Trivially $n^1=n^1$, and everyone knows that $3^2+4^2=5^2$. Denis Serre quoted $3^3+4^3+5^3=6^3$ in a recent MathOverflow question (which prompted this one). Are any other examples known?
$\begingroup$
$\endgroup$
5
asked Jan 24, 2011 at 21:42
-
6$\begingroup$ I've always liked $1^2 + 2^2 + 3^2 + ... + 24^2 = 70^2$. $\endgroup$– J.C. OttemCommented Jan 24, 2011 at 21:47
-
1$\begingroup$ Tito Piezas III has been collecting examples for more general sums at sites.google.com/site/tpiezas/Home . If he doesn't know of an example to your question, it's probably because no computer has found it yet. Gerhard "Ask Me About System Design" Paseman, 2011.01.24 $\endgroup$– Gerhard PasemanCommented Jan 24, 2011 at 22:12
-
$\begingroup$ this one is already there but...: $(-2)^3+(-1)^3+0^3+1^3+ \cdots + 5^3 = 6^3.$ $\endgroup$– Luis H GallardoCommented Jan 24, 2011 at 22:22
-
$\begingroup$ @Luis: Negative integers aren't natural numbers. $\endgroup$– John BentinCommented Jan 24, 2011 at 22:27
-
2$\begingroup$ This page has many examples, and speculates that there are none for exponents greater than 3. mathpages.com/home/kmath147.htm $\endgroup$– JSECommented Jan 24, 2011 at 22:41
Add a comment
|
6 Answers
$\begingroup$
$\endgroup$
1
There is a good discussion at http://www.mathpages.com/home/kmath147.htm along with some nice examples, e.g., $6^3 + 7^3 + \dots + 69^3 = 180^3$, $1134^3 + \dots + 2133^3 = 16830^3$, which apparently are part of an infinite family (starting with $3^3+4^3+5^3=6^3$). There is a table of sums of consecutive cubes equal to a cube, not coming from this infinite family. The author of this page (Kevin Brown, if I'm not mistaken) writes, "If we go on to consider sums of higher powers, it appears that there are no sums of two or more consecutive 4th powers equal to a 4th power, or in general sums of two or more consecutive $n$th powers equal to an $n$th power for any $n\gt3$. Can anyone supply a proof, reference, or counter-example?"
I suspect that any proof will be too big too fit in the margin.
answered Jan 24, 2011 at 22:38
-
$\begingroup$ While the same holds to be true for the Fermat's case. $\endgroup$– awllowerCommented Feb 11, 2011 at 10:25
$\begingroup$
$\endgroup$
Let S be the set of integers k such that there exists a sequence of k consecutive squares whose sum is a square. According to the paper "Squares Expressible as Sum of Consecutive Squares" by L. Beekmans, S is known to be infinite and density 0; the citation is to problem 6552 in the American Math Monthly.
If F(x) is the sum of the first x squares, then you are really asking about integral solutions to the Diophantine equation
(*) F(x) - F(y) = z^2
which is a double cover of the plane branched at a cubic curve (Even a reducible cubic curve, since x-y | F(x) - F(y).) Heuristically, you would expect about N^{1/2} solutions as x and y range over a box of size N. It would be interesting to ask:
a) whether the geometry of this surface is so easy to describe that you can say something about its integral points; and
b) whether (*) has a solution in polynomials in one variable (or, what's the same -- does the surface contain an affine line?)
$\begingroup$
$\endgroup$
2
Well, if we consider n consecutive 4th powers with initial a,
F(a,n) = a^4 + (a+1)^4 + (a+2)^4 + ... + (a+n-1)^4
or, equivalently,
F(a,n) = (n/30)(-1+30a^2-60a^3+30a^4+30a(1-3a+2a^2)n+10(1-6a+6a^2)n^2+(-15+30a)n^3+6n^4)
it is easy to check that F(a,n) = y^4 (or even just y^2) has NO solution in the positive integers with BOTH {a,*n*} < 1000, with the exception of the trivial n = 1. (I had checked this with Mathematica some time back.)
If we relax your question and allow n 4th powers in arithmetic progression d equal to some kth power, then the smallest I found was 64 4th powers with common difference d = 2 starting with,
29^4 + 31^4 + 33^4 + ... + 155^4 = 96104^2
P.S. The closed-form formula for general d is available, but I find it too tedious to include in this post.
answered Feb 11, 2011 at 9:47
-
$\begingroup$ You mean it's easy to check these 1000 Diophantine equations by computer? Even though you can find a closed form of $F(a,n)-y^4$, it is not so easy to see that this has no integer solutions. $\endgroup$ Commented Feb 11, 2011 at 10:14
-
$\begingroup$ Yes, actually its 1000x1000 = 10^6 equations, since you go through two variables {a,n} < 1000. (Disregarding the trivial case n = 1). The closed form is, F(a,n) = (n/30)(-1+30a^2-60a^3+30a^4+30a(1-3a+2a^2)n+10(1-6a+6a^2)n^2+(-15+30a)n^3+6n^4 A simple Mathematica code can show that, with the exception of the trivial case n = 1, F(a,n) = y^2 has no positive integer solution with BOTH {a,n} < 1000. Of course, beyond that is another matter. $\endgroup$ Commented Feb 11, 2011 at 11:44
$\begingroup$
$\endgroup$
2
There are 126 pairs $i\lt x\le 1000$ such that $i^2+(i+1)^2+...+x^2$ is a square. If you fix $i$ then the sum $i^2+...+x^2$ is a cubic polynomial $f_i(x)$ in $x$. So you are looking for integer points on the elliptic curve $y^2=f_i(x)$. For example for $i=3$, the first of these are $(4,5), (580, 8075), (963,17267)$. I hope number theorists here can give more information. See also the comment by JSE below.
-
$\begingroup$ Not quite -- the rational points form a finitely generated group, but only finitely many of these are integral (Siegel's theorem) $\endgroup$– JSECommented Jan 24, 2011 at 22:31
-
$\begingroup$ JSE: yes, I forgot. Thanks! I have edited the answer. $\endgroup$– user6976Commented Jan 24, 2011 at 22:35
$\begingroup$
$\endgroup$
J. C. Ottem's example $1^2 + ... + 24^2 = 70^2$ in the comments is of particular mathematical interest; it is one way to construct the Leech lattice, and is therefore somehow mysteriously related to other appearances of the number $24$ in mathematics (see, e.g. John Baez's thoughts).
answered Jan 24, 2011 at 22:53
$\begingroup$
$\endgroup$
While browsing the site http://sites.google.com/site/tpiezas/Home mentioned in the comments above, I found this on the page for cubes:
"There are many particular cubic equations with this property, one of which is $9^3+13^3+19^3+23^3 = 28^3, (9+23 = 13+19) as well as those in a nice arithmetic progression like,
11^3+12^3+13^3+14^3 = 20^3
31^3+33^3+35^3+37^3+39^3+41^3 = 66^3" .
You might ask Mr. Piezas directly for more examples.
Gerhard "Ask Me About System Design" Paseman, 2011.01.24
answered Jan 24, 2011 at 22:23