1.Introduction

By Tito Piezas[1],it seems that a^4 + b^4 = c^4 + 12d^4 has a follwing parametric solution.
  
(192n^7)^4 + (192n^8-24n^4-1)^4 = (192n^8+24n^4-1)^4 + 12(2n)^4.(Noam Elkies)

And a^4 + b^4 + 4c^4 = d^4 has a parametric solution as
(p^4-2q^4)^4 + (2p^3q)^4 + 4(2pq^3)^4 = (p^4+2q^4)^4.(Carmichael)

It seems that these Elkies and Carmichael type identity have more other solutions.

So,I searched the parametric solutions of x^4 + ay^4 = z^4 + bt^4. 

I show only six solutions.


[1].Tito Piezas:http://sites.google.com/site/tpiezas/updates03


2. Search results

Parametric solutions of x^4 + ay^4 = z^4 + bt^4 where (a,b) < 50.
Smallest examples are chosen about each (a,b).

1.Elkies type

(a,b)=(4,3)
(12x^8 + 6x^4 - 1)^4 + 3(2x)^4    = (12x^8 - 6x^4 - 1)^4 + 4(12x^7)^4

(a,b)=(6,2)
(48x^8 + 12x^4 - 1)^4 + 6(2x)^4   = (48x^8 - 12x^4 - 1)^4 + 2(48x^7)^4

(a,b)=(24,8)
(3x^8 + 3x^4 - 1)^4 + 24(x)^4     = (3x^8 - 3x^4 - 1)^4 + 8(3x^7)^4

(a,b)=(36,27)
(972x^8 + 54x^4 - 1)^4 + 27(2x)^4 = (972x^8 - 54x^4 - 1)^4 + 36(324x^7)^4

2.Carmichael type

(a,b)=(27,12)
(24x^4 + 1)^4 = (24x^4 - 1)^4 + 27(8x^3)^4 + 12(2x)^4

(a,b)=(36,9)
(18x^4 + 1)^4 = (18x^4 - 1)^4 + 36(6x^3)^4 + 9(2x)^4




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