**INTRODUCTION**

*Balmoral Software*

- Two-dimensional shapes
- Area- and perimeter-dominant polygons
- Three-dimensional solids
- Cuboids
- Prisms
- Integer prism dimensions
- Summary
- Heronian polygons
- Generalized polygon inequality
- Area of Heronian isosceles triangle
- Pythagorean fraction limits
- Integer squares of rational numbers
- Pell recursion
- Non-squarefree Pell equations
- Reference

In these pages, we are interested in equable convex plane figures that have integer area and sides. Regular n-sided polygons are excluded because their areas

A = (1/4)nsare irrational, except for the square n = 4, which can be treated as a special case of the quadrilateral.^{2}cot(π/n), n ≥ 3

*Equability*, or the condition of being equable, is clearly an abstraction
only. In any real-world scenario, the fallacy of equating units of length and
area, or of area and volume, would lead to conclusions that are unscalable.

Proof. Equivalently, for positive rational numbers x, y and z, if xy, xz and yz all are integers, then xyz is an integer. LetThe volume of a rectangular prism with rational dimensions is an integer if all of its faces have integer areas.

xy = kfor positive integers k, l and m. Then

xz = l

yz = m,

Let x = p/q for positive integers p and q. We can write the preceding equation as

mxThis is a quadratic equation in x with integer coefficients, so by the Rational Root Theorem, p divides kl:^{2}- kl = 0

for some positive integer n

Then

so xyz is an integer, QED.

V = hA = 2A + hP = S [1]

Therefore, the base shape of an equable right prism must be area-dominant, and thus cannot itself be equable. The prism height h = V/A is rational. It follows from [1] that the perimeter

is also rational.

Suppose there exists an equable solution with integer base area A and rational
base perimeter P and height h. If we multiply all sides of the base by a
positive rational factor α such that α^{2}A is an integer,
the new area and perimeter are

A' = αand the corresponding height to preserve equability is^{2}A ∈ ℤP' = αP,

For a valid solution, we require that the denominator of the right-hand fraction be positive, or

[2]

α > P/AFor example, consider an equable right prism with base area A = 40, base perimeter P = 100/3 and therefore height h = 12. The perimeter-to-area ratio is P/A = 5/6, so let α be a rational number bigger than that and such that α

t = A/P = h/(h - 2),or equivalently,

h = 2t/(t - 1) = 2 + 2/(t - 1)When h is an integer, we have the following results:

The following relationships also hold:

t Solutions for integer h (0,1] none (1,2) h ≥ 5 2 4 (2,3) none 3 3 (3,∞) none

We conclude that

h t ⌊4t ^{2}⌋⌊4t ^{2}⌋ - 4t^{2}+ 13 3 36 1 4 2 16 1 5 5/3 11 8/9 6 3/2 9 1 7 or 8 4/3 or 7/5 7 4/25 or 8/9 9 or 10 5/4 or 9/7 6 19/49 or 3/4 11 to 18 9/8 to 11/9 5 2/81 to 15/16 ≥ 19 1 < t ≤ 19/17 4 ≥ 1/289 where ⌊⌋ is the floor function.

1/289 ≤ ⌊4tfor any values of interest for t.^{2}⌋ - 4t^{2}+ 1 ≤ 1

Many solutions require solving Diophantine equations.

2 dimensions 3 dimensions Equable requirement: Area = Perimeter Volume = Surface Area Integer requirement: Area & dimensions Volume & surface areas

Dimensions (prisms)

The rest of this paper presents definitions and miscellaneous results that will be used in the analysis of specific equable shapes.

max{awhere s = (a_{1},a_{2},...} < s,

Proof. When the inequality is satisfied, we have

afor all i. Then_{i}< s

so side a

Proof. Let the integer legs and base of an isosceles triangle be a and b, respectively. The areaThe area of a Heronian isosceles triangle is a multiple of 3.

of the triangle is an integer. Then

(4A)so (4A,b^{2}+ (b^{2})^{2}= (2ab)^{2},

Lemma. *At least one of the two smallest members of a Pythagorean triple is a
multiple of 3.*

Proof. Let the integer elements of a Pythagorean triple be a, b and c, where

aA square mod 3 cannot be 2, so we have^{2}+ b^{2}= c^{2}

(aIf a^{2}+ b^{2}) mod 3 = c^{2}mod 3 ∈ {0,1}

Proof. We can write

b ^{2}/a > 2 for all a and b[3] b ^{2}/a ≥ 16/3 = 5.333... if a < b[4]

cwhich proves [3].^{2}= a^{2}+ b^{2}> a^{2}c > a

c ≥ a + 1

c

^{2}≥ (a + 1)^{2}> a^{2}+ 2ab

^{2}= c^{2}- a^{2}> 2a,

The Pythagorean triple with the shortest leg is 3,4,5, so a ≥ 3. If b > a, then

b ≥ a + 1which proves [4].b

^{2}≥ (a + 1)^{2}= (a - 3)(a - 1/3) + 16a/3

≥ (3 - 3)(3 - 1/3) + 16a/3 = 16a/3,

Proof. If xIf x is rational and x^{2}is an integer, then x is an integer.

xor as a monic polynomial in x with integer coefficients:^{2}= m, for some integer m,

xBy the integral root theorem, x is an integer. The converse is trivially true.^{2}- m = 0.

