20060216, 08:43  #1 
Bronze Medalist
Jan 2004
Mumbai,India
4004_{8} Posts 
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 
20060218, 21:47  #2 
Jun 2005
Near Beetlegeuse
2^{2}·97 Posts 
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. 
20060218, 22:04  #3  
Jun 2003
1,579 Posts 
Quote:
For question2) let me make it more difficult by saying that cubes must be coprime to each other. 

20060219, 08:29  #4  
Bronze Medalist
Jan 2004
Mumbai,India
100000000100_{2} Posts 
Squares and cubes
Quote:
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 20060219 at 08:32 

20060219, 11:07  #5 
Nov 2005
B6_{16} Posts 
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 20060219 at 11:12 
20060219, 11:14  #6 
Nov 2005
2×7×13 Posts 
Bah typos.
My solution seems to match c^4=c(a^3) which makes a=b=c. Last fiddled with by nibble4bits on 20060219 at 11:16 
20060219, 16:40  #7 
Jun 2003
62B_{16} Posts 
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.

20060220, 04:09  #8  
Bronze Medalist
Jan 2004
Mumbai,India
4004_{8} Posts 
Squares and cubes
Quote:
Solutions can be found for a , b , being coprime. Well I may be wrong, but off hand Im pretty sure of it. Mally 

20060220, 12:15  #9  
Mar 2005
252_{8} Posts 
a query about the other question...
Quote:
thanks Richard 

20060220, 16:06  #10  
"William"
May 2003
New Haven
3×787 Posts 
Quote:
Quote:
http://www.bealconjecture.com/ William 

20060221, 09:08  #11  
Bronze Medalist
Jan 2004
Mumbai,India
804_{16} Posts 
Squares and cubes
Quote:
Since maybe I'm sitting on a goldmine I would prefer you to correspond with me direct to my email 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 

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