THEOREM (WDS using by using idea of E.B. Escott 1900):

In order for a^4+b^4=d*c^2 to be soluble integers (a,b,c), it

is necessary that d be a "congruent number."

PROOF & EXPLANATION:

A "congruent number" is the area of a rational right triangle.

For example 2006 is a congruent number since it is the area of a

right triangle with legs 340/3 and 177/5

and hypotenuse 1781/15.

157 also is a congruent number, but the only known rational triangles

showing this involve very big numbers.

Equivalently a number n is "a congruent number" iff

x^2 + n*y^2 = z^2 and x^2 - n*y^2 = t^2

have simultaneous integer solutions (x,y,z,t).

Equivalently if

x^4 - n^2 * y^4 = t^2 * z^2

is soluble.

See

http://oeis.org/A003273
http://oeis.org/A006991
large table of all small congruent numbers:

http://oeis.org/A003273/b003273.txt
which makes it look like about 57% of positive integers are conguent numbers.

Ronald Alter: The congruent number problem, Amer. Math. Monthly 87 (1980) 43-45.

Jaap Top & Noriko Yui:

http://www.math.rug.nl/~top/19yui.pdf
Conjectured asymptotics for congruent numbers see pages 453-4 of

Henri Cohen: Number Theory I, Tools and Diophantine Equations, Springer-Verlag 2007

based on the Birch and Swinnerton-Dyer conjecture!!

E.B.Escott in 1900 showed that a^4+b^4=d*c^2

MIGHT be soluble if 2*d is a so-called "congruent number."

EB Escott: L'Intermediaire des Math. 7 (1900) 199, replying to 3 (1896) 130.

Let me explain.

Escott's method was to set a=c*K/L to get a quadratic for c^2 namely

c^4 * K^4 + b^4 * L^4 = d * L^4 * c^2.

This has rational roots c^2 iff

d^2 * L^4 - 4*K^4*b^4 = square * L^4

equivalently (upon multiplying by d^2) if

(d*L)^4 - 4*d^2 * (K*b)^4 = (d*M*L^2)^2.

Thus if 2*d is a congruent number, then a rational solution (a,b,c^2)

exists because suitable rational L and K and M

must exist. Then this a,b,c^2 may be converted from rational to integer form

by multiplying by suitable 4th powers of integers.

Escott proposed this as a way to find (a,b,c)... BUT

this solution might not have rational c, it only will have rational c^2.

So this is not a very useful method.

But reversing the logic, a solution (a,b,c) existing with a,b integer and c^2 rational

forces d to be a congruent number.

PROOF COMPLETE. [sorry, it took a while...]

This gets into very deep waters connected to the

"Birch and Swinnerton-Dyer conjecture" which is a Clay prize problem

worth $1,000,000.

R. Alter, T. B. Curtz, K. K, Kubota: Remarks and results on congruent numbers,

Proc. Third Southeastern Conf. on Combinatorics, Graph Theory and

Computing (1972) 2735.

conjectured that if n mod 8 = 5, 6 or 7 then n is a congruent number. This was shown to be true if the weak Birch and Swinnerton-Dyer conjecture is by

Nelson M. Stephens: Congruence properties of congruent numbers,

Bull. London Math. Soc. 7 (1975) 182-184.

This indicates for our problem that a solution (a,b,c) might exist if d=3 mod 8...

but we know from J.Wroblewski's theorem posted earlier that these d NEVER work.

Monsky showed 2*d is a congruent number if d=p*q with p,q prime and

p=1 (mod8) and q=7(mod8) with p nonsquare mod q,

Keqin Feng:Non-congruent Numbers, Odd graphs and the B-S-D Conjecture,

Acta Arith. LXXV 1 (1996) 7183

F. Lemmermeyer: Some families of non-congruent numbers,

Acta. Arith. 110 (2003) 1536

but again we know from J.Wroblewski's theorem that Monsky's d NEVER work.

So it does not seem that this Escott condition is very useful if you try to use it as a sufficient condition -- but it works as a necessary condition!

The congruent numbers up to 2000 had been computed by 1986

G. Kramarz: All congruent numbers less than 2000,

Math. Annalen 273 (1986) 337340.

A huge list of the first TRILLION congruent numbers (claimed to be correct if

Birch & Swinnerton-Dyer conjecture correct) was computed by

Robert Bradshaw, William B. Hart, David Harvey,

Gonzalo Tornaria, Mark Watkins:

Congruent number theta coefficients to 10^12:

http://www.warwick.ac.uk/~masfaw/congruent.pdf .