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Discordance Impedes Square Magic |
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As discussed in a previous section, the terms of a 3x3 magic square of squares can be re-arranged into a 3x3 bilinear array as shown below: |
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The values of 2u and 2v are the vertical and horizontal "partials", respectively. Notice that the differences are even, because we have the conditions |
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which show that if any of the terms a,b,..,i is even, then they ALL must be even, so we can divide each of them by 2, and repeat until all are odd. (By a similar analysis we can show that all the terms a,b,..,i must be congruent modulo 3, so all the squares must be congruent modulo 36.) Therefore, we have |
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Subtracting equation (1b) from (1a) gives |
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Letting B and H denote the two integer factors in parentheses, we see that |
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Similarly from equations (2), (3), and (4) we deduce integers F, D, C, G, A, and I such that |
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These equations together imply that if a magic square of square exists then there must be four rational right triangles on a common hypotenuse with areas A1, A2, A3, A4 such that A1+A2 = A3 and A1–A2 = A4. This can also be expressed by the conditions |
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Just to emphasize how surprising a magic square of squares would be, it's worth noting that even if we drop the right hand requirement, it appears that there do not exist THREE rational right triangles on a common hypotenuse with areas in arithmetic progression (signified by the left hand condition). This would correspond with a set of integers a,b,..,f,H such that |
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No such integers have ever been found. However, if we relax the requirement by replacing H^2 with an arbitrary integer K, then there exist some solutions, although even these are not terribly plentiful. There are only 39 primitive solutions with K less than 52 million, as listed in the table below. |
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K diff a b c d e f |
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-------- -------- ------ ----- ----- ----- ----- ----- |
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25345 5460 8 159 44 153 96 127 |
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32266 6720 15 179 55 171 125 129 |
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36490 6720 3 191 39 187 81 173 |
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99025 15960 48 311 104 297 183 256 |
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99125 9828 23 314 55 310 89 302 |
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325117 61200 69 566 186 539 379 426 |
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419050 87360 21 647 161 627 343 549 |
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743665 143220 96 857 276 817 569 648 |
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1006561 170940 81 1000 260 969 480 881 |
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1229045 83160 439 1018 551 962 758 809 |
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2047786 255360 5 1431 185 1419 375 1381 |
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3129802 235200 661 1641 851 1551 1179 1319 |
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4658425 228228 948 1939 1120 1845 1389 1652 |
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5602945 840840 16 2367 376 2337 768 2239 |
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7805890 779520 357 2771 651 2717 973 2619 |
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8181625 879648 844 2733 1235 2580 1908 2131 |
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9157850 993600 139 3023 473 2989 827 2911 |
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9551777 1096200 604 3031 1001 2924 1484 2711 |
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9699265 999180 672 3041 1036 2937 1473 2744 |
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9887266 549120 1179 2915 1421 2805 1725 2629 |
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14281930 2291520 33 3779 649 3723 1331 3537 |
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14751841 3127320 79 3840 920 3729 2000 3279 |
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18870865 3714480 96 4343 976 4233 2048 3831 |
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19734650 1330560 1741 4087 2183 3869 2921 3347 |
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21651370 2735040 891 4567 1551 4387 2389 3993 |
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22163530 1404480 1629 4417 2023 4251 2513 3981 |
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22884277 4395300 486 4759 1474 4551 2946 3769 |
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23386441 683760 1429 4620 1596 4565 1771 4500 |
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26809445 4241160 778 5119 1679 4898 2911 4282 |
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29713450 2735040 7 5451 511 5427 1029 5353 |
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30002050 2542848 1917 5131 2555 4845 3669 4067 |
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31676033 4781700 1057 5528 2023 5252 3488 4417 |
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33826325 4823172 1223 5686 2185 5390 3698 4489 |
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36133681 7332780 160 6009 1420 5841 3000 5209 |
