Polygonal Numbers
An Honors Thesis
Ruston W. Pennington
Submitted to the Texas A&M UniversityCommerce Honors Committee in partial
fulfillment of the Program of Honors Study leading to the degree of Bachelor of Science
Directed by
Dr. Stuart Anderson
Professor
Mathematics
April 26m, 2012
Approved:
Advisor
Committee Member
~ ~Co.§ZMember
¥ Q . "clq," ::& Dean, College of S ience, Engmeenng, and Agnculture
Polygonal Numbers
The mathematician's patterns, like the painter's or the poet's
must be beautiful· the ideas, like the colors or the words must
fit together in a harmonious way. Beauty is the first test:
there is no permanent place in this world for ugly
mathematics.
G. H. Hardy[Har92)
Contents
Abstract
1 The definition of polygonal numbers
1.1 Overview . . . . . . . . . . . . . . . . . . .
1.2 Introduction and history . . . . . . . . ..
1.3 From triangular to all polygonal numbers .
1.4 Other general formulas . . . . . . .
2 Explorations using polygonal numbers
2.1 Overview ...... . ............... . ..... .
2.2 The sum of some polygonal numbers ............ .
2.3 Sum of cubic numbers and the partition of the odd integers .
2.4 Relations of square and triangular numbers . . . . . . . . . .
Summary
Acknowledgments
Bibliography
1
3
3
3
4
7
9
9
9
11
12
17
19
21
Abstract
Polygonal numbers are figurate numbers with a very old and rich ltistory. T hey
wcr<' first used by the anciC'nt GrC'C'ks to group numbers tog<'th<'r by r<'pr<'S<'nting
them with figures of regular polygons, especially triangles and squares. Since then ,
polygonl numbers have occupied the works of many important mathematicians and
their efforts in the field of number theory. This study begins with the figure of
th<' t riangular numh<'r and from h<'r<' we dC'fin<' all polygonal numh<'rs and C'xplor<'
different ideas using them.
1
1 The definition of polygonal
numbers
1.1 Overview
This chapter focuses on polygonal numbPrs in gPn<'ral. It bPgins with the definition
of the oldest and most basic polygonal number and from here builds up to a general
formula defining any polygonal number. From this first general formula we then
make a table of values and explore other general formulas and relationships between
the polygonal numbers.
1.2 Introduction and history
Polygonal numbers have a history that dates back to the era of ancient Greek mathematics.
Lacking a good number system, the ancient Greeks needed ways to group
numbers to better understand their properties and find inspiration for new ideas.
Therefore, they took objects such as pebbles and grouped them into basic regular
polygons. This is the beginning of the use of polygonal numbers. The Pythagoreans
assigned special properties and superstitions to these numbers, especially triangular
and square numbers. This elevated these numbers to a special status that led to the
further research and developments in elementary number t heory. It was after this
that Hypsicles in 175 B.C. defined all polygonal numbers. This was alluded to by
Diophantus' around 250 B.C. in one of his books which is now lost, titled Polygonal
Numbers. Diophantus is well known for his method of writing mathematics in a syncopated
form that made it easier to understand. This stems from the same need for
better notation that led to polygonal numbers in the first place. [HcalO] Inspired by
Diophantus' work Fermat claimed one of his famous theorems without proof named
the polygonal number theorem stating that all positive integers a re the sum of at
most n, ngonal numbers[Nat96]. Gauss proved the triangular case which is often
called the Eureka theorem after his notes upon discovering the proof. Euler continued
work on the square case and while he did not complete his work, it was used by
Lagrange. Lagrange and Jacobi then discovered the proof independently in 1772. It
was finally proved in its entirety in 1813 by Cau<.:hy[Wcia]. And finally Euler u:>cd
CXl
polygonal numbers in the beautiful expansion of IT (1  xk) which is named the
k ~ l
3
Chapter 1 The definition of polygonal numbers
pentagonal number theorem[Weib]. Because of their simplicity and yet broad relavance
they are still used in many mathematical texts to illustrate elemetary number
theory topics such as arithmetic progressions[Nat96].
1.3 From triangular to all polygonal numbers
This study begins with only the figures of the numbers themselves. Starting at this
point serves as a link between the algebraic nature of modern number theory and
the geometric roots of polygonal numbers. The triangular numbers are the lowest
degree of polygonal numbers and are the building blocks of polygonal numbers in
general.
