Download Puzzles and Pythagoras in the Classroom

“Puzzles and Pythagoras in the Classroom”
Presented by
Professor Spencer Hurd
School of Science and Mathematics
The Citadel, Charleston, SC
Sigma Xi Talk – The Citadel, September 2008
Necessary Background - about that of an entering freshman
Prerequisites - High School Algebra and a Sense of Humor
A little puzzle to start with: Label the edges AB, BC, CD, and DA
with reciprocals of positive whole numbers. Say
AB 
1
,
w
BC 
1
,
x
CD 
1
,
y
DA 
1
.
z
And, do this so that, for each vertex A, B, C, and D, the sum of the two
associated reciprocals is also a reciprocal. This means there must be
positive integers a, b, c, and d so that
1 1 1
  ,
a w z
1 1 1
  ,
b x w
1 1 1
  ,
c x y
1 1 1
  .
d y z
The ancient Egyptians had crude number symbols, essentially
base ten, using repetitions:
│=1
││ = 2
│││ = 3.
∩ = 10
∩∩ = 20
∩∩∩ = 30
And
│││ ∩∩∩ = 33, and so on.
Fractions were indicated by putting a bar (or eye) over
the integer symbols. In effect, all fractions were “unit fractions”
or reciprocals.
In our notation, all fractions were of the form
1
.
n
If they wanted to write 2/5 they had to use
1 1
 .
3 15
This has led to a huge literature in puzzles and theoretical
problems of great difficulty.
How many ways are there to write
might as
2
as the Egyptians
35
2 1 1
  for different positive integers x, y.
35 x y
What is the fewest number of distinct reciprocals so that
they add to
3
. (Ans: it can be done with 3 reciprocals.)
7
Can you find an Egyptian labeling for a square?
An Egyptian graph is defined to be a connected graph
whose edges have different positive Egyptian fraction labels
and whose vertices have different labels also, and such that, for
each vertex, the sum of the labels of the edges incident with a
vertex is the label for that vertex.
A graph is Egyptian, then, if such a labeling is known to
exist. Before going further, can you construct an edge labeling
of, say, a square so that the labeling is Egyptian.
Table 1: Egyptian Labels for the Square
1
2
3
4
5
6
7
8
9
10
AB
6
9
10
12
12
8
9
9
10
10
BC
12
18
15
24
24
24
18
18
15
40
CD
60
36
60
40
72
168
90
144
60
60
DA
30
72
40
60
36
56
72
72
90
90
A
5
8
8
10
9
7
8
8
9
9
B
4
6
6
8
8
6
6
6
6
8
(All entries represent reciprocals.)
C
10
12
12
15
18
21
15
16
12
24
D
20
24
24
24
24
42
40
48
36
36
Pythagoras
Pythagoras of Samos (about 500-600 BCE) was the sage-philosopher
who founded a cult based on his unwritten teachings. He founded a
commune on the coast of Italy which became successful in the sense that the
cult lasted for 200 years, at least. Their philosophy was based on the notion
that “all is number.”
He is credited with discovering (or naming) prime and composite
numbers, odd and even numbers, figurate numbers, polygonal numbers
amicable numbers and perfect numbers.
Figurate numbers are fore-runners of today’s labeled graphs. But
that’s another talk.
His eponymous theorem is a theorem about areas – the area of the two
squares erected on the legs of a right triangle is equal to the area of the
square erected on the hypotenuse.
We call a positive integer triple (a, b, c), which satisfies a 2  b 2  c 2 , a
Pythagorean Triple or PT. If gcd(a, b, c) = 1, then the triple is called
primitive.
A Pythagorean Quadruple, or PQ, is a 4-tuple of integers (a, b, c, x)
such that a 2  b 2  c 2  x 2 .
Teachers for 2500 years have used side lengths with whole number
values to illustrate Pythagoras’s theorem. Pythagoras himself is credited
with discovering an infinite sequence of triples in which the hypotenuse and
the larger leg differ by unity.
a
3
5
7
9
11
c = b+1
B
4
12
24
40
?
c
5
13
25
41
?
“Let’s find one example.” Students find all the examples by trial and error.
Then they look for patterns.
Observe that
Also
So
b + c = a2 .
c = b + 1.
a2 = 2b + 1
Algorithm: select any odd positive integer for a. Then
Find two consecutive positive integers which sum to a2.
