## 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