To make {(x^2 ± x*y + y^2), (x^2 ± x*z + z^2), (y^2 ± y*z + z^2)} squares

 

Find positive integers   x, y, z   such that

(1)

x^2 \; - \; x \, y \; + \; y^2
x^2 \; - \; x \, z \; + \; z^2
y^2 \; - \; y \, z \; + \; z^2

are square numbers.

 
Here are some solutions:

(x, \; y, \; z) \; = \; (3, \; 8, \; 3)

3^2 \; - \; (3) \,(8) \; + \; 8^2 \; = \; 7^2
3^2 \; - \; (3) \,(3) \; + \; 3^2 \; = \; 3^2
8^2 \; - \; (8) \,(3) \; + \; 3^2 \; = \; 7^2

(x, \; y, \; z) \; = \; (5, \; 8, \; 5)

5^2 \; - \; (5) \,(8) \; + \; 8^2 \; = \; 7^2
5^2 \; - \; (5) \,(5) \; + \; 5^2 \; = \; 5^2
8^2 \; - \; (8) \,(5) \; + \; 5^2 \; = \; 7^2

(x, \; y, \; z) \; = \; (5, \; 21, \; 5)

5^2 \; - \; (5) \,(21) \; + \; 21^2 \; = \; 19^2
5^2 \; - \; (5) \,(5) \; + \; 5^2 \; = \; 5^2
21^2 \; - \; (21) \,(5) \; + \; 5^2 \; = \; 19^2

(x, \; y, \; z) \; = \; (7, \; 40, \; 7)

7^2 \; - \; (7) \,(40) \; + \; 40^2 \; = \; 37^2
7^2 \; - \; (7) \,(7) \; + \; 7^2 \; = \; 7^2
40^2 \; - \; (40) \,(7) \; + \; 7^2 \; = \; 37^2

(x, \; y, \; z) \; = \; (8, \; 3, \; 8)

8^2 \; - \; (8) \,(3) \; + \; 3^2 \; = \; 7^2
8^2 \; - \; (8) \,(8) \; + \; 8^2 \; = \; 8^2
3^2 \; - \; (3) \,(8) \; + \; 8^2 \; = \; 7^2

(x, \; y, \; z) \; = \; (8, \; 5, \; 8)

8^2 \; - \; (8) \,(5) \; + \; 5^2 \; = \; 7^2
8^2 \; - \; (8) \,(8) \; + \; 8^2 \; = \; 8^2
5^2 \; - \; (5) \,(8) \; + \; 8^2 \; = \; 7^2

(x, \; y, \; z) \; = \; (8, \; 15, \; 8)

8^2 \; - \; (8) \,(15) \; + \; 15^2 \; = \; 13^2
8^2 \; - \; (8) \,(8) \; + \; 8^2 \; = \; 8^2
15^2 \; - \; (15) \,(8) \; + \; 8^2 \; = \; 13^2

 
 

Can you find positive integers   x, y, z   such that

(2)

x^2 \; + \; x \, y \; + \; y^2
x^2 \; + \; x \, z \; + \; z^2
y^2 \; + \; y \, z \; + \; z^2

are square numbers?

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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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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