So, b + b +1 = 112 and b = 60, c = 61.
Conclusion:
Arrange for a parameter N.
N = 1 corresponds to (3, 4, 5)
N = 2 corresponds to (5, 12, 13)
Solution: take a = 2N + 1
Then b  (a 2  1) / 2  2 N 2  2 N
And c  (a 2  1) / 2  2 N 2  2 N  1
Student Check:
(2 N  1)2  (2 N 2  2 N )2  (2 N 2  2 N  1)2
(i.e., that a 2  b 2  c 2 )
Euclid proved in The Elements (about 300 BCE):
If (a, b, c) is a PPT, then there are two numbers m and n
(one odd and the other even and gcd(n, m) = 1)
so that a = 2mn, b = n2 – m2, and c = n2 + m2.
It is easy to have students check that:
(2mn)2  (n2  m2 )2  (n2  m2 )2
and, thus, such (a, b, c) is always a PT. The hard part was to show
that the existence of the m and n.
The philosopher Plato is credited with discovering a
second infinite sequence of triples in which the hypotenuse and
the larger leg differ by two.
a
4
8
12
16
11
c = b+2
b
3
15
?
?
?
c
5
17
?
?
?
A set of Discovery lessons
A Pythagorean Quadruple, or PQ, is a 4-tuple of integers (a, b, c, x) such
that a 2  b 2  c 2  x 2 .
1. For simplicity, let a = 1. Can someone find a PQ with a = 1?” Someone
will find (1, 2, 2, 3).
Ask her to put it on the board.
“Do you think (2, 4, 4, 6) is a PQ?”
For PTs, multiples of a PT corresponded to similar right triangles and gave
new PTs, is this true for PQs as well?
HMWK
Prove or disprove:
(1) There is a PQ given by (1, 1, b, c) for some integers b and c.
(2) (a, b, c, d) is a PQ if and only if (ka, kb, kc, kd) is a PQ.
2. Remember that 32 + 42 = 52 and 52 + 122 = 132. Put these facts together as
13 2  12 2  5 2  12 2  4 2  3 2 .
HMWK: string together known PTs to make at least a dozen new
PQs. Put them in a chart or organized list. Are there any patterns?
3. Going back to a = 1, the table of sums of squares gives values of k so that
k = m2 + n2. Look for values of k so that k itself is one less than a square.
That is, look for k = x2 – 1.
The table shows 80 = 82 + 42 and 80 = 81 – 1 = 92 – 1. Put together, this gives
4 2  8 2  80  9 2  1 .
And re-arranging, we get
12  4 2  8 2  81  9 2 .
HMWK: Find as many of these as you can, put them in a chart, and see if
you can find any patterns. If yes, state and prove an identity (or 2).
4. Using the PQ (1, 2, 2, 3), one might suppose that there are others in the
form (1, 2n, 2n, x). If so, they satisfy 12 + (2n)2 + (2n)2 = x2, from which we
get x  1  8n 2 . Use a spreadsheet to search for such quadruples. In column
A, beginning with cell A2=0, put the integers from 1 to 100. (Put =1+A2 in
cell A3.) In column B, in B3, put the formula =1+8*A3^2. In cell C3 put
=sqrt(B3). [If the entry in cell C is an integer, we have found a PQ.] Select
A3, B3 and C3 and FILL down to n = 100 at least. Make a list of the
solutions found, if any. Can you find anything?
Note:
(1, 2, 2, 3)
(1, 12, 12, 17)
(1, 70, 70, 99) and
(1, 2378, 2378, 3363) are in this pattern, a rather sparse sequence. Are there
more?
5. Maybe a = 2 is promising. In this case x 2 – 4 = b2 + c2. Lets look in the
table for k = x2 – 4. Adam , you check x = 1,..., 5 and see if x2 – 2 is in the
table, Betty, you check x = 6,..., 10. Cathy, you check x = 11,..., 15.
Continue until each student has some to check. A spreadsheet that extends to
40^2 + 40^2 would get more, but your students should find
(2, 3, 6, 7) and (2, 5, 14, 15).
Can somebody find some more like (2, 7, c, c+1)?
Perhaps they can find these
(2, 7, 26, 27), (2, 9, 42, 43), (2, 11, 62, 63).
