Title: Cliff Puzzle 16: Undulating Squares From: cliff@watson.ibm.com
pickover/pickover.16 Added material
Problem
Title: Cliff Puzzle 16: Undulating Squares
From: cliff@watson.ibm.com
If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information. PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup. I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise. Thanks, Cliff Pickover
* * *
A square number is of the form y=x**2. For example, 25 is a square
number.
Undulating numbers are of the form: ababababab... For example, the
following are undulating numbers: 1717171, 282828, etc.
1. Are there any undulating square numbers?
2. Are there any undulating cube numbers?
Show solution
-------------------------
In article <1992Oct30.175102.142177@watson.ibm.com> you write:
: 1. Are there any undulating square numbers?
11^2 = 121
: 2. Are there any undulating cube numbers?
7^3 = 343
(yes, I know they're short, but they qualify!)
--
Michael Neylon aka Masem the Great and Almighty Thermodynamics GOD!
// | Senior, Chemical Engineering, Univ. of Toledo
\\ // Only the | Summer Intern, NASA Lewis Research Center
\ \X/ AMIGA! | mneylon@jupiter.cse.utoledo.edu /
--------+ How do YOU spell 'potato'? How 'bout 'lousy'? +----------
"Me and Spike are big Malcolm 10 supporters." - J.S.,P.L.C.L
-------------------------
In article <1992Oct30.204134.97881@watson.ibm.com> you write:
>Hi, I was interested in non-trivial cases. Those with greater
>than 3 digits. Award goes to the person who finds the largest
>undulating square or cube number. Thanks, Cliff
343 and 676 aren't trivial (unlike 121 and 484 it doesn't come from
obvious algebraic identities). The chance that a "random"
number around x should be a perfect square is about 1/sqrt(x);
more generally, x^(-1+1/d) for a perfect d-th power. Since
there are for each k only 90 k-digit undulants you expect
to find only finitely many of these that are perfect powers,
and none that are very large. But provably listing all cases
is probably only barely, if at all, possible by present-day
methods for treating exponential Diophantine equations, unless
(as was shown in a rec.puzzles posting re your puzzles on
arith. prog. of squares with common difference 10^k) there is
some ad-hoc trick available. At any rate the largest undulating
power is probably 69696=264^2, though 211^3=9393931 comes
remarkably close.
--Noam D. Elkies
-------------------------
In article <1992Oct30.175102.142177@watson.ibm.com>, you write...
>1. Are there any undulating square numbers?
>
Other than the obvious 11**2, 22**2, and 26**2, there is 264**2
which equals 69696.
>2. Are there any undulating cube numbers?
>
Just 7**3 as far as I can tell, though I'm limited to IEEE computational
reals.
PauL M SchwartZ (-Z-) | Follow men's eyes as they look to the skies
v206gb6c@ubvms.BitNet | the shifting shafts of shining
pms@geog.buffalo.edu | weave the fabric of their dreams
pms@acsu.buffalo.edu | - RUSH -
Additional material
I have proved that there are no base 10 undulating squares other than those already known (the 1- and 2-digit squares plus 121, 484, 676, and 69696.) After reducing the problem to a Pell equation plus a constraint of the form y=10^x, a result of Baker & Wuestholz on linear combinations of logarithms is used to bound the size of a possible solution, and small solutions are ruled out by using a lattice-reduction algorithm. A URL for the proof is http://xraysgi.ims.uconn.edu/dmoews/und.tex (or, use .dvi or .ps instead of .tex). -- David Moews <dmoews@xraysgi.ims.uconn.edu>