dfrac 2 3 + \sqrt {8 \paren {\dfrac n m} - 8} } {\dfrac 1 2 + \sqrt {6 \paren {\dfrac n m} - 3} }$

Then:

For $\dfrac n m \ge 116$, the length of $I$ is greater than $4$.


Proof

We need to show that $\paren {\dfrac 2 3 + \sqrt {8 \paren {\dfrac n m} - 8} } - \paren {\dfrac 1 2 + \sqrt {6 \paren {\dfrac n m} - 3}} > 4$ when $\dfrac n m \ge 116$.

Let $x = \dfrac n m - 1$.

Then:

\(\ds \paren {\frac 2 3 + \sqrt {8 \paren {\frac n m} - 8} } - \paren {\frac 1 2 + \sqrt {6 \paren {\frac n m} - 3} }\) \(>\) \(\ds 4\)
\(\ds \leadstoandfrom \ \ \) \(\ds \sqrt {8 x} - \sqrt {6 x + 3}\) \(>\) \(\ds \frac {23} 6\)

To simplify calculations, we consider:

\(\ds \sqrt {8 x} - \sqrt {6 x + 3}\) \(>\) \(\ds 4\) which is greater than $\dfrac {23} 6$
\(\ds \leadstoandfrom \ \ \) \(\ds 8 x + 6 x + 3 - 2 \sqrt {48 x^2 + 24 x}\) \(>\) \(\ds 16\) squaring Both Sides
\(\ds \leadstoandfrom \ \ \) \(\ds \sqrt {48 x^2 + 24 x}\) \(<\) \(\ds 7 x - \frac {13} 2\)
\(\ds \leadstoandfrom \ \ \) \(\ds 48 x^2 + 24 x\) \(<\) \(\ds 49 x^2 - 91 x + \frac {169} 4\) squaring Both Sides
\(\ds \leadstoandfrom \ \ \) \(\ds x^2 - 115 x + \frac {169} 4\) \(>\) \(\ds 0\)
\(\ds \leadstoandfrom \ \ \) \(\ds x \paren {x - 115} + \frac {169} 4\) \(>\) \(\ds 0\)

which is true when $x \ge 115$.

Thus this condition is satisfied when $\dfrac n m \ge 116$.

$\blacksquare$


Sources