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2006-02-19, 16:24   #12
xilman
Bamboozled!

"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across

1068210 Posts

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

2006-02-19, 18:41   #13
Numbers

Jun 2005
Near Beetlegeuse

6048 Posts

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

2006-02-19, 19:23   #14
ewmayer
2ω=0

Sep 2002
República de California

2×5×1,163 Posts

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."

2006-02-19, 19:26   #15
xilman
Bamboozled!

"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across

2×72×109 Posts

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

 2006-02-19, 19:27 #16 alpertron     Aug 2002 Buenos Aires, Argentina 54C16 Posts 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.
 2006-02-19, 19:48 #17 Citrix     Jun 2003 1,579 Posts What is the solution for the number=2 and 125. Anyone know one? Are these all the numbers upto infinity or under 500? Citrix
 2006-02-19, 20:48 #18 Numbers     Jun 2005 Near Beetlegeuse 38810 Posts 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.
2006-02-19, 20:55   #19
Numbers

Jun 2005
Near Beetlegeuse

22·97 Posts

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

 2006-02-20, 00:55 #20 alpertron     Aug 2002 Buenos Aires, Argentina 22·3·113 Posts The trivial cases are: Odd numbers: $2n+1 = (n+1)^2 - n^2$ Numbers multiple of 4: $4n = (n+1)^2 - (n-1)^2$
2006-02-20, 04:01   #21
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

22·33·19 Posts
why not?

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

 2006-02-20, 20:02 #22 alpertron     Aug 2002 Buenos Aires, Argentina 22·3·113 Posts 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