Given any triangle with sides of lengths 2A, 2B, 2C, the lengths of the medians to the midpoints of those sides, respectively, are (M_A)^2 = - A^2 + 2B^2 + 2C^2 (M_B)^2 = 2A^2 - B^2 + 2C^2 (M_C)^2 = 2A^2 + 2B^2 - C^2 An "automedian triangle" is defined as one whose three medians are proportional to the three sides. Obviously the middle edge (in terms of length) must map to the middle-sized median, whereas the largest edge must map to the smallest median, and vice versa. Thus we have three equations of the form 3C^2 = - A^2 + 2B^2 + 2C^2 3B^2 = 2A^2 - B^2 + 2C^2 (1) 3A^2 = 2A^2 + 2B^2 - C^2 Each of these equations reduces to simply A^2 + C^2 = 2B^2, so any solution of this equation in integers gives an automedian triangle with integer sides. Of course, if we want a strictly real triangle we also have to impose the requirement that all three lengths satisfy the triangle inequality. For example, the smallest integer solution of (1) is {A,B,C} = {1,5,7}, which doesn't make a real triangle, but there are infinitely many others, such as {7,13,17}, {17,25,31}, {31,41,49}, {23,37,47}, etc. that do give real automedian triangles with integer sides. Equations (1) have a very interesting "duality", which can be seen if we allow the variables on the left side to be any three quantities (not necessarily some permutation of A,B,C). In other words, we have the equations 3I^2 = - A^2 + 2B^2 + 2C^2 3H^2 = 2A^2 - B^2 + 2C^2 (2) 3G^2 = 2A^2 + 2B^2 - C^2 If we solve these three linear equations for the squares of A,B,C in terms of the squares of I,H,G we get 3A^2 = - I^2 + 2H^2 + 2G^2 3B^2 = 2I^2 - H^2 + 2G^2 (3) 3C^2 = 2I^2 + 2H^2 - G^2 which is formally identical to (2). Thus, the triples {A,B,C} and {I,H,G} are duals of each other. Now, if we restrict the variables to integer values, it turns out that almost the only solutions of (2) give {I,H,G} as some permutation of {A,B,C}, so they are automedian triangles (with integer sides). This raises the question of whether there exist pairs of *distinct* triples of integers {A,B,C} and {I,H,G} satisfying equations (2) and (3). These would correspond to integer triangles whose medians are proportional to *each others* sides. One reason this is an interesting question is that if no such pair of "amicable triangles" exists, it follows that there does not exist a 3x3 magic square of (distinct) square numbers, which is an open problem in number theory. In case the connection isn't obvious, consider a 3x3 magic square consisting of distinct squares A^2 B^2 C^2 D^2 E^2 F^2 G^2 H^2 I^2 By assumption the sum of each row, column, and main diagonal is the same, and it's easy to see that this sum must be 3E^2. Therefore, we immediately have the following necessary conditions A^2 + I^2 = 2E^2 B^2 + H^2 = 2E^2 (4) C^2 + G^2 = 2E^2 A^2 + B^2 + C^2 = 3E^2 Subtracting twice the 4th equation from three times each of the previous three equations immediately gives equations (2). Obviously if {I,H,G} is a permutation of {A,B,C} the entries of the magic square are not distinct, so all the "self-dual" solutions can be ruled out, leaving us only with the question of whether there exist any "amicable" solutions in all distinct numbers. It turns that "amicable" solutions exist, but they aren't very common. The first several primitive solutions are A B C I H G E^2 factorization of E^2 ---- ---- ---- ---- ---- ---- ---------- ----------------------- 49 421 541 559 371 149 157441 (29)(61)(89) 191 763 785 887 491 455 411625 (5^3)(37)(89) 361 941 1159 1201 829 479 786361 (37)(53)(401) 137 1123 1523 1543 1067 283 1199809 (13)(17)(61)(89) 931 1541 1691 1789 1301 1099 2033641 (41)(193)(257) 487 3647 4783 4903 3313 1183 12138289 (17^2)(97)(433) 53 4597 6223 6317 4333 1087 19953649 (61)(109)(3001) 5323 9005 10489 10861 8075 6023 73147825 (5^2)(73)(149)(269) 2447 11777 15487 15823 10847 4063 128177569 (13)(17^2)(109)(313) 10271 14939 16331 17071 13181 11411 198456241 (61)(1409)(2309) 8837 15467 16783 17923 12653 10847 199663249 (17)(73)(349)(461) 11215 16387 17041 18185 13709 12887 228235225 (5^2)(53)(281)(613) 6419 19283 19585 22133 12619 12145 265536625 (5^2)(53)(149)(269) 5071 20815 21997 24553 13975 12029 314282425 (5^2)(29)(41)(97)(109) 15953 23393 24553 26113 19727 18263 468193489 (89)(97)(193)(281) 3391 30085 30737 35063 18325 17209 620456425 (5^2)(53)(197)(2377) 7021 32653 39745 41803 27029 14735 898392625 (5^3)(13)(17^2)(1913) 2503 33827 41177 43487 27443 14207 948692089 (17)(149)(353)(1061) 11039 43727 51395 54727 34711 21805 1558452025 (5^2)(13)(29)(37)(41)(109) 17791 53675 61313 65737 41915 29641 2318936425 (5^2)(13)(17)(29)(41)(353) 34285 54173 61939 64205 48611 38227 2648871625 (5^3)(37)(41)(61)(229) 1523 53077 73693 74147 51797 8333 2750048569 (29)(37^2)(113)(613) 19177 56039 72755 74161 52223 23965 2933805625 (5^4)(1361)(3449) 9419 70651 74909 83899 46219 38941 3563879881 (17)(353)(401)(1481) 44405 61243 73751 73955 60749 44743 3720573025 (5^2)(13^2)(113)(7793) 22207 69395 77801 84151 52525 39007 3787270825 (5^2)(13)(149)(197)(397) 45653 68873 77593 80507 61823 50447 4282786729 (13)(29)(101)(137)(821) 17977 77635 84161 92911 54115 43273 4477813225 (5^2)(7561)(23689) 3001 67741 95041 95279 67069 7369 4543546921 (89)(173)(269)(1097) 42125 69283 86569 87205 67669 43417 4689613825 (5^2)(13)(41)(353)(997) 11659 80771 87871 97219 55349 43199 4793733121 (73)(113)(701)(829) 28729 72401 93881 95369 68401 33271 4960300801 (241)(2633)(7817) 41513 73195 90541 91991 69485 44587 5092836625 (5^3)(17)(61)(101)(389) 34391 77087 93395 96863 67991 42925 5282590825 (5^2)(37)(53)(277)(389) 27265 79717 94981 100015 66269 41533 5373190225 (5^2)(13^2)(521)(2441) 16885 84049 103343 108325 70393 36601 6009704425 (5^2)(17^3)(113)(433) 11095 83821 112447 114335 78547 23479 6597795625 (5^4)(17)(613)(1013) 20587 107309 110735 125341 67963 62225 8067095425 (5^2)(17^2)(41)(113)(241) 3167 107881 124295 134369 80167 51145 9032529025 (5^2)(13)(3181)(8737) 6889 110123 123635 135127 78611 54965 9153382225 (5^2)(13^2)(41)(53)(997) 30415 108683 128081 136025 87269 54983 9713936425 (5^2)(17)(29)(373)(2113) 82427 101483 111503 113533 96877 85153 9841976209 (13)(17^2)(37)(101)(701) 43409 107309 130151 135431 93331 57329 10112948521 (13)(37)(41)(193)(2657) 30395 119993 122599 138965 76399 72143 10117563625 (5^3)(13)(1933)(3221) 52303 116971 119125 132929 81997 78835 10202861425 (5^2)(73)(557)(10037) 57001 106171 133111 135071 101099 61439 10746644521 (29)(41)(101)(109)(821) 11095 119701 137407 148655 88843 57799 11110704025 (5^2)(17)(29)(53)(73)(233) 34577 132287 149177 161567 98993 71033 13649732209 (29)(73)(109)(149)(397) 34841 132685 171137 175663 120275 52759 16035692425 (5^2)(17^2)(1249)(1777) 7331 147239 171731 184651 111671 68251 17074867681 (73)(193)(509)(2381) 44375 164297 170621 191695 108271 98003 19358056825 (5^2)(17)(37)(157)(7841) 95851 157901 181969 188771 141029 108209 22410952321 (17)(53)(61)(173)(2357) 47905 196189 198623 226265 122477 118489 26745369625 (5^3)(13^2)(29)(149)(293) 49511 189449 224789 238321 152831 93371 29624119081 (109)(409)(557)(1193) 81121 200569 210541 232759 143281 128179 30378684361 (17)(41)(97)(613)(733) (Some of these don't satisfy the triangle inequality, so they don't give "real" geometrical triangles.) Does this mean we have found 3x3 magic squares of squares? No, because we eliminated the central number E^2 from the magic square