## 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   $n > 0$

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

$A_0 = a$   … $B_0 = b$   …   $C_0 = c$

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

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

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

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

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

…………………………….
…………………………….

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