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37425389 5242860 1358 5965 2365 5642 4067 4570 |
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37661026 2914560 1901 5835 2499 5605 3251 5205 |
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38878705 4084080 1063 6144 1776 5977 2592 5671 |
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42639466 7512960 875 6471 2135 6171 4021 5145 |
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50205361 6846840 1280 6969 2360 6681 3769 6000 |
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For a conjecture related to this table, see the note entitled “No Four Rectangles in Line?” |
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Suppose we place each of our three putative right triangles with their common hypotenuses on the x axis as shown below. (Only the a,b triangle edges are shown.) |
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We want the areas of our three rational right triangles to be in arithmetic progression, and since the triangles are on a common base, it's clear that their heights must be in arithmetic progression. Also, if the edges of the right triangle are rational, their areas and therefore their heights are rational, so the y coordinates of their upper vertices are rational. We can also show that the x coordinates of those vertices are rational because, referring to the above figure, we have |
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Therefore, if we re-scale the figure so that the radius of the circumscribing circle is unity (which we can do by dividing each length by half the length of the hypotenuse), the three upper vertices have rational x,y coordinates on the unit circle. Now, the usual way of parameterizing rational points on a unit circle is with the mapping |
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(It follows that t = y/(1+x).) So, letting r,s,t denote the rational parameters of our three points on the circle, the requirement for their heights (i.e., their y coordinates) to be in arithmetic progression gives the following necessary condition for a complete solution of (5) |
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It's important to note that although this is necessary, it is not sufficient, because while the rationality of the edges implies rationality of the x,y coordinates, the converse is not true. We can have rational solutions of (7) that don't correspond to triangles with rational edge lengths, even though the vertices have rational coordinates. The table below shows the first 41 primitive sets of three rational points on the unit circle with heights in arithmetic progression, which correspond to three rational right triangles on a common hypotenuse with areas in arithmetic progression. |
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H del y x1 y1 x2 y2 x3 y3 rs r/s |
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---- ----- ---- ---- ---- ---- ---- ---- ------ ------ |
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65 -4 25 60 33 56 39 52 0.3333 1.3333 |
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185 44 175 60 153 104 111 148 0.0833 0.3333 |
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205 39 200 45 187 84 164 123 0.0370 0.3333 |
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221 55 204 85 171 140 104 195 0.1200 0.3333 |
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425 -104 119 408 297 304 375 200 0.1875 3.0000 |
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725 161 696 203 627 364 500 525 0.0612 0.3333 |
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901 -15 424 795 451 780 476 765 0.3333 1.0800 |
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1105 -256 264 1073 744 817 952 561 0.2138 2.8739 |
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1105 -276 1020 1020 817 744 1001 468 0.1067 2.1600 |
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1469 185 1456 195 1419 380 1356 565 0.0133 0.3333 |
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1769 -260 319 1740 969 1480 1281 1220 0.3333 2.0833 |
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1885 -23 312 1859 427 1836 516 1813 0.6389 1.1206 |
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1885 -231 464 1827 1003 1596 1300 1365 0.3333 1.8148 |
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1885 -343 1643 1643 1365 1300 1624 957 0.1270 1.7076 |
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2405 -68 483 2356 741 2288 925 2220 0.5439 1.2237 |
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2425 -264 679 2328 1273 2064 1625 1800 0.3333 1.6875 |
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2465 376 2431 408 2337 784 2175 1160 0.0208 0.3333 |
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2465 729 2340 775 1953 1504 1044 2233 0.1026 0.2535 |
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3445 -484 1749 2968 2387 2484 2805 2000 0.1829 1.7857 |
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3485 -117 236 3477 925 3360 1276 3243 0.6365 1.3718 |
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3625 -129 1276 3393 1577 3264 1820 3135 0.3986 1.2024 |
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3965 649 3900 715 3723 1364 3416 2013 0.0248 0.3333 |
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4505 -1036 689 4452 2937 3416 3825 2380 0.2449 3.0000 |
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4745 976 4599 1168 4233 2144 3575 3120 0.0469 0.3333 |
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4745 -329 2144 4233 2697 3904 3120 3575 0.2793 1.3518 |
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5069 -455 1644 4795 2619 4340 3256 3885 0.3333 1.5306 |
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5185 -243 144 5183 1575 4940 2196 4697 0.6189 1.5284 |
