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Old 2006-02-19, 16:24   #12
xilman
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Quote:
Originally Posted by Numbers
I am clearly missing something here. I make 3^5 - 15^2 = 18

Is that not a solution?
I think you may be missing that 18 is not in the series " N = 6,14,50, ..... etc.".

Paul
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Old 2006-02-19, 18:41   #13
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Quote:
Originally Posted by R.D. Silverman
Numbers that are 2 mod 4 are the problem.
Quote:
Originally Posted by xilman
I think you may be missing that 18 is not in the series " N = 6,14,50, ..... etc.".
So in what sense is 18 not congruent 2 mod 4 ?

Or, alternatively, can you please define (6, 14, 50, ...etc) a little more precisely?

I know a certain member of this forum who would describe "...etc" as "hand waving nonsense. So lacking in precision as to be mathematically useless". LOL

Many thanks,

Numbers
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Old 2006-02-19, 19:23   #14
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Quote:
Originally Posted by Numbers
So in what sense is 18 not congruent 2 mod 4 ?
If I read Bob's post correctly, he meant that "all problematic numbers are == 2 mod 4," not "all numbers that == 2 mod 4 are problematic."
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Old 2006-02-19, 19:26   #15
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Quote:
Originally Posted by Numbers
So in what sense is 18 not congruent 2 mod 4 ?

Or, alternatively, can you please define (6, 14, 50, ...etc) a little more precisely?

I know a certain member of this forum who would describe "...etc" as "hand waving nonsense. So lacking in precision as to be mathematically useless". LOL

Many thanks,

Numbers
You failed to quote all of Bob's post.

He claimed that numbers which are congruent to 2 mod 4 are the problem. He did not say that all numbers in that residue class lead to difficult problems, only that the other three residue classes mod 4 have trivial solutions. He did say that some, including 6, 14 and 50, do lead to difficult and, indeed, presently unsolved problems.

Bob could and should have phrased himself much more clearly, IMO.

Paul
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Old 2006-02-19, 19:27   #16
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See the sequence at OEIS: http://www.research.att.com/~njas/sequences/A074981 : 6, 14, 34, 42, 50, 58, 62, 66, 70, 78, 82, 86, 90, 102, 110, 114, 130, 134, 158, 178, 182, 202, 206, 210, 226, 230, 238, 246, 254, 258, 266, 274, 278, 290, 302, 306, 310, 314, 322, 326, 330, 358, 374, 378, 390, 394, 398, 402, 410, 418, 422, 426, ...

According to that page it is conjectured that none of these numbers can be a difference of two powers.
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Old 2006-02-19, 19:48   #17
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What is the solution for the number=2 and 125. Anyone know one?
Are these all the numbers upto infinity or under 500?

Citrix
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Old 2006-02-19, 20:48   #18
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Thank you, all three of you. Now it makes sense. And I note that none of you were able to define N (which is not a criticism, just a point that I will come back to).

What actually interested me about this problem is not a desire to solve it, but a conundrum that arises out of my suspicion that you would not in fact be able to define N.

Now, let f(x) = x^a - y^b, so that the problem can be stated as: When does f(x) = N? The conjecture referred to in Alpertron's post says that f(x) never = N for N in A074981.

For obvious reasons, A074981 is a finite representation of what is presumably an infinite sequence. So there are values of N that should be in the sequence that are not at OEIS, or any other list of the sequence. Let's say M is a number that should be in N, but is not on any recorded list of N.

Joe Nobody finds that f(x) = M.

Since no one can define N, how does Joe even prove that M is in N and that he has found a counter-example to the conjecture?

