A Generalization of Fermat’s Last Theorem

CraigD · 02-10-2009

Using the general formula
$b^n = \sum_{i=1}^m a_i^n$
where all terms are positive integers, Fermat’s Last Theorem can be written “if $m=2$ , then $n \le 2$ ”.

Considering other values of $m$ appears interesting. For example,
$5^3+4^3+3^3=6^3$
proves that, for $m=3$ , there exist $a_1$ , $a_2$ , $a_3$ and $b$ such that $n = 3$ .

Because $2^n = \sum_{i=1}^{2^n} 1^n$ , we know that there exist $a$ s and $b$ for $m = 2^n$ for any $n$ . It appears possible for $m$ to be much smaller than $2^n$ in many cases, for example
$4^4+4^4+3^4+2^4+2^4=5^4$ and $14^4+9^4+8^4+6^4+4^4=15^4$ for $n=4$ and $m=5$ , and
$11^5+9^5+7^5+6^5+5^5+4^5=12^5$ for $n=5$ and $m=6$

The best my clunky initial searches have managed for $n=6$ is $m=56$ , proven by
$5^6 \, +4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6$
$+3^6+3^6+3^6+3^6+3^6+3^6+3^6+3^6+3^6+3^6$
$+2^6+2^6+2^6+2^6+2^6+2^6+2^6+2^6$
$+1^6+1^6+1^6+1^6+1^6+1^6+1^6+1^6+1^6+1^6+1^6+1^6+1^6+1^6 = 7^6$

Lots of questions come to mind, such as

Is there always a solution such that $m=n$ ? (proven above for 1, 2 and 3, but not 4, 5, or 6)
If not, is there some other function $f(n) \ge m$ for all $n$ ?
Is there any $m$ such that and $m<n$ ? Fermat’s Last Theorem states that there is not for $m=2$ . Is there some $m$ for which a “generalized FLT” is false? Edit: Yes – see “one question answered”.
Is there a computationally efficient way to generate examples for a given $n$ and $m$ ?
Is there an elementary proof of [1], [2], or [3]? (The only known valid proof or FLT is not elementary)

I’m working on a less clunky approach to [4], and frankly intimidated by [5], and invite you number-crunch and proof-aholics out there to try answering some of these questions, or proposing new ones related to this FMT generalization.

Also, if anyone knows of any literature on this generalization, please share it. Though it seems much too obvious a generalization to be so new, the only mention of it I’ve found is this 2005 arXiv preprint, a 1-page invitation similar to but even briefer than this post.

Turtle · 02-10-2009

Quote:

Originally Posted by CraigD

Using the general formula
$b^n = \sum_{i=1}^m a_i^n$
where all terms are positive integers, Fermat’s Last Theorem can be written “if $m=2$ , then $n \le 2$ ”.

Considering other values of $m$ appears interesting. For example,
...
The best my clunky initial searches have managed for $n=6$ is $m=56$ , proven by
$5^6 \, +4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6+4^6$
$+3^6+3^6+3^6+3^6+3^6+3^6+3^6+3^6+3^6+3^6$
$+2^6+2^6+2^6+2^6+2^6+2^6+2^6+2^6$
$+1^6+1^6+1^6+1^6+1^6+1^6+1^6+1^6+1^6+1^6+1^6+1^6+1^6+1^6 = 7^6$

...invite you number-crunch and proof-aholics out there to try answering some of these questions, or proposing new ones related to this FMT generalization.

Interesting questions C. Just a couple observations.

On $n=6$ is $m=56$ , m & n are both multiples of Perfect Numbers.

Your generalized approach put me in mind of Euler & his proofs on figurate numbers, and that Squares belong to that group. Maybe there's something in that you can use here, since Squares are your exception?

History of the Theory of Numbers - Google Book Search

Qfwfq · 02-11-2009

Amusing quest Craig!

Mot much time these day but I try to find some for this too...

modest · 02-11-2009

That's fascinating. I wonder if such a generalization has been tried before.

Quote:

Originally Posted by CraigD

Is there any $m$ such that and $m<n$ ? Fermat’s Last Theorem states that there is not for $m=2$ . Is there some $m$ for which a “generalized FLT” is false?

