Re: Diophantine eq. a^4+b^4=d*c^2 and specialness of d=17
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:
ft+gs 784 28
gt-fs 25 5
d 2129 289
as my 2nd example
--- In email@example.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).
> (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.
- This will probably read better as Yahoo will not respect my lining up.
ft+gs 784 (28^2)
gt-fs 25 (5^2)
sum 615281 divided by 289 gives
as my 2nd example
- This perhaps suggests the conjecture that the set of allowed d's
is "multiplicative" -- i.e. if there is a solution with d=X, and another with d=Y,
then there automatically will be a solution with d=X*Y?
But a counterexample to that seems to be
which apparently is not allowed.
> This will probably read better as Yahoo will not respect my lining up.
> f 8
> g 15
> t 23
> s 40
> ft+gs 784 (28^2)
> gt-fs 25 (5^2)
> 28^4 614656
> 5^4 625
> sum 615281 divided by 289 gives
> d 2129
> as my 2nd example
> So, 5^4+28^4=(23^2+40^2)*(8^2+15^2)=2129^2*17^2
- 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).
x^4 - n^2 * y^4 = t^2 * z^2
large table of all small congruent numbers:
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
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:
- It appears that the following papers (most of which are difficult for me to obtain,
and I haven't) likely have something important to say about this diophantine equation:
H.C.Pocklington: Some diophantine impossibilities, Proc Cambridge Philos Soc 17 (1914) 110-118.
Same title & author: 17 (1913) 108-121.
Trygve Nagell: Sur l'impossibilite de quelques..., Norsk Mat. For Skriftern I #13 Kristiana 1923.
C.E.Lind: Untersuchungen :uber die rationalen Punkte der ebenen kubischen Kurven von Geschlecht,
Dissertation Uppsala (1940).
> THEOREM (WDS using by using idea of E.B. Escott 1900):typo correction:
> 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."
THEOREM (WDS using by using idea of E.B. Escott 1900):
In order for a^4+b^4=d*c^2 to be soluble in integers (a,b,c), it
is necessary that 2*d be a "congruent number."
> H.C.Pocklington: Some diophantine impossibilities, Proc Cambridge Philos Soc 17 (1914) 110-118.I obtained Pocklington (only the 2nd paper actually exists; the first was repeated cites by numerous authors who'd obviously never seen it but kept citing it anyway). It contains J.Wroblewski's and P.Carmody's theorems, proven in a less-clear way than JW proved it, but no new results for us aside from that.
> Same title & author: 17 (1913) 108-121.
- --- In firstname.lastname@example.org,
"WarrenS" <warren.wds@...> wrote:
> H.C.Pocklington: Some diophantine impossibilities,...
> Proc Cambridge Philos Soc
> 17 (1913) 108-121.A scanned version is freely available:
- --- In email@example.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.