• ••• • ••• ••••• ••• •••• 7"1  1 T~ = ~ T.! .. T, 10 • ••• •••••••• • ••• • ••• ••••• •• • •••••••• 1·, ):, h = .!l
Figure 1.1: [Mel]
Examining the figure we observe a few characteristics that define polygonal numbers.
First we sec that the first degree triangular number is one. All polygonal numbers
begin with one dot. Secondly, each triangular figure is a regular Lriangle. Similarly,
all polygonal numbers an• the arangemrnt of dots into regular polygons. The final
observation is that each successive triangular number is the previous figure' in addition
to the number of dots needed to make a regular triangle with a side length of
one more dot. This pattern also carries over to all other polygonal numbers. These
obscrvd.tions give us the needed geomctxic properties to define a progression that we
can evaluate for any value of n , the number of dots of a side of the polygon. Being
the first of the polygonal numbers, triangular numbers arc the simplest. Each successive
triangular number only needs one row of dots one dot greater than the last.
From this we can define t he sequence {1, 2, 3, 4, ... a,}where a, = n. To evaluate the
sum of this series we observe that this is the sum of the integers 1 to n. From this
we use t he anecdote of young Gauss who observed when adding the integers 1 to
100 that the terms can be paired into (1 + 100), (2 + 99), (3 + 98), ... , (50+ 51) which
all add to 101. Therefore, he simplified the process to the sum of the pairs which is
1.3 From triangular to all polygonal numbers
50 * 101 since there are 50 pairs. This generalizes to the formula
This is the formula for the nthtriangular number, denoted Tn.
An obvious formula can be based on the structure of the square numbers as each
square number is n rows of dots by n columns of dots so the n th square number is
n * n = n2 . We can define the square numbers r<'Cursivcly as we did the t riangular
numbers. Each successive square number is an added row of n  1 dots and an added
column of n  1 dots with 1 more added for the corner.
F ig ure 1.2:
n
+1
This leads to the recursive definition as S1 = 1, and Sn = Sn1 + 2(n  1) + 1 =
Snl + 2n  1. Since each successive square number is the sum of the previous square
number and t he next odd, Sn is also the sum of the odd numbers from 1 to 2n 1.
For the next stage, we seck a r<'cursivc defini tion of the successive pentagonal numbers.
We also need a more effici ent way of notating t he polygonal numbers. From
here on, any polygonal number other t han the triangular or square numbers will
be represented by the furlction P(x, n) where x is the number of sides and n is the
order of the polygonal number.
+1
F igure 1.3:
5
Chapt<'r 1 The dcfinition of polygonal numb<'rS
Now, examining the figure of the pentagonal number, the growt.h can be expressed
in terms of n for the nth pentagonal number. Each successive polygonal number is
the addition of a number of dots to the previous number, in order to make a regular
polygon with a length of n dots. From figure 1.3 we can sec that by adding 3 sides
of (n 1) dots you connect 2 sides of the previous figure and we find that we have
1 dot left to add (sec figure 1.3.) So the difi"crencc of t.hc P(5, n 1) and P(5, n) is
3(n  1) + 1  (3(n 2) + 1) = 3. Since th<' differ<'nc<' is a constant, W<' ran labd
this series an arithmetic series. An arithmetic series can be denoted as:
S = a1 +(at+ d)+ (a1 + 2d) ... + (a1 + (n 2)d +a,,
where a 1 is the initial term, S is the sum of the series, d is the constant difference
and a,. = a1 + (n 1)d is the nth term. It can also be denoted as:
S = (an (n 1)d) +(an (n 2)d) .. . +(a,. 2d) +(a,. d)+ an
Since (a,. (n 1)d) = a 1 and because the portion of each term multiplied by dis
the negation of the same term in the other equation, then the addition of these two
equations yields
28 = na1 +nan
S _ na1 +nan
 2 .
In the case of the pentagonal numbers, we have already shown d = 3 and in all
polygonal numbers a1 = 1 so then an = 1 + (n 1)3 = 3n  2. Hence P(5, n) =
n+n(3n 2) _ 3n2  n
2  2 .
