- Warren,

As a reply to your personal email to me, here is a further explanation. My typo below caused the confusion.

*So (a^2)^2+(b^2)^2 = ((s^2)^2+(t^2)^2)*c^2* is a typo caused by a false cut and paste and should have read

*So (a^2)^2+(b^2)^2 = ((s^2)+(t^2))*c^2*

My examples can probably be understood from the following general identity.

(f^2+g^2)*(s^2+t^2) = (ft+gs)^2 + (gt-fs)^2

So to put figures to it using an 8,15,17 right angled triangle with f=8,g=15,c=17 we get:

f 8

g 15

t 23

s 40

ft+gs 784 28

gt-fs 25 5

28^4 614656

5^4 625

sum 615281

d 2129 289

as my 2nd example

So, 5^4+28^4=(23^2+40^2)*(8^2+15^2)=2129^2*17^2--- In primenumbers@yahoogroups.com, "John" <mistermac39@...> wrote:

>

>

>

> 17, 68, 82, 97, 113, 153, 193, 257, 272, 274, 328, 337, 388, 425, 433,

> 452, 514, 577, 593, 612, 626, 641, 673, 706, 738, 772, 833, 873, 881,

> 914, 1017.

>

> I take it you have noticed that these are all of the form s^2+t^2.

>

> So (a^2)^2+(b^2)^2 = ((s^2)^2+(t^2)^2)*c^2

>

> Now, if c is the hypotenuse of an integer right angled triangle, all sides mutually prime, then you get c^2 = f^2+g^2 say, so it is easy to get an a^2 = (f*t+g*s) and b^2 = abs(f*s-g*t).

>

> Example:

> (8,13,17) with d = 113, where s=7 and t=8, f=8, g=15, giving b=8, a=13, and c=17. [triangle used was (8,15,17)

>

> So you might try to derive by keeping c=17 (5,28,17) using Triangle (8,15,17) and get a d = 2129.

>

> Whether there is a fancier way of going about it, I do not know, but at least this way is fun for an amateur.

> - --- In primenumbers@yahoogroups.com, "John" <mistermac39@...> wrote:
>

Upon looking in Henri Cohen: Number Theory volume I, Springer 2007 GTM 239, pages 392-395,lo and behold, he has a discussion of the diophantinea*X^4+b*Y^4=c*Z^2leading to a complete solution of our problem!

> This is a copy of a post sent to me by Warren.

>

>

>

>

This 2-volume book by Cohen is packed with very powerful modern stuff and dispenses with a lot of old gunk. Specialized to our case a^4+b^4=d*c^2,here is the result.

THEOREM [H.Cohen].For integer d>=3 fixed, a^4+b^4=d*c^2either has an infinite number of inequivalent nonzero solutions(a,b,c), or no nonzero solutions.

For infinite solutions d must have squarefree part divisible only by 2 and primes 8*k+1. Then the question is related to this elliptic curve EC:y^2 = x*(x^2 + d^2)and we have infinite solutions if and only ifthe class of d modulo squares belongs to the image of EC's 2-descent map (described in section 8.3 of Cohen).Here is how the problem maps to EC:x=d*a^(-2)*b^2, y=d*d*b*c*a^(-3).

Under the Birch/Swinnerton-Dyer conjecture, the rank of EC willalways be even. Cohen p395 gives the following table, computed with MAGMA and MWRANK using above theorem, listing every d with 3<=d<=10001 such that infinite solutions exist.

17, 82, 97, 113, 193, 257, 274, 337, 433, 514, 577, 593, 626, 641, 673, 706, 881, 914, 929, 1153, 1217, 1297, 1409, 1522, 1777, 1873, 1889, 1921, 2129, 2402, 2417, 2434, 2482, 2498, 2642, 2657, 2753, 2801, 2833, 2897, 3026, 3121, 3137, 3298, 3329, 3457, 3649, 3697, 4001, 4097, 4129, 4177, 4226, 4289, 4481, 4546, 4561, 4721, 4817, 4993, 5281, 5554, 5617, 5666, 5729, 5906, 6002, 6353, 6449, 6481, 6497, 6562, 6577, 6673, 6817, 6866, 7057, 7186, 7489, 7522, 7537, 7633, 7762, 8017, 8081, 8737, 8753, 8882, 8962, 9281, 9298, 9553, 9586, 9649, 9778, 9857, 10001. Cohen notes that d=1513=17*89 is not there and notes d=2801 was a difficult case for his software.