We can generalise the concept of Taxicab numbers, by denoting
x11p + x12p + … + x1tp =
x21p + x22p + … + x2tp =
xm1p + xm2p + … + xmtp =
S = rp
as Taxicab(p,t,m). Generalised Taxicab numbers are equal sums of like powers, with an equal number of terms. Using this notation, conventional Taxicab(m) numbers are Taxicab(3,2,m). Here are known results, along with some results of computer searches I have performed myself.
The minimal solutions, and some upper bounds, are:
The upper bounds were calculated by a non-exhaustive method described here.
The minimal co-prime solutions are:
The minimal solutions are:
No solutions are known.
I have exhaustively searched this to a radius of 19,034. The minimal solutions are here. Solutions with some terms zero and with Taxicab(4,2,n) being a subset (e.g. x11 = x21) were allowed in the search.
I have exhaustively searched this to a radius of 52,586. 15,904 solutions with multiplicity 2 have been found, and none with multiplicity 3. These solutions include one with a zero term, 141325 + 2205 = 140685 + 62375 + 50275, which was first found by Bob Scher & Ed Seidl in 1997.
I have exhaustively searched this to a radius of 33,225. 550 solutions with multiplicity 2 have been found, and none with multiplicity 3.
I have exhaustively searched this to a radius of 34,197. No solutions have been found.
We can extend the concept of Cabtaxi numbers, denoting
x11p ± x12p = x21p ± x22p = … = xm1p ± xm2p = S = rp
by Cabtaxi(p,m). Using this notation, conventional Cabtaxi(m) numbers are Cabtaxi(3,m).
The (exhaustive) search algorithm I have used is based on that by D. Bernstein, which you'll need to understand to make sense of some of what follows. In essence it uses a priority queue to enumerate tuples in order of their sums, whilst testing for equal sums. I have extended the algorithm to powers higher than 3, and the transformation of xp ± yp is different - I have used x, y = ½ (a±b) with a ≡ b (mod 2) as this gives greater cancellation of terms, and speeds up computation of the sum. Here are known results, along with some results of computer searches I have performed myself.
I have exhaustively searched this to a radius of 2,394,889. The minimal solutions, and some upper bounds, are:
The upper bounds are calculated by a non-exhaustive method described here.
The minimal co-prime solutions are:
I have exhaustively searched this to a radius of 489,506. The minimal solutions are:
No solutions to Cabtaxi(5,2) have been found.
The algorithm can be used to search for solutions to x15 ± x25 ± x35 ± x45 = 0. Inspection of this equation modulo 25 shows that the terms sum in pairs to 0 (mod 5). We can re-arrange any solution of this equation into Cabtaxi(5,2) form, with the two terms of a tuple summing to 0 (mod 5). Since a, b = x±y, this means we can force a in the search algorithm to be divisible by 5, significantly reducing the queue size and speeding up the search. Furthermore, the tuples are generated in order, and instead of looking for equality of tuples, we can look for their difference being a 5th power. This gives solutions of x15 ± x25 ± x35 ± x45 ± x55 = 0, albeit where x5 is relatively small (and divisible by 5 because of the imposed modulo constraint).
The search has been taken to a radius of 5,270,658, with this constraint on a. As a search for Cabtaxi(5,2) it is messed up a bit, but still exhaustive (after rearrangement of terms). As a search for four 5th powers summing to zero it is exhaustive, and as a search for five 5th powers summing to zero is quite effective, although not exhaustive. It finds 141325 + 2205 = 140685 + 62375 + 50275 (Bob Scher & Ed Seidl, 1997) and 853595 = 852825 + 289695 + 31835 + 555 (Jim Frye, 2004) in about a minute and an hour respectively on a 1.8GHz PC, which is much faster than the original exhaustive searches. It does not find the only other known solution of five 5th powers summing to zero, 1445 = 1335 + 1105 + 845 + 275 ( L.J. Lander & T.R. Parkin, 1966), since the term divisible by 5 is relatively big in this case. I find it (a little) curious that in all 3 known solutions, it is the same term that is divisible by 5 and 11, and it is also small (in absolute terms).