
COMMENTS

These are solutions to equations of the form Sum[(x_i)^2, i:1 to k] = Sum[(x_i)^2, i:k+1 to 50].
Solutions after alignment of LHS and RHS to positive and negative terms (or vice versa) in Sum[(+/)(x_i)^2, i:1 to 50]=0 where the plus and minus sign counts are p and m. (See Example.)
A Minkowski signature over 50 variables (Mink50) where all possibilities are considered. This idea is useful because the examples given below render associated equations over various x^2 as zero, and so the metric tells us that the point and trivial transforms are origins. From this we can define the unit 50sphere as points that return 1 instead of 0.
The sequences can be read either forwards or backwards  a solution to 50k/k is a solution to k/50k if the solution is reversed.
The nonstandard entries are at 2, 3, 20, 23, 24, 26, 27, 30, 47 and 48.
Entries with more than 1 solution are 2, 5, 6, 7, 9, 10, 12, 13, 16, 34, 37, 38, 40, 41, 43, 44, 45 and 48.
Entries in which the LHS of a solution has no elements in common with a RHS solution are 1, 2, 3, 5, 9, 41, 45, 47, 48 and 49.
Conjecture: If a vector (x1,...,x50) is a zero to a signature S, then it is only also a zero for S' and no other signature T. {S' is the complement signature of S  all the signs are reversed, so if S=(+,,,+,) then S' is (,+,+,,+)}.
This is trivially false  for example consider 1,1,1,1,2,1,1,1,1,2 on Mink10  satisfies both ++++++++
and +++++ but these are not complementary. However we argue that they are locally complementary and hence are trivial counterexamples.
We therefore modify the conjecture to say that if we define Z_S as the set of integer zeros for S, and Z_t as the collection of integer zeros for all signatures T that are not S or S' (i.e., belong to t), then Z_S is NOT a subset of Z_t. The opposite  for some S, Z_S is a subset of Z_t  implies S has no function.
The conjecture is obviously true for k=1, and also for k=1 there are no 'shared zeros'. For k=2, pick one of the 2 squares so that it cannot be equaled by some combination of the other 48 squares, therefore it must be combined with its companion square and so must be a unique integer zero for S(2). For k=3 we require that 2 of the 3 available squares are not matchable, and nor is their sum. This method can be extended for higher k, however the basic existence of such a zero in the first place becomes increasingly less provable.
There are 3 spaces with trivial solutions, namely 25/25, 40/10 and 45/5. With 25/25 take any 25 integers and place on the +'s. The same 25 integers placed on the 's results in a zero. With the other 2, consider 10 and 5 squares respectively. Then by making 4 copies on the LHS and multipling each RHS entry by 4 (9 with 45/5) we get the desired result. Note that these are far from all the solutions, but just readily generatable solutions.
The first nontrivial solutions for k=25, 40 and 45 are 40, 40 and 45.
Other spaces, but not all, have simplifiable, or shortened, solutions. These also do not cover all solutions, but do provide an 'easy way in'. For example, any solution to a^2 + b^2 + c^2 = d^2 + e^2 (for example, 4 + 4 + 9 = 1 + 16) can be used to find zeros for both 48/2 and 30/20 spaces. For 48/2 make 16 copies of the LHS and multiply the RHS elements by 16 (this is denoted 'Cx16'). For 30/20 make 10 copies of both the LHS and RHS (denoted 'CC10'). A full list of shortened solutions is in the Example section. If the operation is Cx or xC then the number must be a square  if the operation is CC then it may be any number.
Another example is 1 square into 6 squares (1:6). Again this has 2 solutions and again they are 48/2 and 30/20. For 48/2 we first CC2 to get (2:12) and xC by 4. For 30/20, CC by 5 and xC by 4. (2:3) and (1:6) are the only shortened solutions that go to 2 different signatures.
Entries with at least 1 shortened solution are 2, 4, 5, 6, 8, 10, 12, 14, 15, 16, 18, 20, 22, 23, 24, 26, 27, 28, 30, 32, 34, 35, 36, 38, 40, 42, 44, 45, 46 and 48.
The number of shortened solutions for these values are 4, 1, 1, 2, 3, 3, 2, 3, 1, 3, 6, 5, 2, 1, 3, 3, 1, 2, 5, 6, 3, 1, 3, 2, 3, 3, 2, 1, 1 and 4.
See 'Links' for a list of shortened solutions.


EXAMPLE

(p/m means: the left sum consists of p squares and the right sum has m squares, for example, look at 39/11. The LHS is just 39 1's, and so a sum of 39 squares which equals 39, and the RHS is 5 1's, 4 4's and 2 9's, giving a total of 11 squares which when summed give 39.)
%49/1 : 49*1 = 49
$%*>48/2 : 44*1 + 4 + 4 + 4 + 9 = 65 = 1 + 64 = 16 + 49
%>47/3 : 46*1 + 4 = 50 = 9 + 16 + 25
$46/4 : 46*1 = 1 + 4 + 16 + 25
$%*45/5 : 45*1 = 1 + 1 + 9 + 9 + 25 = 9 + 9 + 9 + 9 + 9
$*44/6 : 44*1 = 4*1 + 4 + 36 = 1 + 1 + 4 + 4 + 9 + 25
*43/7 : 43*1 = 1 + 1 + 4*4 + 25 = 3*1 + 4 + 4 + 16 + 16
$42/8 : 42*1 = 4*1 + 4 + 9 + 9 + 16
%*41/9 : 41*1 = 7*1 + 9 + 25 = 4*1 + 4 + 4 + 4 + 9 + 16
$*40/10 : 40*1 = 10*4 = 8*1 + 16 + 16
39/11 : 39*1 = 5*1 + 4*4 + 9 + 9
$*38/12 : 38*1 = 5*1 + 6*4 + 9 = 9*1 + 4 + 9 + 16
*37/13 : 37*1 = 9*1 + 4 + 4 + 4 + 16 = 12*1 + 25
$36/14 : 36*1 = 10*1 + 4 + 4 + 9 + 9
$35/15 : 35*1 = 10*1 + 4*4 + 9
$*34/16 : 34*1 = 10*1 + 6*4 = 14*1 + 4 + 16
33/17 : 33*1 = 15*1 + 9 + 9
$32/18 : 32*1 = 15*1 + 4 + 4 + 9
31/19 : 31*1 = 15*1 + 4*4
$>30/20 : 28*1 + 4 + 4 = 36 = 18*1 + 9 + 9
29/21 : 29*1 = 20*1 + 9
$28/22 : 28*1 = 20*1 + 4 + 4
$>27/23 : 26*1 + 9 = 35 = 19*1 + 4*4
$>26/24 : 24*1 + 4 + 4 = 32 = 23*1 + 9
25/25 : 25*1 = 25*1
> indicates a nonstandard result.
* indicates at least 2 results.
% indicates LHS intersect RHS is null.
$ indicates a shortened solution.
The smallest nontrivial solution for n=25 is:
40 = 24*1 + 16 = 20*1 + 5*4

Some solutions for 49/1 spaces:
64 = 48*1 + 16
81 = 47*1 + 9 + 25 = 45*1 + 4*9
100 = 47*1 + 4 + 49 (not complete)