This same logic can be applied to every polygonal number. Each polygonal number
will need to add (x  2) sides (where x is the number of sides) of (n  1) dots then
add 1 final dot to finish the polygon to from the previous figurc. So the diffcr<'DC<'
of P(x, n) and P(x, n 1) is
(x 2)(n 1) + 1  ((x 2)(n 2) + 1) = x 2.
So then the evaluation of any polygonal number is, where x is the number of sides
and n is the order of the polygonal number is the sum of another arithmetic series
P(x, n) = n + n(1 + (n
2
 1)(x 2))
= n + n(1 + xn  2n  x + 2) n + 3n + xn2
 2n2  xn
2
P(x, n) = 4n + xn2
; 2n2
 xn
we then simplify this by factoring out common terms and we have a much more
intuitive formula
P(x,n) = (x 2)n2
; (4 x)n.
6
1.4 Other general formulas
1.4 Other general formulas
Now that we have a general equation we can compile some numerical data to find
patterns. Using the polygonal number formula, the following x by n table results.
1 2 3
3 1 3 6
1 4 9
1 5 12 22
1 6 15 28
1 7 18 34
1 21 40
15
25
35
45
55
65
1 9
1 10
1 11
1 12
24 46 75
27 52 85
30 58 95
33 64 105
A few patterns can be observed. First it appears that every polygonal number is the
sum of the previous polygonal number and the previous order triangular number.
That is, P(x, n) = P (3, n  1) + P(x  1, n). To prove this we start with the general
formula for P(x, n).
P(x n) = (x 2)n2 + (4  x)n = xn2  2n2 + 4n xn = xn2
 3n2 + n2 + 5n n xn
' 2 2 2
(x 3)n2 +(5  x)n n2 n ((x 1)2)n2 +(4 (x 1))n (n1)2+(n1)
= 2 +2 = 2 + 2
= P(x 1, n) + P(3, n 1)
as desired.
The second pattern we notice is related to the first. Every polygonal number also
seems to be a sum of a multiple of the previous order triangular number and the
triangular number of the same order, otherwise denoted as P(x, n) = aP(3, n 1) +
P(3, n) Where a is an natural number.
Proof:
)
_ (x 2)n2 + (4 x)n _ xn2  2n2 + 4n xn _ xn2  3n2 + n2 + 3n + n xn
P(x, n  2  2  2
(x 3)(n2 n) + n2 + n = (x3) (n 1)2 + (n 1) + n2 + n = (x3)P(3, n1)+P(3, n).
2 2 2 2
Hence, every polygonal number is the sum of a triangular number of the same order
and (x 3) triangular numbers of the previous order.
7
2 Explorations using polygonal
numbers
2.1 Overview
Since a thorough understanding of polygonal numbers in general has been established,
this chapter turns to an exploration of specific relations using polygonal
numbers. It begins with the sums of polygonal numbers and the sum of cubes.
After these we generalize and expand on some other results found in certain mathematical
texts.[Nat96, Bur09, RK03]
2.2 The sum of some polygonal numbers
A topic commonly found in mathematical texts concerning polygonal numbers is
their sum. The sum of triangular numbers, denoted tn, are known as tetrahedral
numbers and the sum of squares, denoted sn, are often called pyramidal numbers.
tn = n___(:.n. + 1)(n + 2) __:....;____~
6
n(n + 1)(2n + 1)
Sn= 6
These two examples arc well known and therefore their mention is sufficient for
t his paper. We explore this idea with the evaluation of t he sum of the pentagonal
numbers from 1 to P(5, n). We start out by listing the sums when n = 1, 2, 3, 4, and
5.
I 2 3 L P(5, i) = 1, L P(5, i) = 6, L P(5, i) = 1
i=l i=l i1
4 5 L P(5, i) = 40, L P(5, i) = 75
i=l
9
Chapter 2 Explorations using polygonal numbers
Looldng back at the table in chapter one (p. 7), we can see that these numbers are
the diagonal starting at P(5,1) continuing downwards. From this observation the
following theorem can be proved.
n
2.::P(5, i) = P(4 + n, n)
i=l
Proof:
Let n = 1, then I:J=1 P(5, i) = 1 = P(5, 1) = P(4 + n , n) as desired, so the base
case is true. If the statement is true for some n then,
n+l L P(5, i) = P(4 + n, n) + P(5, n + 1) =
i=l
[(4 + n) 2Jn2 + [4 (4 + n)Jn 3(n + 1)2
 (n + 1)
2 + 2 =
1
2
[(n + 1){n2 + 3n + 2)] =
1
2
[(n2 + 4n + 3)(n + 1) + (  1  n)(n + 1))] =
1
2[{n + 3)(n + 1)(n + 1) + ( 1 n)(n + 1)] =
1
2[(n + 3)(n + 1)2 + ( 1  n)(n + 1)] =
1
2{[(5 + n) 2](n + 1)2 + [4 (5 + n)](n + 1)} =
P(5 +n,n + 1)
as desu.ed. Thus, by mathematical induction the conjecture is proved.
