Some Quickies

Part III of my (never-ending) series “A Means to an End” should be up soon, but I wanted to record this little proof now before I forget it.  Incidentally, these observations will be relevant to the to-be-had discussion in my upcoming posts, but for now I’ll treat it as stand-alone material.

Recall that a pair of whole numbers (m, n) is said to be an isoperimetric pair if

\displaystyle m + n = \left\lceil\,\frac{mn}{\left\lceil\,\sqrt{mn}\,\right\rceil}\,\right\rceil + \left\lceil\,\sqrt{mn}\,\right\rceil.

This definition arises from the work of my REU students.  Specifically, a rectangular arrangement of mn squares will require the smallest perimeter possible if and only if (m, n) is an isoperimetric pair.

Also, if you take the expression on the right-hand side and replace + with \cdot, we stumble across our previously celebrated “next pronic or square” function SP(mn).  Something else to point out is that the formula

SP(x) = \left\lceil\,\frac{x}{\left\lceil\,\sqrt{x}\right\rceil}\right\rceil\,\cdot\,\left\lceil\,\sqrt{x}\,\right\rceil

presents the output as a product of its “square” or “pronic” factors.  In particular

\left\lceil\,\sqrt{x}\,\right\rceil =\left\lceil\,\frac{x}{\left\lceil\,\sqrt{x}\right\rceil}\right\rceil \text{ or }\left\lceil\,\frac{x}{\left\lceil\,\sqrt{x}\right\rceil}\right\rceil + 1

These observations are helpful in proving the following neat, little facts.

Proposition 1: All pairs of the form (n, n+2) are isoperimetric.  Moreover, whenever n > 1, all pairs of the form (n, n+3) are also isoperimetric.

Proof (sketch): For the first claim, one uses the fact that SP(n(n+2)) = (n+1)^2 for all n \geq 1.  This fact itself is easily verified; high-school algebra confirms that n(n+1) < n(n+2) < (n+1)^2.  That is, numbers of the form n(n+2) are always sandwiched in between the pronic n(n+1) and the square (n+1)^2.  by definition of SP(x), it follows that SP(n(n+2)) = (n+1)^2.  Using the remark above, one readily verifies that (n, n+2) is isoperimetric.

For the second claim, one uses the fact that n > 1 to conclude that (n+1)^2 < n(n+3) < (n+1)(n+2).  This immediately implies that SP(n(n+3)) = (n+1)(n+2).  Comparing both sides of our isoperimetric pair equation, we indeed confirm that n + (n+3) = (n+1) + (n+2).  Therefore (n, n+3) is isoperimetric.  \square

Corollary 1: Every number of the form k^2 - 1 is hydropronic.

Proof: Note that k^2-1 = (k-1)(k+1).  The pair of factors (k-1) and (k+1) are isoperimetric since they differ by 2.  \square.

Corollary 2: For k>2 every number of the form k(k+1) - 2 is hydropronic.

Proof: Again, we show that such numbers have isoperimetric factors.  This follows since k(k+1) - 2 = (k-1)(k+2) and these factors differ by 3.  \square.

These corollaries appear to be the tip of the iceberg.  My REU students have the following conjectures that, if true, generalizes these results:

Conjecture: Every number of the form n^2 - k^2 is hydropronic (provided k is not too big).  Also, every number of the form n(n+1) - k(k+1) is hydropronic (provided k is not too big).  Finally, all hydropronics are of these forms.

I should point out that k being “not too big” is an interesting condition that is somewhat difficult (at the moment) to make precise.  The bottom line is that provided n^2 - k^2 is not less than the previous pronic n(n-1), you’re good (and a similar statement holds for the other case).

Interesting, no?

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