NUMERICAL FULCRUMS

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NUMERICAL FULCRUMS
Richard Zucker Professor of Mathematics Irvine
Valley College rzucker_at_ivc.edu
RJ Liljestrom Senior
Woodbridge High School RJsHouse_2000_at_yahoo.com
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Six has an interesting property
In the list of consecutive integers from 1 to 8,
the numbers less than 6 have the same sum as the
numbers greater than 6.
Lets call six and all integers with this
property Numerical Centers.





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What is a Numerical Center?
n is a Numerical Center if it belongs to a finite
list of consecutive integers beginning with 1,
and in this list the integers smaller than n have
the same sum as the integers greater than n
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Some examples, please
6 is a Numerical Center because 12345 78
35 is a Numerical Center because 134 3649
204 is a Numerical Center because 1203
205288
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Is there another way to think about Numerical
Centers?
?
1(n-1) (n1)m
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TRIANGULAR NUMBERS
Nth Triangular Number
Sum
Diagram
N
1
1
1
3
12
2
6
123
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Formula for the Nth Triangular Number
T(n) (1/2)(n)(n1)
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Formula for the sum of the numbers from a to b,
inclusive
ab T(b) T(a-1)
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Using this formula with Numerical Centers, we get
1(n-1) (n1)m
T(n-1) T(m) T(n)
T(n-1) T(n) T(m)
(1/2)(n-1)(n) (1/2)(n)(n1) T(m)
(1/2)n2 (1/2)n (1/2)n2 (1/2)n T(m)
n2 T(m)
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So, is a number n a Numerical Center if and only
if its square is not only a square number but
also a Triangular Number?
YES!!!
Almost
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Numerical Centers
6, 35, 204, 1189, 6930, ...
... they have a nice recursion formula!
a1 6 a2 35
ak 6ak-1 ak-2
For example
6930 61189 - 204
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It is even possible to derive the following
general formula for ak, the kth Numerical Center
but we wont try to prove that here
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Numerical Centers sure are exciting!!!
(smile and nod)
but is there possibly a way to expand the
definition of Numerical Centers, so that more
integers can join the party?
Hmmmm
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A Numerical Center was defined as a number that
belonged to a finite list of consecutive integers
starting with 1 such that the sum of the numbers
below n is equal to the sum of the numbers
greater than n.
What if the list could start at any positive
integer, not just 1?
Lets call integers that fit this more general
definition Numerical Fulcrums.
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Definition
n is a Numerical Fulcrum if it belongs to a
finite list of consecutive positive integers in
which the sum of the integers less than n is
equal to the sum of the integers greater than n
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Some more examples, please
4 is a Numerical Fulcrum because 23 5
6 is a Numerical Fulcrum because 12345 78
9 is a Numerical Fulcrum because 678
1011 -and- 345678 101112
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We know that a Numerical Center is an integer
whose square is also triangular.
Can we think about Numerical Fulcrums in a
similar way?
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We can! Just as we called the upper limit to the
higher sum m, lets call the lower limit of the
lower sum (b1).
Why start at b1 instead of just b?
Youll see in a moment that it will make our
result algebraically simpler.
Were also going to need this formula again
ab T(b) T(a-1)
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Using this formula with Numerical Fulcrums, we
get
(b1)(n-1) (n1)m
T(n-1) T(b) T(m) T(n)
T(n-1) T(n) T(b) T(m)
(1/2)(n-1)(n) (1/2)(n)(n1) T(b) T(m)
(1/2)n2 (1/2)n (1/2)n2 (1/2)n T(b)
T(m)
See how nicely it turned out?
n2 T(b) T(m)
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In order for this formula to make sense
(b1)(n-1) (n1)m
we need to make the following restrictions
b 1lt n
n lt m
b 1 lt m
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This all leads to
An integer n is a Numerical Fulcrum if there
exist non-negative integers b and m such that n2
T(b) T(m), b1 lt m
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What does the set of Numerical Fulcrums look like?
How do these Numerical Fulcrums behave?
Is there a simple way to determine if an integer
n is a Numerical Fulcrum?
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Lets put this information into a graph that we
can understand.
If we have integers n, b, and m such that n2
T(b) T(m) where b1 lt m, then lets plot

• point (n,b) the color blue

• point (n,m) the color green

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Numerical Fulcrums Graph
n2 T(b) T(m)
(n,b) (n,m)
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We know that 9 is a Numerical Fulcrum in two
different ways are there other numbers that are
Numerical Fulcrums in two ways?
Three ways?
More?
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Number of ways that n is a Numerical Fulcrum
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25
21
30
1
27
5
8
11
4
14
2
20
24
9
10
16
23
12
6
13
15
32
19
31
7
3
18
22
17
26
28
29
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Percentage of Numerical Fulcrums lt n
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Is there a simple way to determine if a given
number n is a Numerical Fulcrum or not?
There is! Youll be amazed to find out that
Numerical Fulcrums are closely related to prime
numbers.
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nNF
Composite Number
4
65 513
9
325 52 13
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Prime Number
5
101 (prime)
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577 (prime)
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The Numerical Fulcrum Theorem
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How is it possible to relate n2 T(b) T(m)
and 4n2 1?
n2 T(b) T(m)
n2 (1/2)(b)(b1) (1/2)(m)(m1)
n2 (1/2)b2 (1/2)b (1/2)m2 (1/2)m
4n2 2b2 2b 2m2 2m
4n2 1 2b2 2b ½ 2m2 2m ½
2(4n2 1) 4b2 4b 1 4m2 4m 1
2(4n2 1) (2b 1)2 (2m 1)2
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PROOF OUTLINE
Prove that 2(4n21) x2y2 implies n2T(b)T(m)
Introduce Beilers formula for the number of ways
an integer can be expressed as the sum of two
squares
Introduce the consecutive solution
Prove that n is a Numerical Fulcrum iff 4n21 is
composite
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STEP 1 Prove that 2(4n21) x2y2 implies
n2T(b)T(m)
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Lets say that 2(4n21)x2y2. What types of
numbers must x and y be?
Obviously, x2y2 is even. If x2y2 is even, then
either both x and y are even or they both are odd.
Lets assume that they are even, and that x2X
and y2Y
2(4n21)(2X)2(2Y)2
2(4n21)4X24Y2
4n212X22Y2
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Now we know that, if 2(4n21)x2y2 then both x
and y must be odd. But can they equal each other?
Assume that xy
2(4n21)x2x2
2(4n21)2x2
4n21x2
(2n)21x2
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Both x and y are odd (of the form 2j1)
x cannot equal y