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5525 511 5500 525 5427 1036 5304 1547 0.0068 0.3333 |
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5525 -324 1131 5408 2163 5084 2805 4760 0.4643 1.4219 |
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5525 961 5084 2163 4557 3124 3720 4085 0.0901 0.4614 |
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6205 679 5244 3317 4747 3996 4080 4675 0.1317 0.6374 |
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6565 -1551 1300 6435 4387 4884 5656 3333 0.2231 3.0000 |
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6641 1420 6409 1740 5841 3160 4809 4580 0.0533 0.3333 |
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6929 -920 1521 6760 3729 5840 4879 4920 0.3333 1.9200 |
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7085 2759 7076 357 6363 3116 3960 5875 0.0134 0.0474 |
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7565 1079 7476 1157 7227 2236 6800 3315 0.0178 0.3333 |
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7565 -1344 2236 7227 4756 5883 6052 4539 0.2458 2.2121 |
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8621 -2135 8155 8155 6171 6020 7696 3885 0.1157 2.0417 |
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9425 -49 776 9393 1233 9344 1560 9295 0.7791 1.0882 |
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9425 1587 8372 4329 7337 5916 5704 7503 0.1206 0.4905 |
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9881 -2360 2169 9640 6681 7280 8569 4920 0.2133 3.0000 |
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In none of these cases are the edge lengths of the triangles rational, so they don't represent solutions of (5). This raises the question of whether we can prove that a rational solution of (7) (which is a necessary condition for (5)) can never give rational edge lengths. If we could prove this, then it would follow that no solution of (5) exists, and therefore no magic square of squares exists. I don't have a proof, but it's interesting that we can rule out a large class of possible solutions by means of a certain "discordant form", as explained below. |
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Note that (7) is a quadratic in t, and if we solve for t we get an expression containing a square root of the quantity |
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so a sufficient condition is for this to be a rational square. Thus we have a rational number q such that |
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Dividing both sides by the right hand side, we can put this into a form that closely resembles the usual parametric representation of a circle |
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where Q is an arbitrary rational number. One obvious class of solutions to (8), in view of (6), is given by identifying the quantity (r-s)/(1+rs) with the rational parameter T, and then setting the quantity (r+s)/(1+rs) to either 1–T2 or 2T, equating the large fraction in the above expression with x or y (respectively) of equations (6). |
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In the first case, we set the quantity (r+s)/(1+rs) to 1–T2, leading to a quadratic equation in r and s which can be solved for either variable to give an expression involving the square root of a number of the form 5q2 + 2q + 5 where q is rational, and we require this to be a rational square. Thus there must be integers m,n,k (not all divisible by 3) such that |
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but this is impossible, as can be seen by inspection modulo 9. |
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In the second case, after identifying (r-s)/(1+rs) with the rational parameter T, we set quantity (r+s)/(1+rs) to 2T, which implies the condition r = 3s, or, by symmetry, r = s/3 or r = 1/(3s) or r = 3/s). This obviously gives an infinite family of solutions, i.e., sets of three rational points on a unit circle with heights in arithmetic progression, and in fact we see that 22 of the 41 sets listed above are of this form. In addition, three more are of the form s = 1/(3r), which is another infinite family of solutions, as shown in the note “Bi-Rational Substitutions Giving Squares”. |
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However, we can show that the edge lengths corresponding to r,s are never rational if r = 3s or 1/(3s). To prove this, note that for any given vertex parameter t, corresponding to the vertex x,y, one of the edge lengths is given by |
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Substituting the expressions from (6) into this equation E2 as a function of the parameter t |
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Consequently, for the two parameters r and s, with r = 3s, we require both of the quantities 4s2/(1+s2) and 36s2/(1+9s2) to be rational squares. Inverting, and noting that the numerators are already squares, we see that we must have 1+s2 and 1+9s2 both equal to rational squares for some rational s. Clearing fractions, we need integers m,n such that |
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This is a problem of Concordant Forms, and it's known that there are no integers m,n that make both m2 + n2 and m2 + kn2 for many specific values of k, including 9. Thus there can be no solution to these equations. An almost identical argument works for the case r = 1/(3s). |
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Therefore, none of the rational solutions of (7) given by setting rs or r/s equal to 1/3 can possibly give three rational right triangles with areas in arithmetic progression. This shows an interesting relation between concordant forms and the obstacles to constructing a magic squares of squares. |
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