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• ## Diophantine eq. a^4+b^4=d*c^2 and specialness of d=17

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• Fermat in about 1640 famously proved a^4+-b^4=c^2 has no positive integer solutions, thus proving Fermat s last theorem in the case of exponent=4. Also
Message 1 of 24 , Dec 12, 2011
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Fermat in about 1640 famously proved a^4+-b^4=c^2 has no positive
integer solutions, thus proving "Fermat's last theorem" in
the case of exponent=4.

Also a^4-b^4=2*c^2 has no positive integer solutions.
These results are shown in T.Nagell:
Intro to number theory, Wiley 1951, pp.227-230.

However, a^4+b^4=2*c^2 trivially has a ton of solutions a=b, c=a^2.
If we exclude the trivial by demanding b>a>0 though, I don't know of
any solutions. [For a^4+b^4=2*c^2 it is helpful to
let a^2=(x-y) and b^2=(x+y) and then x^2+y^2=c^2
whose solutions are completely known.]

Let us generalize.
a^4 + b^4 = d*c^2
with a,b,c,d integers and b>a>0.

My computer looked for solutions.
Search I: a^4+b^4=d*c^2 with 0<a<b<65536 and 0<d<65.
Result: a ton of solutions for d=17, but zero solutions for any other d.
a=1, b=2, d=17, c=1
a=2, b=4, d=17, c=4
a=2, b=13, d=17, c=41
a=3, b=6, d=17, c=9
a=4, b=8, d=17, c=16
...

Well, that's interesting: 17 is special.

Search II: a^4+b^4=d*c^2 with 0<a<b<=1024 and 0<d<=1024.
Result: The following d are "special":
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.
[This sequence is not in OEIS.]

Phil Carmody pointed out that really, we should only worry about squarefree d.
[The existence of an integer solution (a,b,c) for d=D
guarantees the existence of a RATIONAL
solution for d=D*k^2, namely use c/k not c.
If we then multiply a times k, b times k, and c/k times k^2
we get INTEGERS.] But the subsequence
d = 17, 82, 97, 113,...
of squarefree "special" d also is not in the OEIS.

If d is fixed, then any solution (a,b,c)
yields a boring infinite set of solutions (a*k, b*k, c*k^2).
So we really only should worry about "primitive" solutions
with gcd(a^2, b^2, c)=1.

Here are primitive solutions with d=17:
a=1, b=2, d=17, c=1
a=2, b=13, d=17, c=41
a=38, b=43, d=17, c=569
a=314, b=863, d=17, c=182209
a=859, b=1186, d=17, c=385241
a=2297, b=12134, d=17, c=35732401

What is going on here?
I do not really know, but an initial speculation is this.
When d=17, there is the easy solution
a=1, b=2, d=17, c=1.

But here is the thing: if there is one solution, then it can usually
be used to generate an infinite set of solutions. I got that from
Tito Piezas's online book
where he attributes it to A.Desboves (probably about 1880?).
So there are "special" d with an infinity of solutions, and there are
"usual" d with zero solutions (and conceivably there are a few "crazy"
d with a finite nonzero number of
solutions because the below Desboves method for generating more
solutions happens merely to regenerate old ones, which would be an
amazing miracle).

How to generate more solutions from one:
Piezas says Desboves said (as edited by me):
if
p^4+q^4 = d*r^2,
then
x^4+y^4 = d*z^2
where
x = p*(4*m^2-3*(m+n)^2),
y = q*(4*n^2-3*(m+n)^2),
z = r*(4*(m+n)^4-3*(m-n)^4)
where m=p^4 and n=q^4.

So for example starting from p=1, q=2, d=17, r=1
we get
x=863, y=314, z=182209
and sure enough
863^4 + 314^4 = 17*182209^2
is another solution.

More generally, if both b and d are fixed, then
a^4+b^4=d*c^2
is an "elliptic curve" enabling you to generate an infinite set of
rational solutions
(a,c) from just a few (indeed usually only one)
starter solutions by using the elliptic curve group.
This observation is much more powerful than Desbove's method of getting
new solutions from old.
Then by multiplying by appropriate denominators we get an infinite set of
integer solutions now with b not fixed.

