Squares and Cubes:

mfgoode · 2006-02-16, 08:43

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

Numbers · 2006-02-18, 21:47

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.

Citrix · 2006-02-18, 22:04

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.

mfgoode · 2006-02-19, 08:29

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

nibble4bits · 2006-02-19, 11:07

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).

nibble4bits · 2006-02-19, 11:14

Bah typos.

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

Citrix · 2006-02-19, 16:40

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.

mfgoode · 2006-02-20, 04:09

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

Richard Cameron · 2006-02-20, 12:15

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

wblipp · 2006-02-20, 16:06

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

mfgoode · 2006-02-21, 09:08

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

2006-02-16, 08:43	#1
mfgoode Bronze Medalist Jan 2004 Mumbai,India 2²×3³×19 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

2006-02-19, 11:07	#5
nibble4bits Nov 2005 2×7×13 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 2006-02-19 at 11:12

2006-02-19, 11:14	#6
nibble4bits 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 2006-02-19 at 11:16

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2006-02-18, 21:47	#2
Numbers Jun 2005 Near Beetlegeuse 388₁₀ 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.

2006-02-19, 16:40	#7
Citrix Jun 2003 1,579 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.