10
2.3 Sum of cubic numbers and the partition of the odd integers
2.3 Sum of cubic numbers and the partition of the
odd integers
Another result that involves polygonal numbers is their connection to the sum of
perfect cubes.[Bur09] This stems from the beautiful fact that the odd numbers can
be partitioned into sets whose summation of the terms in each set is the cube of the
cardinality of the set. Shown mathematically:
1 = 13
, 3+5 = 23
, 7+9+11 = 33
, 13+ 15 + 17 + 19 = 43
, ....
This result is proven as follows and will be used as a lemma for the sum of the cubes.
Lemma 1: j 3 = [j(j  1) + 1] + [j(j  1) + 3] ... +[j(j  1) + (2j  1)
Proof of Lemma 1:
We know from the square numbers that the sum of the odd integers 1 to 2j  1 is
P so then the series conjectured can be written as
j[j(j  1)] + i 2 = i(i  j) + l = l  l + i = i 3
as desired.
ow by Lemma 1 we can show this patt ern for the sum of the cubes.
33 = 7 + 9 + 11
43 = 13 + 15 + 17 + 19
Examining the shape of these terms arranged in this way, we see that this is a
triangular number of terms. Also we see that these terms are the series of odd
numbers 1 to 2n  1 which is n2 . This can also be shown algebraically. The last
term of this series of odds can be rewritten as such:
2(n2 + n)
n(n  1) + 2n  1 = n2  n + 2n 1 = n2 + n  1 =  1
2
Observe that if we let (n
2
:n) = k then the sum of the series, since it is a sum of
consecutive odds starting with one, is k2
. Finally, we see that k is also of the form
of a triangular number of order n or
k = (n2+n) = Tn
2
so then the sum of the cubes from 1 to n3 is (Tn)2 = P(4, T,1) .
11
Chapter 2 Explorations using polygonal numbers
2.4 Relations of square and triangular numbers
The next few results were inspired by three exercises in chapter two section three of
Burton's History of Mathematics[Bur09]. The first exercise asked for a proof of the
statement that the square of the product of three and any odd positive integer was
a difference of two triangular numbers. Denoted:
(3(2n + 1))2 = T9n+4 T3n+1
We begin with the goal of generalizing this statement with the very broad search for
a square of the product of any two numbers being the difference of two triangular
numbers. This would yield the conclusion thaL any square is the difference of two
Lriangular numbers. To find this we usc a Lable of values given by Lhe square of the
product of two numh<'rs. 'Irying to find a pattern manually proved to he incffici<'nt
and revealed no immediate resul ts. upon that discovery, we simplified the search
to the square of the product of two odds being the difference of two triangular
numbers. NoLing when the square of Lhe inner product of five and another odd
match<'d a diff<'renc<' of two triangular numbers, we compile a seqn<'nce of numbers
in which to find a pattern. using th<:> smallest of these r<'SUlts a lin<'ar pattern
was quickly observed. From this the conjecture can be made that the square of
Lhe product of five and anoLhcr positive odd was lhe difference of two triangular
numbers denoted:
(5(2n + 1))2 = T1sn+1 Tsn+2
To prove this conjecture we use algebra for the proof that follows:
12
(5(2n + 1))2 = 100n2 + lOOn + 25
200n 2 + 200n + 50 = 2
225n2 + 210n + 49 + 15n + 7  (25n2 + 20n + 4 + 5n + 2)
=
(15n + 7)2 + (15n + 7)
2
= Tt5n+7  Tsn+2
2
(5n + 2)2 + (5n + 2)
2
2.4 Relations of square and triangular numbers
Using a similar method we found and proved a very similar equation for the square
of the product of seven and another positive odd denoted:
(7(2n + 1)? = T21n+IO T7n+3
From the examination of the subscripts of the two cases now found with the one mentioned
by Burton, the following conjecture was made for the square of the product
of any two odds:
[(2k + 1)(2n + 1W = T(2k+l)3n+3k+l  T(2k+l)n+k
The proof is algebraic.
