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How can I find the set of distinct positive integers $\mathbb S=\{a, b, c, d, e, f, g, h, i\}$ such that:

  1. $a+b$, $b+c$, $a+c$ are squares,
  2. $d+e$, $e+f$, $d+f$ are squares,
  3. $g+h$, $h+i$, $g+i$ are squares,
  4. $a+b+c = d+e+f = g+h+i$ is square?

If there are several solutions then choose with minimal $a+b+c$.

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    @MarkDominus, edited: yes, I want them to be distinct.2012-04-05

5 Answers 5

1

I used essentially the same strategy as @ronald: Let $(I, J, K)$ be perfect squares, and then if $I+J+K\over 2$ is also a perfect square, we have found a triple $(a,b,c) = ({I+J-K\over 2}, {J+K-I\over 2}, {K+I-J\over 2})$ that satisfies the conditions. We accumulate many such $(I,J,K)$ triples and see if the $I+J+K\over 2$ values match. Often, they do. When we find three or more matching sets, we have solved the problem.

# Python 3.6 from collections import defaultdict                                                                                                             N = 1000                                                                                                                                       is_twice_square = { s*s*2 for s in range(1, N+1) }                                                                                              solutions = defaultdict(list)                                                                                                                   for i in range(N+1):                                                                                                                               I = i*i                                                                                                                                        for j in range(i+1, N+1):                                                                                                                          J = j*j                                                                                                                                        for k in range(j+1, N+1):                                                                                                                          K = k*k                                                                                                                                        if I+J <= K or J+K <= I or K+I <= J:                                                                                                               continue                                                                                                                                   if I+J+K in is_twice_square:                                                                                                                       a, b, c = (I+J-K)//2, (J+K-I)//2, (K+I-J)//2                                                                                                   z = (I+J+K)//2                                                                                                                                 solutions[z].append((a, b, c))                                                                                                                 if len(solutions[z]) > 2:                                                                                                                          for triple in solutions[z]:                                                                                                                        print(" + ".join(map(str, triple)), "=", z)                       

Solutions are abundant. For example, all of the following add up to $901^2$:

1801 + 722400 + 87600 8985 + 488040 + 314776 14352 + 606592 + 190857 17920 + 679305 + 114576 21480 + 440920 + 349401 23257 + 694152 + 94392 25032 + 564792 + 221977 39160 + 515865 + 256776 44425 + 721200 + 46176 49672 + 423672 + 338457 63576 + 682920 + 65305 70480 + 577545 + 163776 70480 + 667401 + 73920 97776 + 601120 + 112905 99465 + 426160 + 286176 99465 + 485760 + 226576 101152 + 494832 + 215817 106201 + 413640 + 291960 107880 + 432345 + 271576 112905 + 519120 + 179776 117912 + 471912 + 221977 134472 + 533017 + 144312 141040 + 492576 + 178185 182952 + 360217 + 268632 192432 + 365577 + 253792 197145 + 403480 + 211176 

Addendum: I should explain how I came up with this strategy so that it doesn't seem like I pulled it out of nowhere. I first tinkered around with various sums and products of $a+b$, $a+c$, and $b+c$ looking for something that seemed promising. All three of these sums are squares, and if you add them all together, you get $2a+2b+2c$, which must be twice a square. So I knew I was looking for three squares that added up to twice a fourth square, and searching over squares is faster than searching over integers. So I let $(I, J, K) = (a+b, a+c, b+c)$ and did the search on $(I, J, K)$ instead of on $(a,b,c)$. I think there's some general principle at work here about looking for lost wallets under lampposts but I'm not quite sure what it is.

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    That's what I did too.2012-04-05
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You can find 'triples' a, b, and c by taking any three square numbers, e.g. ${1, 4, 9}$ and then solving for a, b and c simultaneously, such that: $a+b = 1, a+c=4, b+c=9$

In this case: $a=-2, b=3, c=6$

This case actually doesn't work because $a = -2$ but nonetheless this gives you a general method. If your chosen three numbers are sufficiently far away from 0 (close to each other), negative numbers shouldn't be a problem. In fact, to avoid negatives, I think we need to choose three squares which satisfy the triangle inequality, that the sum of the two smaller squares is bigger than the largest square.

Now note that $2(a+b+c) = (1+4+9)$ So, then you need to find three sets of squares that sum to the same square number, to give your three sets. The big square also has to be even, or you'll get non-integer results. You also have to ensure that you get integer results :S

The 'minimal (a+b+c)' condition is covered by choosing the smallest square number that has this property, so we can stop when we find one solution.

So, we need the smallest even square that can be split into three distinct squares, in three distinct ways, where each way satisfies the triangle inequality!

