Set {a,b,c,d} such that the product of any two of them increased by 1 is a square — Part 3

 
 
1 + a \,b = A^2   ………………….   1 + b \,c = D^2
1 + a \,c = B^2   ………………….   1 + b \,d = E^2
1 + a \,d = C^2   ………………….   1 + c \,d = F^2

 

a = 3
b = 3 \, n^2 - 4 \, n + 1
c = 3 \, n^2 + 2 \, n
d = 4 \, (27 \, n^4 - 18 \, n^3 - 12 \, n^2 + 5 \, n + 2)

1 + a \,b = (3 n - 2)^2   ……………………   1 + b \,c = (3 \, n^2 - n - 1)^2
1 + a \,c = (3 \, n + 1)^2   ……………………   1 + b \,d = (18 \, n^3 - 18 \, n^2 - 2 \, n + 3)^2
1 + a \,d = (18 \, n^2 - 6 \, n - 5)^2   ………..   1 + c \,d = (18 \, n^3 - 8 \, n - 1)^2

diophantus-6

 
Or
 

a = 3
b = 3 \, n^2 - 2 \, n
c = 3 \, n^2 + 4 \, n + 1
d = 4 \, (27 \, n^4 + 18 \, n^3 - 12 \, n^2 - 5 \, n + 2)

1 + a \,b = (3 \, n - 1)^2   ……………………   1 + b \,c = (3 \, n^2 + n - 1)^2
1 + a \,c = (3 \, n + 2)^2   ……………………   1 + b \,d = (18 \, n^3 - 8 \, n + 1)^2
1 + a \,d = (18 \, n^2 + 6 \, n - 5)^2   ………..   1 + c \,d = (18 \, n^3 + 18 \, n^2 - 2 \, n - 3)^2

diophantus-7

 
                                 …………………………………………………………..    
 
a = 4
 

a = 4
b = n^2 + n
c = n^2 + 5 \, n + 6
d = 4 \, (4 \, n^4 + 24 \, n^3 + 45 \, n^2 + 27 \, n + 5)

1 + a \,b = (2 n + 1)^2   ……………………………..   1 + b \,c = (n^2 + 3 \, n + 1)^2
1 + a \,c = (2 n + 5)^2   ……………………………..   1 + b \,d = (4 \, n^3 + 14 \, n^2 + 10 \, n + 1)^2
1 + a \,d = (8 n^2 + 24 n + 9)^2   ………………….   1 + c \,d = (4 \, n^3 + 22 \, n^2 + 34 \, n + 11)^2

diophantus-5

 
                                 …………………………………………………………..    
 
a = 5
 

a = 5
b = 5 \, n^2 - 8 \, n + 3
c = 5 \, n^2 + 2 \, n
d = 4 \, (125 \, n^4 - 150 \, n^3 + 27 \, n + 4)

1 + a \,b = (5 \, n - 4)^2   ……………………….   1 + b \,c = (5 \, n^2 - 3 \, n - 1)^2
1 + a \,c = (5 \, n + 1)^2   ……………………….   1 + b \,d = (50 \, n^3 - 70 \, n^2 + 14 \, n + 7)^2
1 + a \,d = (50 \, n^2 - 30 \, n - 9)^2   ………….   1 + c \,d = (50 \, n^3 - 20 \, n^2 - 16 \, n - 1)^2

diophantus-3
 
10 \, (1 + b \,c) \; + \; 1 \; = \; 1 \; + \; a \,d

 
Or
 

a = 5
b = 5 \, n^2 - 2 \, n
c = 5 \, n^2 + 8 \, n + 3
d = 4 \, (125 \, n^4 + 150 \, n^3 - 27 \, n + 4)

1 + a \,b = (5 \, n - 1)^2   ……………………….   1 + b \,c = (5 \, n^2 + 3 \, n - 1)^2
1 + a \,c = (5 \, n + 4)^2   ……………………….   1 + b \,d = (50 \, n^3 + 20 \, n^2 - 16 \, n + 1)^2
1 + a \,d = (50 \, n^2 + 30 \, n - 9)^2   ………….   1 + c \,d = (50 \, n^3 + 70 \, n^2 + 14 \, n - 7)^2

diophantus-4
 
10 \, (1 + b \,c) \; + \; 1 \; = \; 1 \; + \; a \,d

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

Advertisements

About benvitalis

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

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