I checked m=3, n=4 $a_1$ , $a_2$ , $a_3$ = {1-100} and I'm currently running through m=4, n=5, $a_1$ ... $a_4$ = {1-100} with nothing found so far.

Very exciting

~modest

CraigD · 02-11-2009

A slightly smarter checking program found
$72^5 = 67^5 +47^5 +46^5 +43^5 +19^5$ , $n=5$ , $m=5$ .

After an hour or two, the best it found for $n$ = 4, 6, and 7 are
$280^4 = 224^4 +208^4 +176^4 +168^4 +32^4$ , $m=5$
$72^6 = 71^6 +40^6 +37^6 +37^6 +35^6 +23^6 +5^6 +5^6 +1^6$ , $m=9$
$54^7 = 49^7 +47^7 +39^7 +27^7 +24^7 +21^7 +5^7 +4^7 +3^7 +2^7 +1^7$ , $m=11$

The program outputs $a_i$ m-tuples for a given $n$ for increasing values of $b$ , avoiding increases in $m$ . Here’s its MUMPS code:

Code:

k A1,B s (B,A1,B1)=0 ;XFMTG2(1): initialize
s B=B+1,R=B**N,B(R)=B zt:R+1=R  k A s A=0 f  s A=A+1,F=$g(A(A)),R=R+F s:'F F=$s($g(A(A-1),R)<R:A(A-1)+1,1:A>1+R) s F=$o(B(F),-1),R=R-F s:$s(F:R/F+A>A1&A1,1:0) R=R+F,F=0 s A(A)=F s:'F A=A-2 q:A<0  i F,'R k A1 m A1=A s B1=B ;XFMTG2(2): check next B, update A1 least A count
w !,N," ",B," ",B1," ",A1,":" f I=1:1:A1 w " ",B(A1(I)) ;XFMTG2(3): display

s N=3 x XFMTG2(1) f  x XFMTG2(2),XFMTG2(3) r R

And its first 10 lines of output for $n=3$ :

Code:

3 1 0 0:
3 2 2 8: 1 1 1 1 1 1 1 1
3 3 3 6: 2 2 2 1 1 1
3 4 4 5: 3 3 2 1 1
3 5 4 5: 3 3 2 1 1
3 6 6 3: 5 4 3
3 7 6 3: 5 4 3
3 8 6 3: 5 4 3
3 9 9 3: 8 6 1
3 10 9 3: 8 6 1

I’m hunching toward conjecturing there’re $b$ and $a_i$ m-tuples such that $m = n$ for all $n$ , but am far from either an efficient program for finding examples, or ideas for proofs.

Don Blazys · 02-11-2009

$30^4+120^4+272^4+315^4=353^4$ , $m=4, n=4$ .

Don.

CraigD · 02-12-2009

Quote:

Originally Posted by Don Blazys

$30^4+120^4+272^4+315^4=353^4$ , $m=4, n=4$ .

, Don!

Not only is this a solution, the program in post #5 confirms that it’s the first solution – that is, 353 is the least $b$ for which a $n=m=4$ solution exists.

How did you find it? It took my program nearly 23 million operations (nearly 5 minutes on my clunky old 2004 laptop) to find this (it’s a lot faster when told to ignore $m>4$ than the hour or so it took to find a $m=5$ solution ignoring only $m$ greater than it has already found).

Unless you’re using a computer program, or you’re a neuroatypical savant, you’ve got to be using some techniques (or intuitions) much better than my program’s, as humans can’t check millions of numbers overnight!

Inquiring minds want to know how you did it.

Turtle · 02-12-2009

is this on the right track?

