Sequence Sums that are Squares

Initializing live version
Download to Desktop

Requires a Wolfram Notebook System

Interact on desktop, mobile and cloud with the free Wolfram Player or other Wolfram Language products.

With we can ask for which and is a square.

[more]

Somewhat surprisingly, in the table that lists all , many of the squares appear connected by straight lines and at regular distances to each other.

The challenge is to find all such lines and come up with a formula that describes the entries on these lines. This Demonstration shows eleven such lines (click "lines" to see them) and describes two of them in detail.

We will come across some surprises on the way...

[less]

Contributed by: Karl Scherer (December 2013)
Open content licensed under CC BY-NC-SA


Snapshots


Details

Introduction

The Demonstration shows the table that lists all sequential sums , where and , n <= m.

The sums that are squares are colored orange. You might see right away that several square entries are lined up on a straight line of - or so it seems - infinite length. Select "short" from the "lines" popup menu to see all lines.

Two lines in detail

Pink line: starts at {2,4} with increment {1,3}

The pink square entries covered are 9, 25, 81,... This suggests (without formal proof) that the associated sums S(m, 3 m-2) are a square for . It is easy to prove by induction that indeed we have , (1), which is a surprisingly simple formula. In the table, the cases associated with equation (1) are connected by a straight line.

Green line: starts at {1,8} with increment {1,17}

The green square entries covered are 36, 324, 900,...

The entries highlighted by green dots suggest that is a square for . It is easy to prove by induction that indeed we have . (2)

More lines

Altogether there are eleven lines shown in the graph:

Start pos. sum Increment {x,y} that leads to the next square

--------------------------------------------------------------------------------------

{1,1} 1 {1,3} {1,8} 36 {1,17} {4,4} 4 {7,9} {4,5} 9 {7,11} {4,21} 225 {7,43} {9,9} 9 {17,19} {9,16} 100 {17,33} {12,13} 25 {23,27} {12,20} 144 {23,41} {16,16} 16 {31,33} {24,25} 49 {47,51}

We give an example of how to read this table:

At position x=4, y=21 an entry with the sum 225 can be found . The next square in the (red) line is 7 steps to the right and 43 steps up, leading to the entry at {11, 64}. This entry carries the sum 2025, calculated from S(n,m)=(m*m + m - n*n +n)/2, which is indeed a square: 2025 = 45*45.

Can you find the associated increments of the other nine lines? Can you prove for these other nine lines that the associated (infinitely many) sums are all squares? Can you find the associated increments some line that are not shown?

Observations

Looking at the graph the following conjectures come to mind:

Observation 1: EVERY sum S(n,m) that is a square seems to lie on one of such a line.

Observation 2: No two lines cross (as long they are not extended to the left). If they are extended to the left, they seem to all intersect at the same point, namely at {.5,-.5}.

Also, it might be interesting to find out whether there is an easy way to calculate the increments from the starting values.

It is time to put the lines data into one mathematical theorem:

Without being able to prove it, the author suggests that the following is true:

Conjectured theorem

Let n and m be are natural numbers, n <= m. Let the sequential sum S(n,m) = n + ... + m be a square. Then there exist natural numbers p and q such that S(n + ip,m + iq) is square for all i=1,2,3,... Furthermore, in the set of pairs {p,q} no two elements are the same.

All lines intersect at {0.5,-0.5}. If P,Q are the first squares on a line, then in vector calculus we have Q-P=2*(P-{.5,-.5}). This means that the distance between two consecutive squares on a line is exactly twice the distance between the first square in a line and the intersection point {.5,-.5}.

Each line contains only one (infinite) set of regularly spaced squares.

Prime Numbers

Click "primes".

The brown points mark all prime numbers in the diagram.

We see that prime numbers only occur in the main diagonal and in the next diagonal above it, but nowhere else.

Can you proof that the are no prime numbers above the second diagonal?

Well, here is the proof:

Let's assume that s = n + (n +1 ) + (n + 2) + ...+ (n + m) is a prime, with m>0.

Then s =n * (m+1) + (1 + 2 + 3 + ... + m) = n * (m+1) + (m + 1) * m/2

If m is even, then m/2 is an integer; hence k = m + 1 is odd and must be a proper divisor of s.

If m is odd, then k = (m + 1)/2 is an integer that divides s. This k is a proper divisor of s if k > 1, hence if and only if m > 1.

Hence in our table prime numbers can only occur for m = 0 or m = 1, i.e. in the main diagonal or in the one above it.

Note:

Looking at the table, you might also notice that each prime number occurs in each of the first two diagonals exactly once!

Can you see why?

Controls

Click any table entry to place a small red frame. The values of (from) and (to) associated with the clicked sum is displayed at the left border. If the table entry is a square, its root will also be displayed at the left border.

Click "squares" to show all squares in the table.

Setter bar "lines none/short/long": to hide/show the lines connecting an infinite number of squares each.

"Short" lines start from a square; option "long" will show the lines starting from the intersection point at {.5,-.5}.

Click "primes": to show all prime numbers in the table.



Feedback (field required)
Email (field required) Name
Occupation Organization
Note: Your message & contact information may be shared with the author of any specific Demonstration for which you give feedback.
Send