Questions:
Q1. Can you find a parameterized family of d=17 solutions?
Q2. Can you understand the set of special d?
Q3. Can you prove there are no "crazy" d (or find them all)
• ... Theorem: Squarefree d cannot have prime factors other than p=2 and p=8k+1. Proof: Those are the only primes p allowing nontrivial solutions to the
Message 2 of 24 , Dec 12, 2011
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> Q2. Can you understand the set of special d?

Theorem: Squarefree d cannot have prime factors other than p=2 and p=8k+1.
Proof: Those are the only primes p allowing nontrivial solutions to
the congruence a^4+b^4=0 (mod p).

Conjecture 1: If squarefree d has odd number of prime factors 2, 16k+9
then there are no solutions.
Argument: The world of numbers is quite regular, so the set of d's
should come from a simple rule. Since we do no see solutions for
numbers like d=34 or d=41, there must be a hidden reason for that.

Conjecture 2. Other d's are not forbidden, hence they have infinitely
many solutions.
Argument: If solutions are not forbidden, then they should appear in
infinite number - this is what I always believe in.

Jarek
• On Tue, Dec 13, 2011 at 1:01 AM, Jaroslaw Wroblewski ... --WDS: Thanks Jarek. Your conjectures would totally solve it, if true. However, I am skeptical,
Message 3 of 24 , Dec 12, 2011
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On Tue, Dec 13, 2011 at 1:01 AM, Jaroslaw Wroblewski
<jaroslaw.wroblewski@...> wrote:
>> Q2. Can you understand the set of special d?
>
> Theorem: Squarefree d cannot have prime factors other than p=2 and p=8k+1.
> Proof: Those are the only primes p allowing nontrivial solutions to
> the congruence a^4+b^4=0 (mod p).
>
> Conjecture 1: If squarefree d has odd number of prime factors 2, 16k+9
> then there are no solutions.
> Argument: The world of numbers is quite regular, so the set of d's
> should come from a simple rule. Since we do no see solutions for
> numbers like d=34 or d=41, there must be a hidden reason for that.

> Conjecture 2. Other d's are not forbidden, hence they have infinitely
> many solutions.
> Argument: If solutions are not forbidden, then they should appear in
> infinite number - this is what I always believe in.
> Jarek

--WDS: Thanks Jarek. Your conjectures would totally solve it, if true.
However, I am skeptical, especially of conj. 2 and your "always believe in
infinite" theory. For example, I do not believe there are an
infinity of Fermat primes
despite no known reason that should be forbidden.

I also note this:

x^4 is always 0 or 1 (mod 16).
And x^2 is always 0,1,4,9 (mod 16).
Hence if a^4+b^4=d*c^2 (mod 16)
then the left hand side will be 0,1, or 2, and the right hand side
will be 0,d,4d, or 9d, all mod 16.
The 0=0 (mod 16) case can only occur if both a and b are even
and then either c is divisible by 4 [in which case the solution is nonprimitive]
or d is divisible by 4.

Consequently I conclude THEOREM:
d must obey d=0,1,2,4,8,9, or 12 mod 16.

It perhaps also is possible that analysis modulo prime powers will
say something interesting.
• ... Fermat primes are not forbidden, but any reasonable heuristic indicates probability of large Fermat numbers to be primes as incredibly low. In the case of
Message 4 of 24 , Dec 12, 2011
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> However, I am skeptical, especially of conj. 2 and your "always believe in
> infinite" theory.   For example, I do not believe there are an
> infinity of Fermat primes
> despite no known reason that should be forbidden.

Fermat primes are not forbidden, but any reasonable heuristic
indicates probability of large Fermat numbers to be primes as
incredibly low.

In the case of the eqaution a^4+b^4=d*c^2 a naive density heuristic
indicates that the number of triples (a,b,c) versus the size of
a^4+b^4-d*c^2 is such that one can expect infinitely many solutions
because of this sum accidentally hitting 0 (although the solutions
should be rather rare) - let me skip the details. Basicly it follows
from the inequality 1/4+1/4+1/2 >= 1 (sum of reciprocals of power
degrees).