Binary Quadratic Forms as Equal Sums of Like Powers

Quote:

Originally Posted by Titus Piezas III

III. 3rd, 4th, 5th Powers: Sum-Product Identities

The most general form for equal sums of like powers is
a₁^k + a₂^k + … a_m^k = b₁^k + b₂^k + … b_n^k

denoted as k.m.n. There are some beautiful parametrizations for the special case of the k.1.k, or k kth powers equal to a kth power given by,

Vieta, 1591:
$a^3(a^3-2b^3)^3 + b^3(2a^3-b^3)^3 + b^3(a^3+b^3)^3 = a^3(a^3+b^3)^3$

Fauquembergue, 1898:
$(4x^4-y^4)^4 + (4x^3y)^4 + (4x^3y)^4 + (2xy^3)^4 + (2xy^3)^4 = (4x^4+y^4)^4$

Sastry, 1934:
$(u^5+25v^5)^5 + (u^5-25v^5)^5 + (10u^3v^2)^5 + (50uv^4)^5 + (-u^5+75v^5)^5 = (u^5+75v^5)^5$
Presented in this manner, it is certainly suggestive what the identity for the next power would look like, though nothing of comparable simplicity is known for sixth powers and higher for a minimum number of terms. (It gets easier the more terms there are.)

Don Blazys · 02-13-2009

To: Craig D,

Quoting Craig D:

Quote:

How'd you do that, Don?!

Not only is this a solution, the program in post #5 confirms that
it’s the first solution – that is, 353 is the least for which a solution exists.

How did you find it? It took my program nearly 23 million operations
(nearly 5 minutes on my clunky old 2004 laptop) to find this
(it’s a lot faster when told to ignore than the hour or so it took to find
a solution ignoring only greater than it has already found).

Unless you’re using a computer program, or you’re a neuroatypical savant,
you’ve got to be using some techniques (or intuitions) much better than my program’s,
as humans can’t check millions of numbers overnight!

Inquiring minds want to know how you did it.

I don't own a computer and never even touched one until recently,
so I can't help with "programs", "computer searches" and the like.
However, I do know how to post in forums, check my e-mails, and "Google search".

Now, if you do a "Google search" on "Diophantine equations fourth powers",
then you will find the solution that I posted, and many other solutions as well!
(Wolfram has seperate articles on "second powers", "third powers", etc.)

For me, the most interesting solutions are those where
no two terms share a common factor, but so far,
the only such solutions that I found are confined to
"second power Diophantine equations".
By the way, I have never ever heard it conjectured that
"absolute co-primality" is a property that is exclusive to
"second power Diophantine equations",
so let's be on the lookout for a counter example.

Don.

CraigD · 02-13-2009

Quote:

Originally Posted by Don Blazys

Now, if you do a "Google search" on "Diophantine equations fourth powers", then you will find the solution that I posted, and many other solutions as well!

Indeed!

From such a search hit, Diophantine Equation--4th Powers -- from Wolfram MathWorld, comes:

Let the notation p.m.n stand for the equation consisting of a sum of m pth powers being equal to a sum of n pth powers. In 1772, Euler proposed that the 4.1.3 equation

$A^4+B^4+C^4=D^4$

had no solutions in integers (Lander et al. 1967). This assertion is known as the Euler quartic conjecture.

So there’s a recognized shorthand notation, “p.m.n”, for equations like
$b^n = \sum_{i=1}^m a_i^n$
(which would be n.1.m or n.m.1 in the shorthand), and 135 years after Fermat, 237 years before this thread, Euler at least conjectured that the answer to one of post #1’s questions

Quote:

Originally Posted by CraigD

3. Is there any $m$ such that and $m<n$ ?

is “no” for the special case of n.1.m, 4.1.3.

From the same mathworld article:

However, the Euler quartic conjecture was disproved in 1987 by N. Elkies, who, using a geometric construction, found

$2692440^4 +15365639^4 +18796760^4 = 20615673^4$

and showed that infinitely many solutions existed (Guy 1994, p. 140). In 1988, Roger Frye found

$95800^4 +217519^4 +414560^4 = 422481^4$

and proved that there are no solutions in smaller integers (Guy 1994, p. 140).

So, in disproving Euler’s quartic conjecture, Elkies answered post #1’s question 3 “yes”, 22 years ago.

My “how did you do that?” question (post #7), however, has merely been deflected from “how did you do that?” to “how did he or she do it?", and the search for an answer to

Quote:

Originally Posted by CraigD

4. Is there a computationally efficient way to generate examples for a given and ?

remains very much still afoot.

Quote:

Originally Posted by Turtle

is this on the right track?

Binary Quadratic Forms as Equal Sums of Like Powers

Maybe. :Shrug: There are a lot of tracks in sight. Which ones are right … well, that’s the rub.