You could probably get quite cute with your semantics and claim that since M should be in N, then Joe cant find that f(x) = M, but I'm sure you know what I mean.
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Old 2006-02-19, 20:55   #19
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Quote:
Originally Posted by Citrix
What is the solution for the number=2 and 125. Anyone know one?
3^3 - 5^2 = 2

15^2 - 10^2 = 125
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Old 2006-02-20, 00:55   #20
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The trivial cases are:

Odd numbers: 2n+1 = (n+1)^2 - n^2

Numbers multiple of 4: 4n = (n+1)^2 - (n-1)^2
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Old 2006-02-20, 04:01   #21
mfgoode
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Quote:
Originally Posted by pacionet
Searching for multimagic squares could be interesting.
Does any software exist ?
I have extensive literature on magic squares of all types. I found that magic squares of primes or consecutive primes are very difficult to compile and these have not been investigated for large numbers.
Try working out a 6X6 magic square from 1 to 36. It shoud take you ages but with a method and suitable programming it should be a wheeze
I also have logic circuits based on magic squares but Im trying to promote them to INTEL, IBM, etc. first. If they are rejected then I can and am willing to collaborate with you.
Mally
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Old 2006-02-20, 20:02   #22
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By the way, the differences of two powers not in A074981 are:

Code:
 3 ^ 3 - 5 ^ 2 = 2
 13 ^ 3 - 3 ^ 7 = 10
 3 ^ 3 - 3 ^ 2 = 18 
 7 ^ 2 - 3 ^ 3 = 22 
 3 ^ 3 - 1 ^ 2 = 26 
 83 ^ 2 - 19 ^ 3 = 30 
 37 ^ 2 - 11 ^ 3 = 38 
 17 ^ 2 - 3 ^ 5 = 46 
 7 ^ 3 - 17 ^ 2 = 54 
 3 ^ 5 - 13 ^ 2 = 74 
 11 ^ 2 - 3 ^ 3 = 94 
 5 ^ 3 - 3 ^ 3 = 98 
 11 ^ 3 - 35 ^ 2 = 106 
 3 ^ 5 - 5 ^ 3 = 118 
 3 ^ 5 - 11 ^ 2 = 122 
 15 ^ 3 - 57 ^ 2 = 126 
 173 ^ 2 - 31 ^ 3 = 138 
 13 ^ 2 - 3 ^ 3 = 142 
 195 ^ 3 - 2723 ^ 2 = 146 
 175 ^ 3 - 2315 ^ 2 = 150 
 111 ^ 2 - 23 ^ 3 = 154 
 3 ^ 5 - 9 ^ 2 = 162 
 7 ^ 5 - 129 ^ 2 = 166 
 59 ^ 3 - 453 ^ 2 = 170 
 7 ^ 3 - 13 ^ 2 = 174 
 23 ^ 2 - 7 ^ 3 = 186 
 39 ^ 2 - 11 ^ 3 = 190 
 3 ^ 5 - 7 ^ 2 = 194 
 15 ^ 2 - 3 ^ 3 = 198 
 49 ^ 2 - 3 ^ 7 = 214 
 3 ^ 5 - 5 ^ 2 = 218 
 7 ^ 3 - 11 ^ 2 = 222 
 3 ^ 5 - 3 ^ 2 = 234 
 3 ^ 5 - 1 ^ 2 = 242 
 11 ^ 5 - 401 ^ 2 = 250 
 7 ^ 3 - 9 ^ 2 = 262 
 39 ^ 3 - 243 ^ 2 = 270 
 25 ^ 2 - 7 ^ 3 = 282 
 23 ^ 2 - 3 ^ 5 = 286 
 7 ^ 3 - 7 ^ 2 = 294 
 19 ^ 3 - 81 ^ 2 = 298 
 7 ^ 3 - 5 ^ 2 = 318 
 7 ^ 3 - 3 ^ 2 = 334 
 3 ^ 7 - 43 ^ 2 = 338 
 7 ^ 3 - 1 ^ 2 = 342 
 61 ^ 2 - 15 ^ 3 = 346 
 15 ^ 3 - 55 ^ 2 = 350 
 131 ^ 2 - 7 ^ 5 = 354 
 27 ^ 3 - 139 ^ 2 = 362 
 85 ^ 2 - 19 ^ 3 = 366 
 11 ^ 3 - 31 ^ 2 = 370 
 25 ^ 2 - 3 ^ 5 = 382 
 9 ^ 3 - 7 ^ 3 = 386 
 1107 ^ 2 - 107 ^ 3 = 406 
 21 ^ 2 - 3 ^ 3 = 414
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