Squares & primes1- At least 2 primes between adjacent squares!!?
- Hi All: Here's my take on showing that there are at least two primes
between adjacent squares. Let me know what you think. My apologies
in advance if the tabular examples are scatterred. Has anyone else
been looking at squares and primes recently?
ARGUMENTS TO DETERMINE THE MINIMUM NUMBER OF PRIMES BETWEEN TWO
ADJACENT PERFECT SQUARES
This article presents a methodology leading to a proof that at
least two prime numbers exist between any adjacent perfect squares.
The basic part of this calculation is elementary. The keys to the
solution are in the observations and assumptions made along the way.
First calculate the number of possible primes between any two
adjacent perfect squares, then calculate the number of possible
primary divisors for that set. Next use the most dense distribution
of divisors occurring anywhere in the complete number field, which is
the "mirror sequence" (ref note 1). Examples are given at the end
of this article to illustrate the methodology presented here.
N^2 = the value of any perfect square, where N is an integer
from 2 to any N.
Squares interval = numbers between but not including N^2 and
Possible primes = all the odd numbers in the squares interval
Divisor = the denominator in any division operation
Primary divisors = unique divisor values that are less than or
equal to N
KE = the number of possible primes in the interval range if N
KD = the number of possible primes in the interval range if N
LE = the number of primary odd divisors less than N, not
including 1, if N is even
LD = the number of primary odd divisors less than and including
N, not including 1, if N is odd
Mirror sequence = that sequence of primary odd divisors that is
centered at 2*the product of all
odd integers to N and spans +/-N from that center.
MS = the number of actual divisors in the mirror sequence
J = the minimum number of primes between any two adjacent
CALCULATIONS equation number
KD = KE-1 =N-1 (ref Note 2)
LE = KE/2 -1 = N/2-1
LD = KD/2 -1 =(N-1)/2-1 (ref Note 2)
For N even:
MS = 2*LE = 2*(N/2-1) = N-2 (5)
J = KE MS = N (N-2) = 2
For N odd:
MS = 2*LD = 2*[(N-1)/2-1] (7)
J = KD MS = N-1 2*[(N-1)/2-1] = N-1 (N-1-2) = 2
1. The most dense distribution of divisors is determined by
examining randomly generated sequences, perturbations of the mirror
sequence, and actual divisor sets. For all variances, the divisor
sequence must be internally mathematically consistent, i.e. every
third odd number must be divisible by 3, every fifth odd number by 5,
etc. For the purposes of this argument, it is not necessary that a
divisor in the mirror sequence be a true divisor of the possible
prime with which it is paired. Using these rules, random sequences
create spaces when the mirror image set is generated if the original
set is not in numerical sequence. Further, perturbations of the
mirror sequence show that any change in the original sequence will
create one more space. It is possible to create a sequence N-4 long
when the divisors are not in numerical order. However the mirror
sequence is N-2 long over the interval N, and thus more dense. These
findings need to be formally presented before this 2 prime proof is
complete. Mirror sequences are illustrated in the examples.
2. KD and LD must be formulated differently than KE and LE to avoid
when calculating J. The reformulation can be made by
observing that, for odd N's, N^2+2N
is the last odd number in the squares interval, and is
divisible by N. Assigning N as divisor
of N^2+2N then diminishes the available possible primes by 1,
and also the available primary
divisors by 1.
Examples of possible primes (KE or KD) and their mirror sequence
EVEN N ODD N
Code: X = no value
N = 12 N = 13
KE = 12 MS = 10 KD = 13 MS = 10
145 11 171 11
147 9 173 9
149 7 175 7
151 5 177 5
153 3 179 3
155 X 181 X
157 X 183 X
159 3 185 3
161 5 187 5
163 7 189 7
165 9 191 9
167 11 193 11
N^2+2N = 195 13*15
N = 2 N = 3
KE = 2 MS = 0 KD = 2 MS = 0
5 X 11 X
7 X 13 X
N^2+2N = 15 3*5
N = 4 N = 5
KE = 4 MS = 2 KD = 4 MS = 2
17 3 27 3
19 X 29 X
21 X 31 X
23 3 33 3
N^2+2N = 35 5*7
N = 1234E10000, or any even N
KE = N MS = N-2
Examples of calculations
N 12 90 4566 1234E10000
KE 12 90 4566 1234E10000
LE 5 44 2282 617E10000
MS 10 88 4564 (1234E10000)-2
J 2 2 2 2
N 13 111 9979 123456789123
KD 12 110 9978 123456789122
LD 5 54 4988 61728394560
MS 10 108 9976 123456789120
J 2 2 2 2
- Could you re-explain what divisors are? And hence what primary divisors
are? It's not clear from what you write, and (3) and (4).
I mean a divisor is defined as "denominator in any division operation".
That could be anything at all, couldn't it?
Presumably you're studying the set J of integers between n^2 and (n+1)^2,
for n even (there's no need to mess things up by studying even and odd
separately throughout - just look at the even). By divisors, I assume
(hope) you mean those integers which divide at least one member of J. Is
that right? Then the primary divisors are just a certain subset of the
Please confirm that, or correct me if Im wrong.
- Hi Andy ; Thanks for taking the time to review my post of 12/30.
Here are the answers to your questions.
Yes, I am studying the set of integers between N^2 and (N+1)^2,
more specifically the odd integers contained therein, as they are the
possible prime numbers in that set.
I agree that we can just look at the evens for now to study the
technique I am presenting.
I would call the KE values (for even N) the set of numbers I am
studying, not J. J is just one number, being equal to the minimum
number of primes between two squares.
The fancy footwork associated with the definition of divisors and
primary divisors is for the purpose of distinguishing between those
divisors of a number in the KE set that are less than or equal to the
square root of that number. For example, if a number in the KE set
were 255, then 3,5 and 17 would be divisors, but only 3 and 5 would
be primary divisors. I hope this helps. Thanks again for your
- Hi Bill,
I had the same confusion about your intention on divisors and now
that I see your reply, another question comes up. If by primary
divisors, you mean those that are < sqrt(x), you may have a problem
with the reasoning because, in the example of x=255, not only are 3 &
5 primary divisors, but also the trivial 1 & the not so trivial 15
would be defined as primary divisors. If by primary divisors you
mean only those divisors that are < sqrt(x) and prime, then the
definition excludes 15, but what then is the impact on your
--- In email@example.com, "billoscarson"
> Hi Andy ; Thanks for taking the time to review my post of 12/30.
> Here are the answers to your questions.
> Yes, I am studying the set of integers between N^2 and (N+1)^2,
> more specifically the odd integers contained therein, as they are
> possible prime numbers in that set.and
> I agree that we can just look at the evens for now to study the
> technique I am presenting.
> I would call the KE values (for even N) the set of numbers I am
> studying, not J. J is just one number, being equal to the minimum
> number of primes between two squares.
> The fancy footwork associated with the definition of divisors
> primary divisors is for the purpose of distinguishing between thosethe
> divisors of a number in the KE set that are less than or equal to
> square root of that number. For example, if a number in the KE set
> were 255, then 3,5 and 17 would be divisors, but only 3 and 5 would
> be primary divisors. I hope this helps. Thanks again for your
- One of these days I'll learn to ignore these posts, but until then...
It's bad enough working out what the mirror sequence is. It's impossible
from the way you've defined it, I had to look at the examples instead.
So you take all the odd numbers between (2n)^2 and (2n+1)^2, of which
there are of course 2n. You then write a sequence of pairs:
You then claim that, of the 2n terms in this sequence, 2n-2 of them
consist of 'actual divisors'? Meaning what? What divides what? It's not