Therefore I believe that what is not forbidden, is allowed in this case.

Jarek
• On Tue, Dec 13, 2011 at 2:10 AM, Jaroslaw Wroblewski ... --exactly, and it could well be that the probabilities for this problem also are just too low. ...
Message 5 of 24 , Dec 12, 2011
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On Tue, Dec 13, 2011 at 2:10 AM, Jaroslaw Wroblewski
<jaroslaw.wroblewski@...> wrote:
>> However, I am skeptical, especially of conj. 2 and your "always believe in
>> infinite" theory.   For example, I do not believe there are an
>> infinity of Fermat primes
>> despite no known reason that should be forbidden.
>
> Fermat primes are not forbidden, but any reasonable heuristic
> indicates probability of large Fermat numbers to be primes as
> incredibly low.

--exactly, and it could well be that the probabilities for this
problem also are just too low.

> In the case of the eqaution a^4+b^4=d*c^2 a naive density heuristic
> indicates that the number of triples (a,b,c) versus the size of
> a^4+b^4-d*c^2 is such that one can expect infinitely many solutions
> because of this sum accidentally hitting 0 (although the solutions
> should be rather rare) - let me skip the details. Basicly it follows
> from the inequality 1/4+1/4+1/2 >= 1 (sum of reciprocals of power
> degrees).
> Therefore I believe that what is not forbidden, is allowed in this case.

--oh. Well... maybe :)

I now think that a computerized exhaustive analysis modulo p^4 for
all small primes p, will tell us useful things about the allowed d.
But I have to sleep now :)
• I did a modular analysis modulo p^4 for small primes p. That is, I exhaustively found every solution of a^4+b^4=d*c^2 (mod p^4) and then listed the allowed
Message 6 of 24 , Dec 13, 2011
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I did a modular analysis modulo p^4 for small primes p.
That is, I exhaustively found every solution of
a^4+b^4=d*c^2 (mod p^4)
and then listed the allowed values of (d mod p^4).
The results are as follows:

(density of allowed ds mod 2^4)=7/16=0.437500
(density of allowed ds mod 3^4)=33/81=0.407407
(density of allowed ds mod 5^4)=233/625=0.372800
(density of allowed ds mod 7^4)=878/2401=0.365681
(density of allowed ds mod 11^4)=5406/14641=0.369237
(density of allowed ds mod 13^4)=10298/28561=0.360562
(density of allowed ds mod 17^4)=35453/83521=0.424480
(density of allowed ds mod 19^4)=48901/130321=0.375235

mod(2^4) analysis: d=0,1,2,4,8,9,12 are the only allowed d mod 16.
[confirming my previous manual analysis.]

mod(3^4) analysis: d=0,1,2,4,5,8,9,11,13,16,17,18,19,26,27,32,36,38,40,41,
43,45,46,54,59,61,63,65,71,72,76,77,80 are the only allowed d mod 81.

I won't print out the other lists since they are long.

Based on this, I would conjecture that the density of allowed d mod p^4 is always<0.5,
and hence that the set of d for which solutions exist has density=0 in the integers...
well, actually, we already knew that from JW's theorem, but this makes the density
zero even within the set JW's theorem had allowed.
• ... --Further such examples are d=5617=41*137: a=157, b=348, d=5617, c=1649 d=6497=73*89: a=7, b=8, d=6497, c=1 d=11521=41*281: a=150, b=419, d=11521,
Message 7 of 24 , Dec 13, 2011
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> a=266, b=675, d=3649=41*89, c=7633
> is a counterexample to J.Wroblewski's "conjecture 1" because both 41
> and 89 are 9 mod 16.

--Further such examples are
d=5617=41*137: a=157, b=348, d=5617, c=1649
d=6497=73*89: a=7, b=8, d=6497, c=1
d=11521=41*281: a=150, b=419, d=11521, c=1649
d=12193=89*137: a=18, b=43, d=12193, c=17

However, maybe none of these are REALLY counterexamples, because JW's precise
wording was

"an odd number of prime factors p=2, p=16*k+9" in d implies (conjecturally) no solutions.

