Re: [math-fun] Prouhet-Tarry-Escott in Magic Square
Paul, Never seen it before, but it surely can't be new???? Let's try it on the math-fun network. R. On Fri, 11 Apr 2008, Paul Muljadi wrote:
Professor Guy,
Have you seen a Prouhet-Tarry-Escott solution in the Lo Shu square?
4 9 2 3 5 7 8 1 6
The two number sets from the magic square are obtained by combining numbers of last rows and last columns in the clockwise and anticlockwise directions: {492, 276, 618, 834} and {294, 438, 816, 672}.
Tarry-Escott solution: 492^1 + 276^1 + 618^1 + 834^1 = 294^1 + 438^1 + 816^1 + 672^1 492^2 + 276^2 + 618^2 + 834^2 = 294^2 + 438^2 + 816^2 + 672^2 492^3 + 276^3 + 618^3 + 834^3 = 294^3 + 438^3 + 816^3 + 672^3.
If you have, who discovered it first?
Paul Muljadi
On Fri, Apr 11, 2008 at 2:56 PM, Richard Guy <rkg@cpsc.ucalgary.ca> wrote:
On Fri, 11 Apr 2008, Paul Muljadi wrote:
Tarry-Escott solution: 492^1 + 276^1 + 618^1 + 834^1 = 294^1 + 438^1 + 816^1 + 672^1 492^2 + 276^2 + 618^2 + 834^2 = 294^2 + 438^2 + 816^2 + 672^2 492^3 + 276^3 + 618^3 + 834^3 = 294^3 + 438^3 + 816^3 + 672^3.
Wikipedia has it but with no citation; Mathworld doesn't have it. J.L.Burchnall & T.W.Chaundy, A type of "Magic Square" in Tarry's problem,Quart. J. Math., 8(1937), 119-130 looks promising, but I don't have that journal; if you have the right kind of subscription you can find it at http://qjmath.oxfordjournals.org/content/volos-8/issue1/index.dtl and see if it does indeed contain this fact. That's about all I could find ... the other 1937 articles on the Tarry-Escott problem seem to cite this one but they don't have this result. --Joshua Zucker
Already seen these formulas, but I can't remember where. Joshua, I have the paper of Burchnall-Chaundy, it does not contain these formulas. Here are some other nice formulas by R. Holmes, The Magic Magic Square, The Mathematical Gazette, December 1970, p.376: 6 1 8 7 5 3 2 9 4 618^n + 753^n + 294^n = 816^n + 357^n + 492^n (rows) 672^n + 159^n + 834^n = 276^n + 951^n + 438^n (columns) 654^n + 132^n + 879^n = 456^n + 231^n + 978^n (diagonals \) 852^n + 174^n + 639^n = 258^n + 471^n + 936^n (diagonals /) True for n = 1 and 2. True again if you delete the first digit in each number. True again if you delete the second digit in each number. True again if you delete the last digit in each number. Christian. -----Message d'origine----- De : math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] De la part de Joshua Zucker Envoyé : samedi 12 avril 2008 00:48 À : math-fun Objet : Re: [math-fun] Prouhet-Tarry-Escott in Magic Square On Fri, Apr 11, 2008 at 2:56 PM, Richard Guy <rkg@cpsc.ucalgary.ca> wrote:
On Fri, 11 Apr 2008, Paul Muljadi wrote:
Tarry-Escott solution: 492^1 + 276^1 + 618^1 + 834^1 = 294^1 + 438^1 + 816^1 + 672^1 492^2 + 276^2 + 618^2 + 834^2 = 294^2 + 438^2 + 816^2 + 672^2 492^3 + 276^3 + 618^3 + 834^3 = 294^3 + 438^3 + 816^3 + 672^3.
Wikipedia has it but with no citation; Mathworld doesn't have it. J.L.Burchnall & T.W.Chaundy, A type of "Magic Square" in Tarry's problem,Quart. J. Math., 8(1937), 119-130 looks promising, but I don't have that journal; if you have the right kind of subscription you can find it at http://qjmath.oxfordjournals.org/content/volos-8/issue1/index.dtl and see if it does indeed contain this fact. That's about all I could find ... the other 1937 articles on the Tarry-Escott problem seem to cite this one but they don't have this result. --Joshua Zucker
Christian Boyer wrote:
Here are some other nice formulas by R. Holmes, The Magic Magic Square, The Mathematical Gazette, December 1970, p.376:
6 1 8 7 5 3 2 9 4
618^n + 753^n + 294^n = 816^n + 357^n + 492^n (rows) 672^n + 159^n + 834^n = 276^n + 951^n + 438^n (columns) 654^n + 132^n + 879^n = 456^n + 231^n + 978^n (diagonals \) 852^n + 174^n + 639^n = 258^n + 471^n + 936^n (diagonals /)
True for n = 1 and 2. True again if you delete the first digit in each number. True again if you delete the second digit in each number. True again if you delete the last digit in each number.
