mersenneforum.org  

Go Back   mersenneforum.org > Fun Stuff > Puzzles

Reply
 
Thread Tools
Old 2006-02-16, 08:43   #1
mfgoode
Bronze Medalist
 
mfgoode's Avatar
 
Jan 2004
Mumbai,India

40048 Posts
Question Squares and Cubes:


Squares and Cubes:
Q 1 : Find two whole numbers which between them make use of each of the 10 digits,
0,1,2,3,4,5,6,7,8 and 9 just once.
The two numbers are respectively the square and the cube of the same number.

Q 2 : Find quickly any two different integers, the sum of whose cubes is the fourth power of an integer.
And after finding one pair of numbers that fill the bill, explain how you could find all such cases.
Mally
mfgoode is offline   Reply With Quote
Old 2006-02-18, 21:47   #2
Numbers
 
Numbers's Avatar
 
Jun 2005
Near Beetlegeuse

22·97 Posts
Default

Question 1.

69^2 = 4761, 69^3 = 328509 using each digit just once.


Question 2.

9^4 = 6561 = 9^3 + 18^3

I can't yet explain a general rule because I have too much information.
For example, 28^4 = 614656 = 28^3 + 84^3, and 35^4 = 1500625 = 70^3 + 105^3
and the relationship between the two integers in each pair is not consistent.
Numbers is offline   Reply With Quote
Old 2006-02-18, 22:04   #3
Citrix
 
Citrix's Avatar
 
Jun 2003

1,579 Posts
Default

Quote:
Originally Posted by mfgoode

Squares and Cubes:
Q 1 : Find two whole numbers which between them make use of each of the 10 digits,
0,1,2,3,4,5,6,7,8 and 9 just once.
The two numbers are respectively the square and the cube of the same number.

Q 2 : Find quickly any two different integers, the sum of whose cubes is the fourth power of an integer.
And after finding one pair of numbers that fill the bill, explain how you could find all such cases.
Mally

For question2) let me make it more difficult by saying that cubes must be coprime to each other.
Citrix is offline   Reply With Quote
Old 2006-02-19, 08:29   #4
mfgoode
Bronze Medalist
 
mfgoode's Avatar
 
Jan 2004
Mumbai,India

1000000001002 Posts
Cool Squares and cubes

Quote:
Originally Posted by Citrix
For question2) let me make it more difficult by saying that cubes must be coprime to each other.
Yeah Citrix,now you are talking fella. The previous examples that are given are trivial. I have an elegant 4 line solution for all numbers.

BTW: coprime and relatively prime are much the same in meaning. I was surprised that I did not find a single word for a relationship of composites other than that one does, or does not go into the other and the vertical line is used. We should coin a word for it or does one already exist?
Mally

Last fiddled with by mfgoode on 2006-02-19 at 08:32
mfgoode is offline   Reply With Quote
Old 2006-02-19, 11:07   #5
nibble4bits
 
nibble4bits's Avatar
 
Nov 2005

B616 Posts
Default

There's cases which I think are trivial: (0,0,0), (0,1,1), and (1,0,1)

Here's one of my solutions to a^3+b^3=c^4

If c=2 and a=b, then (2,2,2) works.
In general, if c is an even constant then (a^3+b^3)/c=c^3 is easy to solve since
you can move the constant into a and b.

Now to make a generalized formula for a>b (use symetry to reduce cases).

Last fiddled with by nibble4bits on 2006-02-19 at 11:12
nibble4bits is offline   Reply With Quote
Old 2006-02-19, 11:14   #6
nibble4bits
 
nibble4bits's Avatar
 
Nov 2005

2×7×13 Posts
Default

Bah typos.

My solution seems to match c^4=c(a^3) which makes a=b=c.

