x^4 + y^4 = z^4 – N with (x,y,z) in arithmetic sequence

 

x^4 \; + \; y^4 \; = \; z^4 \; - \; N   with   z > y > x > 0   and in arithmetic sequence.

 
There are solutions of the form:

1^4 + 2^4 = 3^4 - 4^3              7^4 + 8^4 = 9^4 - 4^3

2^4 + 4^4 = 6^4 - 4^5              14^4 + 16^4 = 18^4 - 4^5

4^4 + 8^4 = 12^4 - 4^7              28^4 + 32^4 = 36^4 - 4^7

8^4 + 16^4 = 24^4 - 4^9              56^4 + 64^4 = 72^4 - 4^9

4^3   ……….    (1,   2,   3),   (7,   8,   9)
4^5   ……….    (2,   4,   6),   (14,   16,   18)
4^7   ……….    (4,   8,   12),   (28,   32,   36)
4^9   ……….    (8,   16,   24),   (56,   64,   72)

4^{11}   ……….   (16,   32,   48),   (112,   128,   144)
4^{13}   ……….   (32,   64,   96),   (224,   256,   288)
4^{15}   ……….   (64,   128,   192),   (448,   512,   576)
4^{17}   ……….   (128,   256,   384),   (896,   1024,   1152)
4^{19}   ……….   (256,   512,   768),   (1792,   2048,   2304)

 

 
the pattern breaks:

5^4 + 10^4 = 15^4 - 200^2                35^4 + 40^4 = 45^4 - 200^2

6^4 + 12^4 = 18^4 - 288^2                 42^4 + 48^4 = 54^4 - 288^2

200 = 2^3 \; \times \; 5^2                          288 = 2^5 \; \times \; 3^2

7^4 + 14^4 = 21^4 - 392^2                49^4 + 56^4 = 63^4 - 392^2

392 = 2^3 \; \times \; 7^2

9^4 + 18^4 = 27^4 - 648^2              63^4 + 72^4 = 81^4 - 648^2

648 = 2^3 \; \times \; 3^4

10^4 + 20^4 = 30^4 - 800^2              70^4 + 80^4 = 90^4 - 800^2

800 = 2^5 \; \times \; 5^2

 
To be continued
 
 
 
 

 
 
 
 
 
 
 
 

 
 
 

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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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