(a,b,c,d) the product of any two positive integers is one less than a square

 
 
Given any four positive integers   a, \; b, \; c, \; d   such that the product of any two is one less than a square number

 
If we let

a \; = \; n
b \; = \; n \; + \; 2
c \; = \; 4 \, n \; + \; 4
d \; = \; 16 \, n^3 \; + \; 48 \, n^2 \; + \; 44 \, n \; + \; 12

then,

a \, b \; + \; 1 \; = \; (n + 1)^2
a \, c \; + \; 1 \; = \; (2 \, n + 1)^2
a \, d \; + \; 1 \; = \; (4 \, n^2 + 6 \, n + 1)^2
b \, c \; + \; 1 \; = \; (2 \, n + 3)^2
b \, d \; + \; 1 \; = \; (4 \, n^2 + 10 \, n + 5)^2
c \, d \; + \; 1 \; = \; (8 \, n^2 + 16 \, n + 7)^2

and

((a + b) - (c + d))^2
= \; (n+n+2-4 \, n-4-16 \, n^3-48 \, n^2-44 \, n-12)^2
= \; (-16 \, n^3 - 48 \, n^2 - 46 \, n - 14)^2
= \; 4 \, (n + 1)^2 \; (8 \, n^2 + 16 \, n + 7)^2

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

((a + c) - (b + d))^2
= \; ((n+4 \, n+4)-(n+2+16 \, n^3+48 \, n^2+44 \, n+12))^2
= \; (-16 \, n^3 - 48 \, n^2 - 40 \, n - 10)^2
= \; 4 \, (2 \, n + 1)^2 \; (4 n^2 + 10 \, n + 5)^2

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

((a + d) - (b + c))^2
= \; ((n+16 \, n^3+48 \, n^2+44 \, n+12)-(n+2+4 \, n+4))^2
= \; (16 \, n^3 + 48 \, n^2 + 40 \, n + 6)^2
= \; 4 \, (2 \, n + 3)^2 \; (4 \, n^2 + 6 \, n + 1)^2

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

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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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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