Last fiddled with by nibble4bits on 2006-02-19 at 11:16
nibble4bits is offline   Reply With Quote
Old 2006-02-19, 16:40   #7
Citrix
 
Citrix's Avatar
 
Jun 2003

62B16 Posts
Default

Prove that there are no coprime solutions to a^3+b^3=c^4? I read a note in a paper some time back that said no solutions to this existed.
Citrix is offline   Reply With Quote
Old 2006-02-20, 04:09   #8
mfgoode
Bronze Medalist
 
mfgoode's Avatar
 
Jan 2004
Mumbai,India

40048 Posts
Lightbulb Squares and cubes

Quote:
Originally Posted by Citrix
Prove that there are no coprime solutions to a^3+b^3=c^4? I read a note in a paper some time back that said no solutions to this existed.
If that is the case then I'm afraid your note in that paper is errroneous.
Solutions can be found for a , b , being co-prime. Well I may be wrong, but off hand Im pretty sure of it.
Mally
mfgoode is offline   Reply With Quote
Old 2006-02-20, 12:15   #9
Richard Cameron
 
Richard Cameron's Avatar
 
Mar 2005

2528 Posts
Default a query about the other question...

Quote:
Originally Posted by mfgoode
Squares and Cubes:
Q 1 : Find two whole numbers which between them make use of each of the 10 digits,
0,1,2,3,4,5,6,7,8 and 9 just once.
The two numbers are respectively the square and the cube of the same number.
I had come across this one before and it wasn't difficult for me to find the answer again that Numbers stated. However I got there by enummerating more than two dozen cases and finding the right one by inspection. I wonder: is there a more elegant means of arriving at the solution?

thanks
Richard
Richard Cameron is offline   Reply With Quote
Old 2006-02-20, 16:06   #10
wblipp
 
wblipp's Avatar
 
"William"
May 2003
New Haven

3×787 Posts
Default

Quote:
Originally Posted by citrix
Prove that there are no coprime solutions to a^3+b^3=c^4? I read a note in a paper some time back that said no solutions to this existed.
Quote:
Originally Posted by mfgoode
Solutions can be found for a , b , being co-prime. Well I may be wrong, but off hand Im pretty sure of it.
If you've got a solution, it's worth $100,000 because it would be a counterexample to the Beal Conjecture

http://www.bealconjecture.com/

William
wblipp is offline   Reply With Quote
Old 2006-02-21, 09:08   #11
mfgoode
Bronze Medalist
 
mfgoode's Avatar
 
Jan 2004
Mumbai,India

80416 Posts
Thumbs up Squares and cubes

Quote:
Originally Posted by wblipp
If you've got a solution, it's worth $100,000 because it would be a counterexample to the Beal Conjecture

http://www.bealconjecture.com/

William
Thanks a million William for this breaking news.

Since maybe I'm sitting on a goldmine I would prefer you to correspond with me direct to my e-mail add. <mfgoode102@ yahoo.com>.

If any thing comes of it, I am willing that GIMPS gets the honour of my membership, provided, if they can protect me as the sole originator. I am also willing to give you a small percentage of the prize money.

I am very close to the Beal conjecture, and more, by applying the exponents to any number >2 though they have a certain relationship.

I can also jet into London, New york. Chicago or LA. or any capital city on our network, though I prefer London, for whatever reasons, if Im required, or meeting you wherever you are.

At the same time it might all turn out to be a pipe dream! Still I wont be disappointed as I can make it into a theorem.
Regards,
Mally
mfgoode is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
Regarding Squares a1call Miscellaneous Math 42 2017-02-03 01:29
Fibonacci number as sum of cubes jux Miscellaneous Math 15 2015-08-30 06:21
Prime cubes! fivemack Puzzles 4 2007-07-04 00:18
squares or not squares m_f_h Puzzles 45 2007-06-15 17:46
Counting Cubes Numbers Puzzles 6 2005-09-03 00:26

All times are UTC. The time now is 16:53.

Fri Apr 16 16:53:51 UTC 2021 up 8 days, 11:34, 0 users, load averages: 3.16, 2.70, 2.53

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.