equations (4). Nothing we've done forces E^2 to be a square integer, and in fact E^2 is not a square for any of the amicable pairs of triangles that I've found. In fact, it's clear that E^2 must be the product of at least three distinct primes of the form 4k+1, and in order for E^2 to be a square it would be necessary for each of those primes to occur to an even power. It's possible that such cases don't exist, which is interesting because equations (4) represent only a subset of the conditions that would have to be met by a complete magic square of square integers. For example, note that D^2 and F^2 don't even appear in those equations, and we haven't imposed the sums on the outer columns. So, since no one has ever been able to find a 3x3 magic square of distinct square integers (nor prove that such is impossible), let me propose the simpler(?) problem of finding a 3x3 magic square of the following form A^2 B^2 C^2 M E^2 N G^2 H^2 I^2 where M and N are integers but need not be squares. Thus I'm only requiring seven of the nine numbers to be squares, but they have to be the seven noted above, not just any seven, because it's known that there exists at least one magic square containing seven squares 23^2 205^2 289^2 373^2 425^2 222121 527^2 565^2 360721 but this doesn't satisfy the "amicable triangle" condition, i.e., it doesn't have both outer rows (or both outer columns) filled with squares, so it doesn't represent a solution of (4). By the way, it's interesting that the "automedian" quality is really just an expression of eigenvectors and eigenvalues. In general, suppose we have the linear system a11 x1 + a12 x2 + a13 x3 = y1 a21 x1 + a22 x2 + a23 x3 = y2 (5) a31 x1 + a32 x2 + a33 x3 = y3 Now suppose we require the values of {y1,y2,y3} be proportional to the values {x3,x2,x1}. In other words, for some constant k we have y1 = k x3 y2 = k x2 y3 = k x1 Inserting these into (5) gives a11 x1 + a12 x2 + (a13-k) x3 = 0 a21 x1 + (a22-k) x2 + a23 x3 = 0 (6) (a31-k) x1 + a32 x2 + a33 x3 = 0 Since the righthand vector is 0, the only non-trivial solutions require a value of k for which the determinant of the coefficient matrix is zero. Obviously the determinant of this matrix is a cubic in k, which has three roots, and those are called the eigenvalues of the system. By selecting any one of those eigenvalues, equations (6) typically give a solution [x1,x2,x3], which is called an eigen vector of the system. Now, in the particular case of the automedian system we have _ _ _ _ _ _ | -1 2 2 | | A^2 | | (Ma)^2 | | | | | | | | 2 -1 2 | | B^2 | = | (Mb)^2 | (3) | | | | | | |_ 2 2 -1 _| |_ C^2 _| |_ (Mc)^2 _| We want the medians to be proportional to the sides, so for some constant k we have (Ma)^2 = k C^2 (Mb)^2 = k B^2 (Mc)^2 = k A^2 (Note that we've chosen to permute the vector components in these relations, but in general we could choose any permutation, including identity.) It follows that the value of k must be such that | -1 2 2-k | | | Det| 2 -1-k 2 | = 0 | | | 2-k 2 -1 | Evaluating the determinant gives the "characteristic equation" k^3 - 3k^2 - 9k + 27 = 0 which has the roots -3, 3, and 3. With k=3 the system degenerates to a single equation in three variables, A^2 + C^2 = 2B^2, corresponding to the automedian triangles. With k=-3 the system degenerates to two conditions, A^2 = C^2 and B^2 + 2A^2 = 0. Obviously the only real solution of the second condition is A=B=0, but if we allow complex values we have the infinite family of solutions A=q, C=+-q, B=sqrt(-2)q. For more on the search for a magic square of squares, see the notes Magic Square of Squares Orthomagic Square of Squares

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