From: jpr2718@aol.com (Jpr2718) Subject: Re: Magic square of squares Date: 22 Feb 1999 00:29:19 GMT Newsgroups: sci.math s2845543@t2.technion.ac.il (Raz Uri ) wrote: > My grandfather has worked on the magic > square of squares presented > in Scientific American. A "near-magic" square is a 3x3 array such that the rows, columns, and one diagonal add to a magic constant. The second diagonal is not required to add to the magic constant. It is not difficult to show that every near-magic square can be decomposed into three three-term arithmetic progressions that have the same difference between terms. If corresponding terms were in arithmetic progression, then the square would be fully magic. Hillyer [Dickson, History of the Theory of Numbers, Volume 2, page 174] gives formulas that can be used to generate triples of arithmetic progressions of squares with the same difference of terms. Pick k, l positive integers, with k>l. Let p, q be one of the three pairs k^2+kl+l^2, k^2-l^2; k^2+kl+l^2, 2kl+l^2; k^2+2kl, k^2+kl+l^2. Then generate a^2, b^2, c^2 in arithmetic progression by taking a=abs(p^2-q^2-2pq), b=p^2+q^2, c=p^2-q^2+2pq, where abs(*) refers to absolute value. So with k=2, l=1, the three p, q are 7, 3; 7, 5; and 8, 7. These give the three arithmetic progressions {2^2, 58^2, 82^2}, {46^2, 74^2, 94^2}, and {97^2, 113^2, 127^2}. Each of these progressions has difference between terms of 3360. The corresponding near-magic square is 113^2 -2^2 94^2 82^2 74^2 -97^2 -46^2 127^2 58^2 or 12769 4 8836 6724 5476 9409 2116 16129 3364 Each row, column, and the upper-left-to-lower-right-diagonal adds to 21609, while the other diagonal adds to 16428. For near-magic squares generated by the Hillyer formulas, the magic constant will always be a perfect square - 21609 = 147^2. This is not true in general for near-magic squares. Andrew Bremner has found other formulas that generate triples of arithmetic progressions of squares with the same difference between terms. The problem of finding near-magic squares is equivalent to the problem of finding three different rational points, each the double of another rational point, on the same elliptic curve y^2=x^3-(n^2)x. Each rational point that is the double of another rational point corresponds to an arithmetic progression of rational squares as follows. Let sqrt(+)=sqrt(2x+2y/sqrt(x)) and sqrt(-)=sqrt(2x-2y/sqrt(x)). Then a=sqrt(+)-sqrt(-), b=2sqrt(x), and c=sqrt(+)+sqrt(-). When [x, y] is the double of another rational point, a, b, and c will all be rational, a^2, b^2, c^2 will be in arithmetic progression, and the difference between terms will be n. Use the three rational points on y^2=x^3-(n^2)x to generate three three-term rational arithmetic progressions of squares. Multiply by the least common multiple of the denominators to make arithmetic progressions of integer squares. For the example given we can take n = 210, and the three points to be [841/4, -1189/8], [1369/4, -39997/8], [12769/16, -1392047/64]. Conversely, all near-magic squares are generated from three such points. John Robertson ============================================================================== From: Jpr2718@aol.com Subject: Re: Magic square of squares Date: Mon, 22 Feb 1999 20:13:33 EST Newsgroups: [missing] To: rusin@math.niu.edu Keywords: Construct a three-by-three magic square with square entries? (open) In a message dated 2/22/99 5:35:26 PM Eastern Standard Time, rusin@math.niu.edu writes: > > >The problem of finding near-magic squares is equivalent to the problem of > >finding three different rational points > > Perhaps you can spare me the calculations and describe what it takes to > have the last diagonal sum to the same as the other -- what other condition > do we need besides having E/ker(2) be noncyclic? > > dave > It's very simple. In terms of ec's, the x-coordinates of the three points have to be in arithmetic progression. In the "arithmetic progression" characterization of the problem, the corresponding terms of the three three- term arithmetic progressions have to themselves be in arithmetic progression. I have an article on this in the October 1996 Mathematics Magazine, page 289. Umm. What does "E/ker(2) be noncyclic" mean? I enjoy your postings and your web pages. John ============================================================================== From: Dave Rusin Subject: Re: Magic square of squares Date: Mon, 22 Feb 1999 20:14:19 -0600 (CST) Newsgroups: [missing] To: Jpr2718@aol.com But "x-coordinates in arithmetic progression" isn't coordinate-independent. I guess I was hoping for something more intrinsic. For example, my cryptic condition "E/ker(2) noncyclic" is a statement about the elliptic curve which is independent of its coordinatization. Actually I see now it's the _wrong_ statement, but I'll clarify it anyway. An elliptic curve defined over Q is a finitely-generated abelian group E, on which the doubling map P -> 2P is a group homomorphism. Its kernel "ker(2)" is just the set of 2-torsion elements. (This is _for example_ the set of points (x,0) on the curve, _if_ the curve is presented in the form y^2=monic cubic in x.) E/ker(2) is the quotient group and as it turns out this is isomorphic as groups to the image of the doubling map. (Write E = Z^r x H x T with H of odd order and T a finite 2-group. Then 2E = (2Z)^r x H x (2T) and E/ker(2)= Z^r x H x (T/ker(2)). Then just use 2Z isomorphic to Z and 2T isomorphic to T/ker(2), the latter immediate from writing T as a product of cyclic 2-groups, for example.) So I was trying to be slick, replacing "...points which are doubles of rational points" with "...elements of E/ker(2)". But I guess I blew it. "E/ker(2) noncyclic" really means "rank(E)>1 or rank(E)=1 and Tors(E) includes elements not of order 1 or 2", each of which implies E has infinitely many points which are doubles of rational points, but the converse is not true. Sorry to have written in haste. I'll have a look for your article. Thanks for replying. dave ============================================================================== From: Jpr2718@aol.com Subject: Re: Magic square of squares Date: Mon, 22 Feb 1999 21:42:49 EST Newsgroups: [missing] To: rusin@math.niu.edu Thanks for the explanation. I'm afraid I do not have a coordinate-independent characterization. You probably know this, but in case it helps, the torsion group for the ec's y^2=x^3-(n^2)x is of order 4 and consists of the point at infinity plus the three points of order 2 (-n, 0), (n, 0), (0, 0) [Koblitz, Intro to EC's and Modular Forms, 2nd ed., page 44]. Bremner told Guy that he thinks if one of these curves has three points as wanted, that it will be of rank at least 1,000. I assume the comment is based more on frustration at finding such points than on some deep analysis. John ============================================================================== 97k:05038 05B15 Robertson, John P. Magic squares of squares. Math. Mag. 69 (1996), no. 4, 289--293. [To journal home page] An open problem [R. K. Guy, Unsolved problems in number theory, Second edition, Springer, New York, 1994; MR 96e:11002 (Problem D15, pp. 170--171)] is to prove or disprove that a three-by-three magic square can be constructed from nine distinct integer squares. There are relationships between magic squares, arithmetic progressions, Pythagorean right triangles, congruent numbers, and elliptic curves. In this paper the author follows this chain and shows that there are three problems which are equivalent to the above-mentioned problem. Reviewed by Rong Si Chen