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I need to solve the system of equations, call it $S_1$, in the integers

$$x_1x_2x_3x_4x_5 = y_1y_2y_3y_4y_5$$

$$x_1^k+x_2^k+\dots+x_5^k=y_1^k+y_2^k+\dots+y_5^k,\;\; k= 2,4,6$$

I used a very primitive code using KSubsets[], Table[], Sort[], and Do[], and with $x_i, y_i\le 50$ found one,

$$2\cdot16\cdot25\cdot45\cdot48 =3\cdot9\cdot32\cdot40\cdot50$$

$$2^k+16^k+25^k+45^k+48^k = 3^k+9^k+32^k+40^k+50^k$$

Using this single example, I managed to find a parametrization, so $S_1$ has an infinite number of primitive solutions. However, I would like to find other small solutions to $S_1$, perhaps $x_i, y_i\le 300$, (which hopefully belongs to other families). Is there an efficient way to do it with Mathematica? The code I used takes forever if I extend the range > 50. :(

P.S. The system $S_1$ leads to an $11$th degree identity, in case one is wondering why I am interested in it.

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Have you looked into FindInstance[], by any chance? –  J. M. May 17 '13 at 2:47
I'm embarrassed to admit I'm still using Mathematica 4. I believe that command began with Mathematica 5. –  Tito Piezas III May 17 '13 at 3:02
Do you want to solve it in Integers or Reals domain? –  Silvia May 17 '13 at 3:05
Oh, well let's tag it appropriately... :) (I used version 5 for a long while before upgrading to 8 myself.) –  J. M. May 17 '13 at 3:05
@TitoPiezasIII Sorry, my comment wasn't clear enough. Mma can't find a solution using that :) –  belisarius May 17 '13 at 4:16
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1 Answer

up vote 4 down vote accepted

This isn't really an answer, but it's too big for a comment.

Here's some code to brute force solve the problem (since there wasn't any code provided in the question). It lists all the possible numbers; finds the ones whose product matches; then of those, finds the ones whose sum of squares match, etc..

Module[{h = 50, r, product, squared, fourth, sixth},
  r = Flatten[
    Table[{a, b, c, d, e}, {a, h}, {b, a + 1, h}, {c, b + 1,h},
        {d, c + 1, h}, {e, d + 1, h}], 4];
  product =
    Select[GatherBy[r, Times @@ # &], Length@# > 1 &];
  squared =
    Flatten[Select[GatherBy[#, Total@(#^2) &], Length@# > 1 &] & /@ product, 1];
  fourth =
    Flatten[Select[GatherBy[#, Total@(#^4) &], Length@# > 1 &] & /@ squared, 1];
  sixth =
    Flatten[Select[GatherBy[#, Total@(#^6) &], Length@# > 1 &] & /@ fourth, 1];

It's quite fast, but runs into memory issues fairly quickly after 50. I've managed to run it up to $1 \leq x_i \leq 60$ and found no more solutions than in the question.

I've got things to be and places to do, but hopefully somebody might find this useful as a starting point. On the other hand, if you want to check up to 300, there will be 20 billion combinations to sift through using this method.


Using a C++ program to check up to 150, I've found the following solutions:

$$\begin{align} 2\cdot 16\cdot 25\cdot 45\cdot 48 &= 3\cdot 9\cdot 32\cdot 40\cdot 50\\ 2\cdot 43\cdot 52\cdot 99\cdot 114 &= 3\cdot 22\cdot 76\cdot 86\cdot 117\\ 4\cdot 32\cdot 50\cdot 90\cdot 96 &= 6\cdot 18\cdot 64\cdot 80\cdot 100\\ 8\cdot 33\cdot 38\cdot 68\cdot 87 &= 12\cdot 17\cdot 57\cdot 58\cdot 88\\ 6\cdot 37\cdot 41\cdot 87\cdot 110 &= 11\cdot 15\cdot 58\cdot 82\cdot 111\\ 8\cdot 50\cdot 51\cdot 100\cdot 123 &= 12\cdot 24\cdot 82\cdot 85\cdot 125\\ 7\cdot 56\cdot 57\cdot 102\cdot 116 &= 8\cdot 38\cdot 84\cdot 87\cdot 119\\ 11\cdot 42\cdot 58\cdot 100\cdot 105 &= 14\cdot 28\cdot 75\cdot 87\cdot 110\\ 14\cdot 44\cdot 56\cdot 93\cdot 123 &= 21\cdot 24\cdot 77\cdot 82\cdot 124\\ 4\cdot 50\cdot 57\cdot 110\cdot 129 &= 6\cdot 25\cdot 86\cdot 95\cdot 132\\ 15\cdot 43\cdot 75\cdot 116\cdot 124 &= 20\cdot 29\cdot 93\cdot 100\cdot 129\\ 6\cdot 48\cdot 75\cdot 135\cdot 144 &= 9\cdot 27\cdot 96\cdot 120\cdot 150 \end{align}$$


