When the product of any two of (a,b,c,d) is one less an integer squared

 
 
 
a, \; b, \; c, \; d   are positive integers such that

1 \; + \; a \, b
1 \; + \; a \, c
1 \; + \; b \, c
1 \; + \; a \, d
1 \; + \; b \, d
1 \; + \; c \, d

are all square numbers.

 
 
Here’s a simple parameterization :

a \; = \; n \; - \; 1
b \; = \; n \; + \; 1
c \; = \; 4 \, n
d \; = \; 4 \, n \, (4n^2 - 1)
 

1 \; + \; a \; b \; = \; 1 + (n - 1) \;(n + 1) = n^2

1 \; + \; a \; c \; = \; 1 + 4 \; n \;(n - 1) = (2 \; n - 1)^2

1 \; + \; b \; c \; = \; 1 + 4 \; n \;(n + 1) = (2 \; n + 1)^2

1 \; + \; a \; d \; = \; 1 + (n - 1) \;(4 \; n \; (4 \; n^2 - 1)) = (4 \; n^2 - 2 \; n - 1)^2

1 \; + \; b \; d \; = \; 1 + (n + 1) \;(4 \; n \; (4 \; n^2 - 1)) = (4 \; n^2 + 2 \; n - 1)^2

1 \; + \; c \; d \; = \; 1 + (4 \; n) \;(4 \; n \; (4 \; n^2 - 1)) = (8 \; n^2 - 1)^2

 

Here’s parameterization #2 ;

a \; = \; m
b \; = \; n \, (m \, n +2)
c \; = \; (n + 1) \, (m \, n + m + 2)
d \; = \; 4 \, (m \, n + 1) \, (m \, n + m + 1) \, (m \, n^2 + m \, n + 2 \, n + 1)

1 + a \, b = 1 + m \, n \, (m \, n + 2) = (m \, n + 1)^2

1 + a \, c = 1 + m \, (n+1) \, (m \, n + m + 2) = (m \, n + m + 1)^2

1 + b \, c = 1 + n \, (m \, n +2) \,(n+1) \, (m \, n + m + 2) = (m \, n^2 + m \, n + 2 \, n + 1)^2

1 + a \, d
= 1 + m \, (4 (m \, n + 1) (m \, n + m + 1) \, (m \, n^2 + m \, n + 2 \, n + 1))
= (2 \, m^2 \, n^2 + 2 \, m^2 \, n + 4 \, m \, n + 2 \, m + 1)^2

1 + b \, d
= 1 + n \, (m \, n +2) \,(4 (m \, n + 1) \, (m \, n + m + 1) \, (m \, n^2 + m \, n + 2 \, n + 1)))
= (2 \, m^2 \, n^3 + 2 \, m^2 \, n^2 + 6 \, m \, n^2 + 4 \, m \, n + 4 \, n + 1)^2

1 + c \, d
= 1 + (n + 1) \, (m \, n + m + 2) \,(4 \, (m \, n + 1) \, (m \, n + m + 1) \, (m \, n^2 + m \, n + 2 \, n + 1))
= (2 \, m^2 \, n^3 + 4 m^2 \, n^2 + 2 \, m^2 \, n + 6 m \, n^2 + 8 \, m \, n + 2 \, m + 4 \, n + 3)^2

 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

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

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