Analysis: 22 + b2 + c2 = (c+1)2 gives, on removing parentheses and
simplifying, c = (b2+3)/2. This shows that b can be any odd number. Put
another way, if b = 2n+1, then c = 2n2 + 2n + 2 and x = c+1 = 2n2 + 2n + 3.
We get identity.
2 2  (2n  1) 2  (2n 2  2n  2) 2  (2n 2  2n  3) 2
HMWK: Put this in a table headed by n, a, b, c, and x. Then prove the
identity.
6. The method above can be used over and over. Let a be any fixed number.
Let x = c+1. For example, a = 3, and we get c = (8+b2)/2. So b has to be
even. We get the sequence
(3, 2, 6, 7), (3, 4, 12, 13), (3, 6, 22, 23), and so on.
In this sequence, a = 3 and b = 2n. Express c and x in terms of n.
One gets, 32  (2n) 2  c 2  (c  1) 2 .
Express c and x in terms of n, make a chart for n = 1,...,5.
n
a=3
b = 2n
c = 4 + 2n2
x = 5 + 2n2
1
3
2
6
7
2
3
4
12
13
3
3
6
22
23
4
3
5
3
HMWH: Prove the identity: 32  (2n) 2  (4  2n 2 ) 2  (5  2n 2 ) 2
HMWK: Let a = 5 and x = c+1 and find a sequence of PQs. Make a chart for
those you find. Determine an identity in terms of a single parameter n and
prove the identity.
7. Let 2n and 2n+1 represent consecutive numbers. Note (2n+1)2 – (2n)2 =
4n2 + 4n + 1 – 4n2 = 4n + 1. This can be viewed backwards – that is, any
number of the form 4n + 1 is a difference of two squares!
Let F1 denote the set {4n+1| n = 1, 2, 3, ...}.
Let F3 denote the set {4n+3| n = 0, 1, 2, ...}.
It can be shown that no member of F3 can be written as a sum of two
squares, but many members of F1 can be so written. [Had you noticed that
EVERY odd number in the list of sums of two squares is from F1 ? NONE
are from F3.]
Pick any odd number from the list. (It is already a sum of two squares,
that is how it got in the list.) Write it as a difference of squares – pick 53 for
example; 53 = 27 + 26 so also 53 = 272 – 262. But, from the list, 53 = 72 +
22. Put together
2 2  7 2  27 2  26 2 which means 2 2  7 2  26 2  27 2 .
Thus, every odd number in the list can be used to create a PQ.
8. For the PQ (a,b,c,x), suppose a and b are consecutive. Are there any
examples like that so far that you have seen? Make a list of them.
HW: find a pattern and express a,b,c,x in terms of a paramenter n.
9. Pick any algebra formula involving squares. Say (n+1)2 = 1 + 2n + n2. If
2n is a square, say, 2n = m2, then
(n+1)2 = 12 + m2 + n2, a new PQ.
HMWK: express m in terms of n and get a one-parameter family of PQs.
HMWK: Choose any other identity involving squaring and create a new
family of PQs..
10. Exploration: let x = c + 2 for a change. What can you find.
11. Exploration: Are there infinitely many PQs with a “leg” equal to 6? (Are
there infinitely many primitive PTs with a leg of 6?)
12. What is the length of the diagonal of a casket (rectangular box) if the
dimensions of the casket are 2-by-3-by-6 ?
Graph Labeling Problems: for which graphs or classes of graphs are
such labelings possible?
1. Label the vertices with colors so that no two vertices which share an edge
also share a color. Use the minimum number of colors. The minimum
number is the chromatic number of the graph. How is the chromatic number
of a graph related to other numbers associated with the graph.
2. Label the edges with plus (+) or minus (-) so that all cycles are “positive”
in the sense that they have an even number of minuses.
3. Label the nodes with distinct non-negative integers. Then label each edge
with the absolute value of the difference of it corresponding vertex labels. If
the graph edge labels then run from 1, 2, …, e, the graph is graceful or
gracefully numbered.
4. Do any chains have graceful labelings?
5. An edge labeling which uses the consecutive integers 1, 2, …, e so
that the vertex sums are consecutive e+1, e+2,…, e+v is super
magic. (A vertex sum is the sum of the labels of the edges incident
with the vertex. )
6. Review hydrocarbon chains in basic chemistry and redescribe
them in graph-theoretic terms.