Solve in rational numbers the system of equations — Part 2

 
 

        x \; + \; y \; + \; z \; + \; x^2 \; = \; a^2

    x \; + \; y \; + \; z \; + \; x^2 \; + \; y^2 \; = \; b^2

x \; + \; y \; + \; z \; + \; x^2 \; + \; y^2 \; + \; z^2 \; = \; c^2

 
 
Solutions:

 
(1)

Integral solutions:     x = 35,   y = 144,   z = 360

        35 + 144 + 360 + 35^2 = 1764 = 42^2

    35 + 144 + 360 + 35^2 + 144^2 = 22500 = 150^2

35 + 144 + 360 + 35^2 + 144^2 + 360^2 = 152100 = 390^2

 
 

(2)

x = 16/5,     y = 48/5,     z = 144/5
 

(16/5) + (48/5) + (144/5) + (16/5)^2 = (36/5)^2

(16/5) + (48/5) + (144/5) + (16/5)^2 + (48/5)^2 = 12^2

(16/5) + (48/5) + (144/5) + (16/5)^2 + (48/5)^2 + (144/5)^2 = (156/5)^2

 
 

(3)

Not only can three numbers be found by this method but also four can be found by means of four square numbers:
 

x = 1295,     y = 31968/7,     z = 79920/7,     w = 79920

x + y + z + w + x^2

1295 + 31968/7 + 79920/7 + 79920 + 1295^2 = 1774224 = 1332^2
 

x + y + z + w + x^2 + y^2

1295 + (31968/7) + (79920/7) + 79920 + 1295^2 + (31968/7)^2
= 1108890000/49
= ( \,33300/7 \,)^2
 

x + y + z + w + x^2 + y^2 + z^2

1295 + (31968/7) + (79920/7) + 79920 + 1295^2 + (31968/7)^2 + (79920/7)^2
= 7496096400/49
= ( \,86580/7 \,)^2
 

x + y + z + w + x^2 + y^2 + z^2 + w^2

1295 + (31968/7) + (79920/7) + 79920 + 1295^2 + (31968/7)^2 + (79920/7)^2 + 79920^2
= 320469210000/49
= ( \,566100/7 \,)^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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