I was hoping for help from The Encyclopedia of Integer Sequences but it only partially covers that situation, at https://oeis.org/A024803, and not well enough to find a suitable square by hand :(

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The equation $x^2+y^2+z^2=2w^2$ has infinitely many solutions. For example, one family is

$x=3m^2+2mn-n^2$

$y=3m^2-2mn-n^2$

$z=4mn$

$w=3m^2+n^2$

5

A computer search gives $a=97$, $b=192$, $c=2112$; $d=720$, $e=801$, $f=880$; and $g=376$, $h=465$, $i=1560$.

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    This obviously yields the related solution 9700 + 19200 + 211200 = 72000 + 80100 + 88000 = 37600 + 46500 + 15600 = $490^2$. But curiously, there is another solution with $9700 + a + b = 490^2$, namely $9700 + 165024 + 65376 = 490^2$, and in fact we can do even better for $490^2$; there are at least 8 pairwise distinct solutions: 9700 + 165024 + 65376 = 26656 + 161700 + 51744 = 19200 + 160576 + 60324 = 37600 + 156000 + 46500 = 30336 + 121764 + 88000 = 55200 + 116196 + 68704 = 44736 + 110500 + 84864 = 72000 + 88000 + 80100 = $490^2$.2012-04-05
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Partial table of results (sorted by $\sqrt{a+b+c}$) :

$\small \begin{array} {r|lllllll} 49&[97, 192, 2112] & [376, 465, 1560] & [720, 801, 880]\\ 71&[*280, 945, 3816] & [*280, 1320, 3441] & [417, 1792, 2832]\\ 79&[312, *1057, 4872] & [616, 2520, 3105] & [*1057, 1752, 3432]\\ 85&[336, 1600, 5289] & [*2041, 2184, 3000] & [984, *2041, 4200]\\ 87&[344, *680, 6545]& [513, 1008, 6048]& [*1169, 1640, 4760]& [*680, *1169, 5720]\\ 91&[360, 3240, 4681]& [1392, 3097, 3792]& [1720, 2505, 4056]\\ 95&[561, 2464, 6000]& [744, 1281, 7000]& [2136, 2625, 4264]\\ 97&[193, 1488, 7728]& [*1128, 2353, 5928]& [*1128, 3633, 4648]\\ 98&[388, 768, 8448]& [1504, 1860, 6240]& [2880, 3204, 3520]\\ 103&[808, *2328, 7473] &[1584, 3040, 5985]& [2688, 3553, 4368]& [1393, *2328, 6888]\\ 105&[1224, 2376, 7425] &[209, *2000, 8816] &[416, *4625, 5984] &[*824, 3800, 6401]& [1616, *4625, 4784]& [*824, *1025, 9176]&[*1025, *2000, 8000]\\ 111&[1296, 5265, 5760]& [1712, 2912, 7697]& [2520, 3105, 6696]\\ 113&[225, 3744, 8800]& [448, 1953, 10368]& [1320, 4305, 7144]& [2568, 4488, 5713]& [2769, 4120, 5880]\\ 115&[*681, 1344, 11200]& [2200, 4200, 6825]& [*681, 4944, 7600]\\ 117&[233, 3488, 9968]& [1145, *3080, 9464]& [*2240, 3689, 7760]& [*3080, 4664, 5945]&[464, *2240, 10985]\\ 119&[472, 6417, 7272]& [*2280, 3345, 8536] &[*2280, 4945, 6936]\\ 121&[480, 3616, 10545]& [*1185, 1416, 12040]& [*1185, 4440, 9016]\\ 123&[2360, 6104, 6665]& [1904, 2585, 10640]& [3465, 4104, 7560]\\ 125&[249, 6976, 8400]& [*1225, 5016, 9384]& [4176, 5625, 5824] &[4600, 4809, 6216]&[984, *1225, 13416]\\ 127&[504, 4680, 10945]& [1968, 3808, 10353]& [3360, 5104, 7665]\\ 129&[512, 7232, 8897]& [1016, 4760, 10865]& [2241, 4320, 10080]\\ 131&[1785, 6136, 9240] &[*2032, 5712, 9417]& [3936, 5280, 7945]&[777, *2032, 14352]\\ 133&[1048, 6873, 9768]& [2313, 7488, 7888]& [4000, 4464, 9225]\\ 135&[*536, 1064, 16625]& [*2600, 5681, 9944]& [*536, *2600, 15089]\\ 137&[273, 2128, 16368]& [*1345, 3144, 14280]& [*1345, 5544, 11880]\\ 141&[3497, 7112, 9272]& [*281, 560, 19040]& [1656, 7560, 10665]& [2457, 6192, 11232]& [2720, 4505, 12656]& [*281, 8000, 11600]\\ \end{array} $

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    @Ronald: Thanks for the observation Ronald! I didn't verify unicity in my computer search and found such solutions of interest too. Anyway I'll update my table and try to indicate the duplicate numbers.2012-04-05