$$\begin{align} 2\cdot 52\cdot 72\cdot 133\cdot 147 &= 3\cdot 28\cdot 98\cdot 117\cdot 152\\ 19\cdot 48\cdot 86\cdot 143\cdot 144 &= 26\cdot 32\cdot 99\cdot 129\cdot 152\\ 4\cdot 41\cdot 86\cdot 144\cdot 159 &= 9\cdot 16\cdot 106\cdot 129\cdot 164\\ 16\cdot 66\cdot 76\cdot 136\cdot 174 &= 24\cdot 34\cdot 114\cdot 116\cdot 176\\ 12\cdot 59\cdot 82\cdot 143\cdot 174 &= 22\cdot 26\cdot 116\cdot 123\cdot 177\\ 11\cdot 75\cdot 100\cdot 148\cdot 183 &= 12\cdot 61\cdot 125\cdot 132\cdot 185\\ 16\cdot 53\cdot 102\cdot 176\cdot 177 &= 24\cdot 32\cdot 118\cdot 159\cdot 187\\ 24\cdot 74\cdot 105\cdot 172\cdot 175 &= 28\cdot 56\cdot 129\cdot 150\cdot 185\\ 3\cdot 32\cdot 112\cdot 182\cdot 183 &= 7\cdot 13\cdot 122\cdot 168\cdot 192\\ 8\cdot 34\cdot 132\cdot 171\cdot 185 &= 11\cdot 24\cdot 148\cdot 153\cdot 190\\ 15\cdot 71\cdot 75\cdot 164\cdot 212 &= 20\cdot 41\cdot 100\cdot 159\cdot 213\\ 8\cdot 64\cdot 100\cdot 180\cdot 192 &= 12\cdot 36\cdot 128\cdot 160\cdot 200\\ 12\cdot 74\cdot 82\cdot 174\cdot 220 &= 22\cdot 30\cdot 116\cdot 164\cdot 222\\ 16\cdot 96\cdot 106\cdot 189\cdot 209 &= 19\cdot 64\cdot 154\cdot 159\cdot 216\\ 26\cdot 74\cdot 123\cdot 177\cdot 220 &= 30\cdot 59\cdot 143\cdot 164\cdot 222\\ 22\cdot 84\cdot 116\cdot 200\cdot 210 &= 28\cdot 56\cdot 150\cdot 174\cdot 220\\ 16\cdot 66\cdot 119\cdot 208\cdot 213 &= 24\cdot 39\cdot 142\cdot 187\cdot 224\\ 4\cdot 86\cdot 104\cdot 198\cdot 228 &= 6\cdot 44\cdot 152\cdot 172\cdot 234\\ 9\cdot 65\cdot 118\cdot 200\cdot 228 &= 20\cdot 25\cdot 152\cdot 177\cdot 234\\ 24\cdot 84\cdot 122\cdot 209\cdot 217 &= 31\cdot 56\cdot 154\cdot 183\cdot 228\\ 28\cdot 88\cdot 112\cdot 186\cdot 246 &= 42\cdot 48\cdot 154\cdot 164\cdot 248\\ 28\cdot 58\cdot 133\cdot 213\cdot 222 &= 37\cdot 42\cdot 142\cdot 203\cdot 228\\ 8\cdot 29\cdot 142\cdot 219\cdot 228 &= 12\cdot 19\cdot 146\cdot 213\cdot 232\\ 16\cdot 100\cdot 102\cdot 200\cdot 246 &= 24\cdot 48\cdot 164\cdot 170\cdot 250\\ 14\cdot 112\cdot 114\cdot 204\cdot 232 &= 16\cdot 76\cdot 168\cdot 174\cdot 238\\ 30\cdot 77\cdot 126\cdot 205\cdot 244 &= 35\cdot 61\cdot 140\cdot 198\cdot 246\\ 10\cdot 80\cdot 125\cdot 225\cdot 240 &= 15\cdot 45\cdot 160\cdot 200\cdot 250 \end{align}$$

share|improve this answer
OP is on version 4, so he won't have GatherBy[] handy. –  J. M. May 17 '13 at 4:46
@Wxffles: The code I came up with used Table[] and KSubsets[] to generate all possible sums $a^k+b^k+c^k+d^k+e^k$ for $k=6$, then used Sort[], then Do[] to filter out those valid for $k=2,4$ as well as being equal products. It was slow, cumbersome, and also ran into memory issues. But thanks so much for the effort though. –  Tito Piezas III May 17 '13 at 14:54
@Wxffles: Wonderful! In general, $S_1$ leads to an multi-grade $11$th deg equality with 16 terms on each side. But with appropriate linear relations among the terms of $S_1$, this can be reduced. Your fourth solution reduces it to just 11 terms on each side, namely {$117, -84, -79, -49, -3, -8, -38, 22, 71, 76, -109$} = {$116, -99, -59, -58, -28, 42, 1, -29, 47, 87, -104$} the kth powers have equal sums for $k = 1,2,3,4,5,6,7,9,11$. Thanks so much! Now I'll inspect this solution if it belongs to a family. –  Tito Piezas III May 17 '13 at 21:58
@Wxffles: Thanks! I found that for your first edit, excepting two solns, the rest fell into 2 families. The 1st can be solved by quadratic forms, the 2nd can be reduced to an elliptic curve. The latter is {$4a−4b,3b−2c,3b+2c,6a,4a+4b$} = {$b−2c,b+2c,4a,−a+d,a+d$} where $−4a^2+5b^2=4c^2,\;25a^2+24b^2=d^2$. The intersection of these two is an elliptic curve and one soln is {$a,b,c,d$}={$29,26,2,193$} which gives the 5th in your list. If you like this programming problem, can you kindly email me at tpiezas@gmail.com as I have a similar one that is simply begging for more solutions. –  Tito Piezas III May 30 '13 at 18:27
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