[(2k + 1)(2n + 1W = (4kn + 2k + 2n + 1)2
= 16k2n2 + 16k2n + 4k2 + 16kn2 + 16kn + 4k + 4n2 + 4n + 1
32k2n2 + 32k2n + 8k2 + 32kn2 + 32kn + 8k + 8n2 + 8n + 2
2
36k2n2 + 36k2n2 + 9k2 + 36kn2 + 36kn + 9k + 9n2 + 2
= ~ 2
4k2n2 + 4k2n + k2 + 4kn + n2 + n + k
2
(6kn + 3n + 3k + 1)2 + (6kn + 3n + 3k + 1)
=~~~~ 2
(2kn + n + k )2 + (2kn + n + k)
2
= T 6kn+3n+3k+l  T2kn+n+k = T (2k+1)3n+3k+J  T(2k+l)n+k
as desired.
The next step we take is to find a generalization that applies to the square of a
product of an even and an odd.
Using a similar method as before several patterns emerged where 2k = 2, 4, 6, and
8. The most immediate results reveals a pattern where each value of k took two
consecutive differences Lo find a common difference. From the methods of finite
differences, this reveals a quadratic description of the sequence. This result is not
13
Chapter 2 Explorations using polygonal numbers
as simple as desired so we return to the data. \Vith a different order we found a
linear pattern for each case. These four linear results were:
[2(2n + 1)]2 = T9n+4 T7n+3
[4(2n + 1W = T33n+16  T3ln+l5
[6(2n + 1W = T13n+36 Tnn+35
[8(2n + 1W = T129n+64 Tmn+63
The constant terms of the subscripts are the square of the even term in the product
and the' square' of th<' ev<'n term k ss on<'. Also the coC'fficients of the suhscripts
have a difference of two between the positive triangular number and the negative
one. Using again the method of finite differences we find that the coefficients of
the subscripts follow a quadratic sequence of specifically 8k2 + 1 and 8k2  2. This
produces a generalized equation to prove.
The proof is algebraic as follows.
[2k(2n + 1W = (4kn + 2k?
= 16k2n2 + 16k2n + 4k2 = 32k2n2 + 32k2n + 8k2
2
64k4n2 + 64k4n + 16k4 + 16k2n2 + 16k2n + 4k2 + n2 + n =  2
64k4n2 + 64k4n + 16k4  16k2n2  16k2n 4k2 + n2 + n
2
(8k2n + n + 4k2)2 + (8k2n + n + 4k2)2 (8k2n n + 4k2
 1)2 + (8k2n n + 4k2
 1)
=~~~~~ 2 2
= Tsk2n+n+4k2  Tsk2nn+4k2 1 = T{Sk2+J)n+{2k)2  T{Sk2 J)n+{2k)2 J
as desired.
lf we assume that n = 0 then the odd term in either product becomes one and,
(2k)2 = T4k2  T4k2  1
Hence the square of any positive odd number (except 1 since T0is not defined) is a
differC'nC'C' of two triangular numbC'rs. For the' S<'rond genC'ral equation we SC'C that
the square of any positive even number is the difference of two triangular numbers.
T herefore, all square numbers greater than 1 can be represented by the difference
of two triangular numbers.
14
2.4 Relations of square and triangular numbers
The final discovery of Lhis study is related solely to triangular numbers. Burton
asks for a proof that 9 times some triangular number plus one is also triangular.
The proof asked for is simple and algebraic hut again is a narrowly defined rdation.
To generalize this we ask for what multiples of a triangular number plus 1 do we
find that the result is also triangular. With a table of calculations we find many
cases and a few patterns. Focusing on t.he multiples we find no patterns for values
lrss than 630 and when rxamining the different triangular numhC'rs only patterns
for T1 and T2 emerge. Thus, we retmn to thC' original qu<'stion to find an altrrnate
phrasing. Examining the question we have the equation
9Tn + 1 = T3n+l·
Looking at the details of t his equation we observe that 9 is a square number and so
is 1. We already know that there are no patterns for multiplying by a square number
up to the 25thsquare number, so we then further examine the equation. Finally, we
leave the idea of some mult.iple of a triangular number plus 1 and try adding dilfcrcnL
values. One is both a square and a triangular number. The immediate results for
adding a square number yield no convention, but when we add triangular numbers
we find a pattern.