So x and y must be unique odd nonnegative
integers. Since they are odd, let us choose b
and m so that x2b1 and y2m1.
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2(4n21)x2y2
2(4n21)(2b1)2(2m1)2
n2T(b)T(m)
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STEP 2 Introduce Beilers formula for the
number of ways an integer can be expressed as the
sum of two squares
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Information from Recreations in the Theory of
Numbers The Queen of Mathematics Entertains
In order for an integer to be the sum of two
squares, it must be of the form
where Q is the product of primes of the form
4k-1, P is the product of primes of the form
4k1, and a0 is a nonnegative integer.
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What kind of prime factors does 4n21 have?
Lets say that p is a prime number that divides
4n21
THEOREM OF EULER -1 is a quadratic residue of
the prime number p iff p is of the form 4k1 and
not of the form 4k-1
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Since all of the prime factors of 4n21 are of
the form 4k1, we can say that 2(4n21) is a
number of the form 2P, where P is the product of
primes of the form 4k1.
This is clearly a special case of numbers of the
form
Now onto the formula
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2(4n21) 2P
Given
2(4n21) can be written as the sum of two unequal
squares in exactly
ways, where x represents the largest integer
less than or equal to x
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NOTE The formula only counts the number of ways
an integer can be expressed as two unequal squares
But we dont need to worry about this, because
weve already proven that an integer of the form
2(4n21) cannot be the sum of two equal squares
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STEP 3 Introduce the consecutive solution
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A positive integer n is a Numerical Fulcrum iff
nonnegative integers b and m exist such that
n2T(b)T(m), where b1ltm
Is there an additional solution that exists if we
allow b and m to be consecutive?
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Let bn-1 and mn
n2T(b)T(m)
n2T(n-1)T(n)
n2(1/2)(n-1)n(1/2)(n)(n1)
n2 ½ n2 ½ n ½ n2 ½ n
n2n2
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Therefore, bn-1 and mn is always a solution of
n2 T(b) T(m). Well call this the
consecutive solution.
The consecutive solution always exists, but it is
not sufficient to make n a Numerical Fulcrum
because it doesnt satisfy the requirement that
b1 lt m.
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STEP 4 Prove that n is a Numerical Fulcrum iff
4n21 is composite
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Those that are a single prime raised to a power
greater than one
Those that are the product of at least two
different prime numbers
Well look at these two cases separately
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CASE 1 4n21 is a single prime raised to a power
greater than one
Can it be a single prime raised to the second
power?
4n21p2
(2n)21p2
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If 4n21 is a single prime raised to a power
greater than one, then it must be raised to at
least the third power.
Lets say that we were given that 4n21 p3.
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Using the formula
We can see that 4n21 can be written as the sum
of two squares in exactly

2 ways
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If 4n21 were a prime raised to a power greater
than three, you would still be able to express it
as the sum of two squares in at least two ways,
because Beilers formula would yield a higher or
equal integer
One of these solutions must be the consecutive
solution
The other solution must NOT be the consecutive
solution
This second solution makes n a Numerical Fulcrum,
given that 4n21 is equal to a prime raised to a
power greater than one
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CASE 2 4n21 is the product of at least two
distinct prime numbers
Lets assume that 4n21 is the product of only
two prime numbers, each raised to the power of one
2(4n21) 2p1p2
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Using the formula
We can see that 4n21 can be written as the sum
of two squares in exactly

2 ways
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If 4n21 were a product of more than two primes,
or primes raised to a power greater than one, you
would still be able to express it as the sum of
two squares in at least two ways, because
Beilers formula would yield a higher or equal
integer
One of these solutions must be the consecutive
solution
The other solution must NOT be the consecutive
solution
This second solution makes n a Numerical Fulcrum,
given that 4n21 is equal to the product of at
least two prime numbers
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Finally, lets see what happens when 4n21 is a
prime number
2(4n21) 2p1
We can see that 4n21 can be written as the sum
of two squares in exactly

1 way
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If 4n21 is the product of at least two different
prime numbers, then n is a Numerical Fulcrum

If 4n21 is a single prime raised to a power
greater than one, then n is a Numerical Fulcrum

If 4n21 is prime, then n is not a Numerical
Fulcrum

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In conclusion, we have proven that
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BIBLIOGRAPHY
Beiler, Albert H. Recreations in the Theory of
Numbers The Queen of Mathematics Entertains.
New York Dover Publications, Inc, 1966.
Weisstein, Eric W. Fermat 4n1 Theorem. Eric
Weissteins World of Mathematics. Wolfram
Research, Inc. lthttp//mathworld.wolfram.com/Ferm
at4 n1Theorem.htmlgt 1999.