To make {(B-A),(C-A),(C-B),(D-A),(D-B),(D-C)} squares

 
 
To find four positive integers   A, B, C, D   such that
B - A,
C - A,     C - B,
D - A,     D - B,     D - C

are all squares.

 
Here are 4 sets of solutions:
 
A \; = \; k_1
B \; = \; k_1 \; + \; 3404^2
C \; = \; k_1 \; + \; 3404^2 \; + \; 1680^2
D \; = \; k_1 \; + \; 6005^2

k_1 \; > \; 0

B \; - \; A \; = \; 3404^2
C \; - \; A \; = \; 3404^2 \; + \; 1680^2 \; = \; 3796^2
C \; - \; B \; = \; 1680^2
D \; - \; A \; = \; 6005^2
D \; - \; B \; = \; 6005^2 \; - \; 3404^2 \; = \; 4947^2
D \; - \; C \; = \; 6005^2 \; - \; (3404^2 + 1680^2) \; = \; 4653^2

 

A \; = \; k_2
B \; = \; k_2 \; + \; 9306^2
C \; = \; k_2 \; + \; 9306^2 \; + \; 3360^2
D \; = \; k_2 \; + \; 12010^2

k_2 \; > \; 0

B \; - A \; = \; 9306^2
C \; - A \; = \; 9306^2 \; + \; 3360^2 \; = \; 9894^2
C \; - B \; = \; 3360^2
D \; - A \; = \; 12010^2
D \; - B \; = \; 12010^2 \; - \; 9306^2 \; = \; 7592^2
D \; - C \; = \; 12010^2 \; - \; (9306^2 + 3360^2) \; = \; 6808^2

 

A \; = \; k_3
B \; = \; k_3 \; + \; 2576^2
C \; = \; k_3 \; + \; 2576^2 \; + \; 1332^2
D \; = \; k_3 \; + \; 8660^2

k_3 \; > \; 0

B \; - \; A \; = \; 2576^2
C \; - \; A \; = \; 2576^2 \; + \; 1332^2 \; = \; 2900^2
C \; - \; B \; = \; 1332^2
D \; - \; A \; = \; 8660^2
D \; - \; B \; = \; 8660^2 \; - \; 2576^2 \; = \; 8268^2
D \; - \; C \; = \; 8660^2 \; - \; (2576^2 + 1332^2) \; = \; 8160^2

 

A \; = \; k_4
B \; = \; k_4 \; + \; 831830^2
C \; = \; k_4 \; + \; 831830^2 \; + \; 742560^2
D \; = \; k_4 \; + \; 1116986^2

k_4 \; > \; 0

B \; - \; A \; = \; 831830^2
C \; - \; A \; = \; 831830^2 \; + \; 742560^2 \; = \; 1115050^2
C \; - \; B \; = \; 742560^2
D \; - \; A \; = \; 1116986^2
D \; - \; B \; = \; 1116986^2 \; - \; 831830^2 \; = \; 745464^2
D \; - \; C \; = \; 1116986^2 \; - \; (831830^2 + 742560^2) \; = \; 65736^2

 
 
Can you find other types of solutions?

 
 
 
 
 
 

 
 

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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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