3 3 1 10
5 5 2 27
7 7 3 52
9 9 4 85
11 11 5 126
It is from these results that we conjecture that
(2n + 1)2
Tm + Tn = T(2n+l}m+n·
The proof follows.
( )2T. T. (4n2 + 4n + 1)(m2 + m) n2 + n
2n + 1 m+ n = 2
+ 
2

1
=  (4n2m2 + 4n2m + 4nm2 + 4nm + m2 + 2n + n2 + n)
2
[{2n + 1)m + n]2 + [(2n + 1)m + n]
= ~~~~~~
2
= T(2n+l}m+n·
Therefore, any triangular number multiplied by an odd postive integer, of the form
2n + 1, squared added to a Lriangular number Tn is also triangular.
15
Summary
This study started with the very basics of polygonal numbers. Working from the
figur<'s that. the Pythagoreans first us<>d, wr havr frw well defined formulas for all
polygonal numbers. The insight gained from each result has led to ideas in several
dillcrent areas ranging from the sum of pentagonal numbers to the difference of
two triangular numbers. In particular, we have found relations among triangular
numbers, and between the square and triangular numbers. In general, we have found
relations between the polygonal numbers. All of these relations and the processes
used to find them have led to the conclusion that the polygonal numbers still have
an important place in number theory and mathematics in general. Despite the
fact that we no longrr m·<·d them because of a lack of an cftid<•nt number syst<·m,
we still desire to use them in the introduction to number theory. This is because
of the simplicity and elegance they contain that made them desirable in the first
place. But we also observe the fact shown by this study that continued effort is
still valuable. We have explored the well known basics thoroughly and found the
inspiration needed to discover new results. Future work is suggested by results
of this study. Since we have shown that the squares are represented by both a.
sum of two individual triangular numbers and the dilference, then do pentagonal
numbers hold a similar relation since they are the sum of one triangular number and
a multipl<' of another? Would th<'y possibly be a diffcrenc<' of a triangular nurnh<'r
and the multiple of another? We can pose this question for all polygonal numbers.
Also, what about the swns of other polygonal numbers? Do their evaluations hold
·orne other pattern? And finally, there is still opportunity to apply the methods of
calculus to these numbers which has been outside the scope of this study.
17
Acknowledgments
I would like to thank and acknowledge these people in the aid of writing and defending
this thesis.
Prof. Stuart Anderson  advisor, mentor, major contributor and inspiration for this
research, and an excellent profes or of mathematics. You have made an impact on
my life as a student, teacher, and mathematician.
Prof. and Dean of Honors College Dr. Raymond Green  committee member, dean
of honors college, and led me through this thesis. Thank you for help and guidance
through this process and through my undergraduate studies.
Prof. Charles Dorsett  committee member and another great math professor.
Thank you for your encouragement and help.
Mrs. Amanda Pennington  Wife and best friend. Thank you for your support,
encouragement, and tokranc<' during times when thLc; task bream<' difficult.
19
Bibliography
[Bur09] D. Burton, History of Mathematics: An Introduction, 7th ed. McGraw
Hill, 2009.
[Har92] G. Hardy, A Mathematician's Apology, ser. A Canto Book Series. Cambridge
University Press, 1992.
[HealO] S. T. L. Heath, Diophantus of Alexandria. University Press, 1910.
[Mel] Melchoi:r. 'Irianglular numbers. [Online] . Available: en.wikipedia.org/wiki/
'Irian gular_ number
[Nat96] M. B. Nathanson, Additive Number Theory: The Classical Bases. Springer,
1996.
[RK03] Robert and E. Kaplan, The Art of the Infinite. Oxfort University Press,
2003.
[Weia] E. W. Weisstein. Fermat's polygonal number theorem. [Online]. Available:
http:// mathworld. wolfram. com/FermatsPolygonalN umberTheorem .html
[Wcib] . Pentagonal number theorem. [Online]. Available: http:/ / maLhworld.
wolfram.com/ FennatsPolygonalNurnberTheorem.html
21