I'd thought by "odd" JW was simply referring to the demand d be squarefree, but
perhaps he meant the total number of primes is odd... in which case note that in
all 5 of my counterexamples the number of prime factors happens to be 2.
• --I re-did the search this time using Carmody s theorem [wlog d is squarefree] my theorem [d=0,1,2,4,8,9,12 mod 16], and Wroblewski s theorem [squarefree d s
Message 8 of 24 , Dec 13, 2011
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--I re-did the search this time using Carmody's theorem [wlog d is squarefree]
my theorem [d=0,1,2,4,8,9,12 mod 16], and Wroblewski's theorem
[squarefree d's only prime factors are p=2 and p=8*k+1] to
restrict the set of d beforehand.

a=266, b=675, d=3649=41*89, c=7633
is a counterexample to J.Wroblewski's "conjecture 1" because both 41
and 89 are 9 mod 16.

It might be that JW was semi-correct in that squarefree d having factors (9 mod 16)
makes solutions (a,b,c) "rarer"... but not necessarily "forbidden."

I am dubious of JW's conjectured "infinity principle" in this case because the heuristic argument JW advanced ("infinite probability") is for our problem right on the borderline. Usually in those kind of arguments it is very clear which side of the borderline you are on, but in this case we are right at the borderline, and the fact that if there is one solution (a,b,c) then there are an infinite set (k*a, k*b, k*k*c) shows clearly that JW's crude probability model is very unrealistic (this is a huge count of solutions, far above what his model would have predicted). So the combination of being right on borderline plus
having an unrealistic probability model, to me means we should be skeptical of
the "infinity principle" in this case. I'm not saying it is wrong, I'm just saying it's
very near the borderline and I don;t have confidence in it.

Some d that might be counterexamples to JW's "conjecture 2" include
d=769=prime, and d=802=2*401 [and probably more too].
For both, no solutions (a,b,c) were found, but of course
there might be larger solutions outside
the range of the search. In most cases I only searched (a,b) up to 8192, but
for these two JW-refuting d, I searched up to 65536 (albeit there would have been some
integer-overflow issues near the top of that range so effectively this search was somewhat smaller).

These eligible d<4096 (shown with their prime factorizations)
were found to yield solutions:
17=17, 82=2*41, 97=97, 113=113, 193=193, 257=257, 274=2*137, 337=337, 433=433, 514=2*257, 577=577, 593=593, 626=2*313, 641=641, 673=673, 706=2*353, 881=881, 914=2*457, 1153=1153, 1297=1297, 1522=2*761, 1777=1777, 1873=1873, 1921=17*113, 2129=2129, 2402=2*1201, 2417=2417, 2434=2*1217, 2482=2*17*73, 2498=2*1249, 2642=2*1321, 2657=2657, 2753=2753, 2897=2897, 3026=2*17*89, 3121=3121, 3137=3137, 3298=2*17*97, 3457=3457, 3649=41*89, 3697=3697,

In all of these cases, a search of max(a,b)<4096 sufficed, EXCEPT
d=3457 required the full search range max(a,b)<8192.
• It seems there are theorems perhaps relevant to this by Aleksander Grelak and Aleksander Grytczuk: Some remarks on matrices and Diophantine equation A*x^2
Message 9 of 24 , Dec 15, 2011
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It seems there are theorems perhaps relevant to this by

Aleksander Grelak and Aleksander Grytczuk:
Some remarks on matrices and Diophantine equation A*x^2 − B*y^2 = C,
Discuss. Math. 10 (1990) 13-27.

A. Grelak, A. Grytczuk:
On the Diophantine equation a*x^2 - b*y^2=c. II
Discussiones Mathematicae 14 (1994) 21-24.

A. Grelak and A. Grytczuk:
On the Diophantine equation a*x^2 − b*y^2 = c, Publ. Math. Debrecen 44 (1994) 1-9.
Or is it 291-299?

[I have not actually seen these papers.]

Grelak & Grytczuk's Theorem:
A*X^2 = C+B*Y^2
[where A*B is not square, A,B are positive integer, C is either 1 or 2, A*C>1,
gcd(C, A*B)=1, and u0, v0 is the least positive integer solution of the Pell
equation U^2=1+A*B*V^2]
is soluble by integers X,Y iff
both (u0+1)*C/(2*A)
and (u0-1)*C/(2*B)
are squared integers.