I don't have the reference by Holmes, but this is also true for n = 1 and 2 if you just take either only the first, second or third digit in each number. In fact, there are then only 2 nontrivial (i.e. different numbers on both sides) equations: {4,9,2}={8,1,6} and {4,3,8}={2,7,6}, which are the outer vertical and horizontal lines in the square above. {a,b,c} denotes here a^n+b^n+c^n (n=1,2). --- Ok, I just found something magic out, which might be not so magic at all, anyway you can get infinitely many equations that way... instead of base 10 you can use any base (smaller or larger than the largest number in the square) for the calculations: in the first above example: (6*b^2 + 1*b + 8)^n + (7*b^2 + 5*b + 3)^n + (2*b^2 + 9*b + 4)^n = (8*b^2 + 1*b + 6)^n + (3*b^2 + 5*b + 7)^n + (4*b^2 + 9*b + 2)^n for (n=1,2). --- Ok, I just played around, and this works also for Duerer's 4x4 magic square 1 15 14 4 12 6 7 9 8 10 11 5 13 3 2 16 in exactly the same form. In all cases I just checked the rows and columns. I hoped that I would work (magically...) also for n=3 in this case, but this isn't true. Just for n=1,2. Also the original case (walk around clock and counter clock wise)
492^1 + 276^1 + 618^1 + 834^1 = 294^1 + 438^1 + 816^1 + 672^1 492^2 + 276^2 + 618^2 + 834^2 = 294^2 + 438^2 + 816^2 + 672^2 492^3 + 276^3 + 618^3 + 834^3 = 294^3 + 438^3 + 816^3 + 672^3.
works as well for any base, and for the 4x4 magic square (for any base) and I conjecture that it works for any nxn magic square. I haven't tried to leave out numbers in other bases and in the 4x4 case. Christoph
Same properties on:
492^1 + 276^1 + 618^1 + 834^1 = 294^1 + 438^1 + 816^1 + 672^1 492^2 + 276^2 + 618^2 + 834^2 = 294^2 + 438^2 + 816^2 + 672^2 492^3 + 276^3 + 618^3 + 834^3 = 294^3 + 438^3 + 816^3 + 672^3.
True again if you delete the first digit in each number. True again if you delete the second digit in each number. True again if you delete the last digit in each number. Christian.
Many thanks, Joshua. Am copying to Paul Muljadi in case he's not on the network. If he, or anyone else, pursues the B & C reference, please let me know. R. On Fri, 11 Apr 2008, Joshua Zucker wrote:
On Fri, Apr 11, 2008 at 2:56 PM, Richard Guy <rkg@cpsc.ucalgary.ca> wrote:
On Fri, 11 Apr 2008, Paul Muljadi wrote:
Tarry-Escott solution: 492^1 + 276^1 + 618^1 + 834^1 = 294^1 + 438^1 + 816^1 + 672^1 492^2 + 276^2 + 618^2 + 834^2 = 294^2 + 438^2 + 816^2 + 672^2 492^3 + 276^3 + 618^3 + 834^3 = 294^3 + 438^3 + 816^3 + 672^3.
Wikipedia has it but with no citation; Mathworld doesn't have it.
J.L.Burchnall & T.W.Chaundy, A type of "Magic Square" in Tarry's problem,Quart. J. Math., 8(1937), 119-130 looks promising, but I don't have that journal; if you have the right kind of subscription you can find it at http://qjmath.oxfordjournals.org/content/volos-8/issue1/index.dtl and see if it does indeed contain this fact.
That's about all I could find ... the other 1937 articles on the Tarry-Escott problem seem to cite this one but they don't have this result.
--Joshua Zucker
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Richard Guy wrote: If he, or anyone else, pursues the B & C reference, please let me know. +++ Richard, as I said, I have the paper. But nothing on the discussed properties. Send me a private message if you want a scan of that paper.
Christoph Pacher wrote: I dropped also a note about the relation to the 3x3 magic square. Christan, if this ok for you, I would/should give credit to you for that... +++ Sure, no problem. And proud! But mention that a similar property (deleting digits) was previously found by R. Holmes in the other formulas.
Paul Muljadi wrote: The Lo Shu is more "magical" than most of us have known for thousands of years! +++ Again on: 6 1 8 7 5 3 2 9 4 Do you know this other one: 6*1*8 + 7*5*3 + 2*9*4 = 6*7*2 + 1*5*9 + 8*3*4 And still another one: If the above magic square is considered as a matrix M, then M*M*M is still a magic square
1101 1077 1197 1221 1125 1029 1053 1173 1149 (magic sum = 3375) In fact, any M^(2k+1) yield a magic square. (and M^2k are semi-magic squares) Christian.
Joshua Zucker wrote:
On Fri, Apr 11, 2008 at 2:56 PM, Richard Guy <rkg@cpsc.ucalgary.ca> wrote:
On Fri, 11 Apr 2008, Paul Muljadi wrote:
Tarry-Escott solution: 492^1 + 276^1 + 618^1 + 834^1 = 294^1 + 438^1 + 816^1 + 672^1 492^2 + 276^2 + 618^2 + 834^2 = 294^2 + 438^2 + 816^2 + 672^2 492^3 + 276^3 + 618^3 + 834^3 = 294^3 + 438^3 + 816^3 + 672^3.
Wikipedia has it but with no citation;
I just dropped a note at the discussion page http://en.wikipedia.org/wiki/Talk:Prouhet-Tarry-Escott_problem of the wikipedia page, http://en.wikipedia.org/wiki/Prouhet-Tarry-Escott_problem asking for the reference. As the page was created only 1 month ago there is a good chance that we get an answer soon. I dropped also a note about the relation to the 3x3 magic square. Christan, if this ok for you, I would/should give credit to you for that... Anyway, a really fascinating peace of math-fun!! Christoph
participants (4)
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Christian Boyer -
Christoph Pacher -
Joshua Zucker -
Richard Guy