where x and y are positive integers and D ≥ 2 is a positive nonsquare integer. Solutions to [5] for a particular value of D are an infinite integer sequence {x

x ^{2}- Dy^{2}= 1[5]

where (p,q) is an initial solution for (x,y) in [5] that depends on D. The initial solution is generally selected with positive p and q, and so x

Note that αβ = p

αThe sequence 2x^{n}± β^{n}= (α + β)(α

^{n-1}± β^{n-1}) - α^{n-1}β ∓ αβ^{n-1}= (α + β)(α

^{n-1}± β^{n-1}) - αβ(α^{n-2}± β^{n-2})= 2p(α

^{n-1}± β^{n-1}) - (α^{n-2}± β^{n-2})

2xso its recursive solution is the second-order linear difference equation_{0}= α^{0}+ β^{0}= 22x

_{1}= α + β = 2p,

The sequence is also of this form (with - sign), and

x _{n}= 2px_{n-1}- x_{n-2}, x_{0}= 1, x_{1}= p[6]

so its recursive solution is the same second-order linear difference equation, but with different initial conditions:

Consider the

y _{n}= 2py_{n-1}- y_{n-2}, y_{0}= 0, y_{1}= q[7]

(afor given constants a_{1}x_{n}+ b_{1})^{2}- D(a_{2}y_{n}+ b_{2})^{2}= 1

By [6],

aand analogously,_{1}x_{n}+ b_{1}= 2p(a_{1}x_{n-1}+ b_{1}) - (a_{1}x_{n-2}+ b_{1})x

_{n}= 2px_{n-1}- x_{n-2}+ 2b_{1}(p - 1)/a_{1},

ywith suitably-chosen initial conditions._{n}= 2py_{n-1}- y_{n-2}+ 2b_{2}(p - 1)/a_{2},

We can write [5] as

(xso_{n}^{2}- 1)/y_{n}^{2}= D

xand_{n}/y_{n}→ D as n → ∞

xthen for a given positive integer m,^{2}- Dy^{2}= 1

has a solution sequence (x'

x' ^{2}- m^{2}Dy'^{2}= 1[8]

Proof. Let

Then

k = smallest n such that m | y _{n}[9]

xand (x'_{k}^{2}= (y_{k}/m)^{2}m^{2}D + 1

For example, take D = 6 and m = 11. Then we can see that k = 3 by examining the
solution sequences for D and m^{2}D:

D = 6 mA table of values for k is easily created using [7] and [9], as described in the Python code for OEIS sequence A298210:^{2}D = 726 p = 5 p = 485 q = 2 q = 18 ---------------------------------------------- ---------------------------------------------- n x y x y n 0 1 0 1 0 0 1 5 2 2 49 20 3 485 198 485 18 3 4 4801 1960 5 47525 19402 6 470449 192060 470449 17460 6 7 4656965 1901198 8 46099201 18819920 9 456335045 186298002 456335045 16936182 9 10 4517251249 1844160100 11 44716177445 18255302998 12 442644523201 180708869880 442644523201 16428079080 12 13 4381729054565 1788833395802 14 43374646022449 17707625088140 15 429364731169925 175287417485598 429364731169925 15935219771418 15 16 4250272665676801 1735166549767840 17 42073361925598085 17176378080192802 18 416483346590304049 170028614252160180 416483346590304049 15457146750196380 18

Table of k values m: 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 OEIS D p q 2 3 2 1 2 2 3 2 3 4 6 3 6 2 7 3 6 8 4 6 10 6 A298210 3 2 1 2 3 2 3 6 4 4 9 6 5 6 6 4 3 8 9 18 5 6 A298211 5 9 4 1 2 1 5 2 4 2 2 5 5 2 7 4 10 4 3 2 3 5 A298212 6 5 2 1 3 2 2 3 4 4 3 2 3 6 7 4 6 8 9 3 9 2 7 8 3 2 1 2 3 2 7 2 3 6 6 2 7 14 3 2 3 6 9 6 8 3 1 2 2 4 3 2 3 8 6 6 6 4 7 6 6 16 4 6 10 12 10 19 6 1 1 2 5 1 4 4 3 5 6 2 3 4 5 8 9 3 2 10 11 10 3 2 1 2 2 2 3 4 3 2 11 2 7 6 2 8 9 6 3 2 12 7 2 1 3 2 3 3 2 4 9 3 5 6 3 2 3 8 9 9 5 6 13 649 180 1 1 1 1 1 4 2 1 1 2 1 10 4 1 4 4 1 38 1 14 15 4 1 2 1 2 2 7 2 6 2 5 2 6 7 2 4 9 6 10 2 15 4 1 2 3 2 5 6 3 2 3 10 5 6 7 6 15 4 8 6 10 10 17 33 8 1 2 1 3 2 4 1 6 3 2 2 3 4 6 2 40 6 9 3 18 17 4 1 3 1 3 3 3 2 9 3 3 3 7 3 3 4 2 9 5 3 19 170 39 2 1 2 2 2 4 4 3 2 3 2 1 4 2 8 2 6 21 2 20 9 2 1 2 2 5 2 4 4 2 5 5 2 7 4 10 8 3 2 3 10

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