Integers (a,b,c) whose squares form an A.P. — Part 2

 

In part 1, I started a discussion on the ordered triples of integers   (a, b, c)   whose squares form an arithmetic progression.

In other words,
b^2 \; - \; a^2 \; = \; c^2 \; - \; b^2 \; = \; D^2,     or equivalently
2 \, b^2 \; = \; a^2 \; + \; c^2

The solutions are of the form

a \; = \; \pm k \, (m^2 \; - \; n^2 \; - \; 2 \, m \, n)
b \; = \; \pm k \, (m^2 \; + \; n^2)
c \; = \; \pm k \, (m^2 \; - \; n^2 \; + \; 2 \, m \, n)

for any integers   m   and   n

The ordered triple   (a, b, c)   is a primitive arithmetic progression triple if   k = 1

 

If   (a, \; b, \; c)   is a primitive arithmetic progression triple
so are   ( \,c, \; 3 \, b - 2 \, a, \; 4 \, b - 3 \, a \,)

(4 \, b - 3 \, a)^2 - (3 \, b - 2 \, a)^2 \; = \; (5 \, a - 7 \, b) \, (a - b)

OR
((m + 3 \, n)^2 - 2 \, n^2)^2 - ((m + 2 \, n)^2 + n^2)^2 \; = \; 4 \, n \, (m + n) \,(m + 2 \, n) \, (m + 3 \, n)

(3 \, b - 2 \, a)^2 - c^2
2 \, b^2 \; - \; a^2 \; = \; c^2
(3 \, b - 2 \, a)^2 - (2 \, b^2 - a^2)
= (5 \, a - 7 \, b) \, (a - b)

OR
((m + 2 n)^2 + n^2)^2 - (m^2 - n^2 + 2 \, m \, n)^2 = 4 \, n \, (m+n) \, (m + 2 \, n) \, (m + 3 \, n)

So, if   (a,b,c)   is a primitive arithmetic progression triple
so are   (c, \; 3 \, b - 2 \, a, \; 4 \, b - 3 \, a)

Similarly, if   (a,b,c)   is a primitive arithmetic progression triple
so are   (4 \, \; b - 3 \, c, \; 3 \, b - 2 \, c, a)

 

Prove that if   (a,b,c)   is a primitive arithmetic progression triple so are

A_{n} \; = \; 2 \,n \,(n - 1) \,b \; - \; (n^2 - 1) \,a \; - \; n \,(n - 2) \,c

B_{n} \; = \; (2 \,n^2 + 1) \,b \; - \; n \,(n + 1) \,a \; - \; n \,(n - 1) \,c

C_{n} \; = \; 2 \,n \,(n+1) \,b \; - \; n \,(n + 2) \,a \; - \; (n^2 - 1) \,c

are also a primitive arithmetic progression triples.

 
and recursively, for   n > 0

A_{n} \; = \; C_{n-1}
B_{n} \; = \; 3 \, B_{n-1} \; - \; 2 \, A_{n-1}
C_{n} \; = \; 4 \, B_{n-1} \; - \; 3 \, A_{n-1}

A_0 = a   … B_0 = b   …   C_0 = c

A_1 \; = \; c,
B_1 \; = \; ( \,3 \, b \; - \; 2 \, a \,)
C_1 \; = \; ( \,4 \, b \; - \; 3 \, a \,)

A_2 \; = \; ( \,4 \, b \; - \; 3 \, a \,)
B_2 \; = \; ( \,-6 \, a \; + \; 9 \, b \; - \; 2 \, c \,)
C_2 \; = \; ( \,-8 \, a \; + \; 12 b \; - \; 3 \, c \,)

A_3 \; = \; ( \,-8 \, a \; + \; 12 \, b \; - \; 3 \, c \,)
B_3 \; = \; ( \,-12 \, a \; + \; 19 \, b \; - \; 6 \, c \,)
C_3 \; = \; ( \,-15 \, a \; + \; 24 \, b \; - \; 8 \, c \,)

A_4 \; = \; ( \,-15 \, a \; + \; 24 \, b \; - \; 8 \, c \,)
B_4 \; = \; ( \,-20 \, a \; + \; 33 \, b \; - \; 12 \, c \,)
C_4 \; = \; ( \,-24 \, a \; + \; 40 \, b \; - \; 15 \, c \,)

A_5 \; = \; ( \,-24 \, a \; + \; 40 \, b \; - \; 15 \, c \,)
B_5 \; = \; ( \,-30 \, a \; + \; 51 \, b \; - \; 20 \, c \,)
C_5 \; = \; ( \,-35 \, a \; + \; 60 \, b \; - \; 24 \, c \,)

…………………………….
…………………………….

( \,A_0, \; B_0, \; C_0 \,) \; = \; ( \,-1, \; 5, \; 7 \,)
( \,A_1, \; B_1, \; C_1 \,) \; = \; ( \,7, \; 17, \; 23 \,)
( \,A_2, \; B_2, \; C_2 \,) \; = \; ( \,23, \; 37, \; 47 \,)
( \,A_3, \; B_3, \; C_3 \,) \; = \; ( \,47, \; 65, \; 79 \,)
( \,A_4, \; B_4, \; C_4 \,) \; = \; ( \,79, \; 101, \; 119 \,)
( \,A_5, \; B_5, \; C_5 \,) \; = \; ( \,119, \; 145, \; 167 \,)
…………………………………………..
…………………………………………..

 

 
 
 

 
 
 
 
 
 
 

Advertisements

About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

2 Responses to Integers (a,b,c) whose squares form an A.P. — Part 2

  1. paul says:

    Using the An, Bn, Cn formulae doesnt generate squares in Ap, the results are in Ap with the same common difference as the original Ap though. I use the method to generate squares in Ap from Pythagorean triples. Lets look at the (3, 4, 5) PT, from this we can generate the (1, 25, 49) squares in Ap with difference 24, by this method.

    1 = (4 – 3)^2
    25 = 5^2
    49 = (4 + 3)^2

    or for the (12, 35, 37) PT generates

    (529, 1369, 2209) or (23^2, 37^2, 47^2) CD 840.

    This method works for all PT’s

    Paul

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s