Example: one special case relevant to us is A=2, B=C=1:
2*X^2 = 1 + Y^2
we find u0=3, v0=2
then (u0+1)*C/(2*A)=4/4=1 and (u0-1)*C/(2*B)=2/2=1.
Hence it is claimed to have a solution.
It does: X=Y=1; X=5, Y=7, and X=29, Y=41.

Another example: A=3, B=C=1:
3*X^2 = 1 + Y^2
we find u0=2, v0=1
then (u0+1)*C/(2*A)=3/6=1/2 and (u0-1)*C/(2*B)=1/2.
Hence it is claimed to have no solution.
[It indeed doesn't with max(X,Y)<=100.]

Why this may be relevant to us:
A*X^2 = 1+Y^4
has a rational solution (X,Y), iff
A*X^2 = Y^4+Z^4
has a rational solution (X,Y,Z) iff
A*X^2 = Y^4+Z^4
has an integer solution (X,Y,Z).
• Dickson: History of theor of numbers, 1919, vol II, chapter XXII discusses. L.Euler (1738?): see Opera Omnia (1) II, 47 proved if d=2 there are no solutions
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Dickson: History of theor of numbers, 1919, vol II, chapter XXII
discusses.

L.Euler (1738?): see Opera Omnia (1) II, 47
proved if d=2 there are no solutions with a>b>=0.

E.B.Escott 1900 reduced solving our equation to solving the system
p^2 + 2*d*q^2 = r^2
p^2 - 2*d*q^2 = s^2
for (p,q,r,s).

RD Carmichael "Diophantine analysis" 1915 treated pp.77-79.
Apparently, if there is a solution (a,b,c) for squarefree d, then
d=s^4+t^4. For example 17=2^4+1^4 and 82=3^4+1^4 and 97=3^4+2^4
explaining our first three "special d."
Furthermore, then one solution is
a = s*p - 2*s*(s^8 - t^8)
b = t*p + 2*t*(s^8 - t^8)
c = p^2 + 16*s^4*t^4*(s^4 - t^4)^2
where
p = (s^4 + t^4)^2 + 4*s^4*t^4.
Wow, this seems to be a COMPLETE SOLUTION of the problem
(but I have not checked it).

Actually, I must say, this was one example where the OEIS
"online encyclopedia of integer sequences" actually hurt me!!
I figured OBVIOUSLY if the special d sequence had a very simple
description like Carmichael's it MUST already be in the OEIS,
hence I did not look for the simple description.
Oops.
• ... (8,13,17) seems to satisfy the equation for d=113, but it isn t of the form d=s^4+t^4. On the other hand, if d=s^4+t^4, the triple (t,s,1) obviously
Message 11 of 24 , Dec 15, 2011
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> RD Carmichael "Diophantine analysis" 1915 treated pp.77-79.
> Apparently, if there is a solution (a,b,c) for squarefree d, then
> d=s^4+t^4. For example 17=2^4+1^4 and 82=3^4+1^4 and 97=3^4+2^4
> explaining our first three "special d."

(8,13,17) seems to satisfy the equation for d=113, but it isn't of the
form d=s^4+t^4.

On the other hand, if d=s^4+t^4, the triple (t,s,1) obviously satisfies
the equation; and one can generate further solutions from this starting
point.

Peter
• ... --quite. OK, there s only 1 slight problem with Carmichael s wonderful complete solution from 1915 as recounted bY Dickson: it s wrong and d=113 is a
Message 12 of 24 , Dec 15, 2011
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--- In primenumbers@yahoogroups.com, Peter Kosinar <goober@...> wrote:
>
> > RD Carmichael "Diophantine analysis" 1915 treated pp.77-79.
> > Apparently, if there is a solution (a,b,c) for squarefree d, then
> > d=s^4+t^4. For example 17=2^4+1^4 and 82=3^4+1^4 and 97=3^4+2^4
> > explaining our first three "special d."
>
> (8,13,17) seems to satisfy the equation for d=113, but it isn't of the
> form d=s^4+t^4.

--quite. OK, there's only 1 slight problem with Carmichael's wonderful
"complete solution" from 1915 as recounted bY Dickson:
it's wrong and d=113 is a counterexample.

It perhaps would be useful to look in Carmichael's book to see exactly
where he went wrong. Project Gutenberg re-typeset book:
http://www.gutenberg.org/files/20073/20073-pdf.pdf
Pagination now different, now see pp.64 EQ 2.

It looks to me that what Carmichael actually said was valid, but
it isn't as useful as what I thought he said, since in cases where d is
not of this form it does not tell you how to find a solution nor
does it tell you in a very useful way which d are special...
• ... --quite. OK, there s only 1 slight problem with Carmichael s wonderful complete solution from 1915 as recounted bY Dickson: it s wrong and d=113 is a
Message 13 of 24 , Dec 15, 2011
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--- In primenumbers@yahoogroups.com, Peter Kosinar <goober@...> wrote:
>
> > RD Carmichael "Diophantine analysis" 1915 treated pp.77-79.
> > Apparently, if there is a solution (a,b,c) for squarefree d, then
> > d=s^4+t^4. For example 17=2^4+1^4 and 82=3^4+1^4 and 97=3^4+2^4
> > explaining our first three "special d."
>
> (8,13,17) seems to satisfy the equation for d=113, but it isn't of the
> form d=s^4+t^4.

--quite. OK, there's only 1 slight problem with Carmichael's wonderful
"complete solution" from 1915 as recounted bY Dickson:
it's wrong and d=113 is a counterexample.

It perhaps would be useful to look in Carmichael's book to see exactly
where he went wrong. Project Gutenberg re-typeset book:
http://www.gutenberg.org/files/20073/20073-pdf.pdf
Pagination now different, now see pp.64 EQ 2.

It looks to me that what Carmichael actually said was valid, but
it isn't as useful as what I thought he said, since in cases where d is
not of this form it does not tell you how to find a solution nor
does it tell you in a very useful way which d are special...
• 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
Message 14 of 24 , Dec 17, 2011
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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 afancier way of going about it, I do not know, but at least this way is fun for an amateur.
• ... --J.Wroblewski s theorem posted earlier that squarefree d must be divisible only by primes 2 and 8*k+1 implies that d must be a sum of 2 squares. Proof:
Message 15 of 24 , Dec 17, 2011
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--- 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 d are all of the form s^2+t^2.

--J.Wroblewski's theorem posted earlier that squarefree d must be divisible only by
primes 2 and 8*k+1 implies that d must be a sum of 2 squares.
Proof:
It is known that those have squarefree part divisible only by primes 2 and 4*k+1;
and conversely all such numbers are sums of 2 squares.
QED
• 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 =
Message 16 of 24 , Dec 18, 2011
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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.
>
• 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
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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
• 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
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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
d=17*82=1394
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
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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) 2735.
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) 7183
F. Lemmermeyer: Some families of non-congruent numbers,
Acta. Arith. 110 (2003) 1536
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) 337340.

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 .
• 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
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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).
• ... typo correction: 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
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> 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."

typo correction:

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."
• ... 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
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> H.C.Pocklington: Some diophantine impossibilities, Proc Cambridge Philos Soc 17 (1914) 110-118.
> Same title & author: 17 (1913) 108-121.

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.
• ... A scanned version is freely available: http://www.math.harvard.edu/~knill/various/eulercuboid/pocklington/index.html David
Message 23 of 24 , Dec 19, 2011
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"WarrenS" <warren.wds@...> wrote:

> H.C.Pocklington: Some diophantine impossibilities,
> Proc Cambridge Philos Soc
...
> 17 (1913) 108-121.

A scanned version is freely available:

http://www.math.harvard.edu/~knill/various/eulercuboid/pocklington/index.html

David
• ... 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
Message 24 of 24 , Dec 20, 2011
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--- In primenumbers@yahoogroups.com, "John" <mistermac39@...> wrote:
>
> This is a copy of a post sent to me by Warren.
>
>
>
>
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 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.
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