Equal Sums of Two Distinct Like Powers

Samuel S. Wagstaff, Jr. Center for Education and Research in Information Assurance and Security, and Department of Computer Sciences Purdue University West Lafayette, IN 47907-1398 USA

                     ssw@cerias.purdue.edu

Abstract

We study the equation a^{j} +b^{j} = c^{k}+d^{k} for positive integers a, b, c, d, and 2 < j < k. Two heuristic arguments correctly predict the cases in which the equation has primitive solutions.

1 Introduction

Let (S_{n}^{(k)})_{n≥1} denote the sequence of sums of two positive k-th powers of integers. We consider mostly the sequences (S_{n}^{(k)})_{n≥1} with 3 ≤ k ≤ 6: \underline{A004999}, \underline{A003336}, \underline{A003347}, and \underline{A003358} in the OEIS [12]. For integers 2 < j < k, we investigate the numbers common to the two sequences (S_{n}^{(j)})_{n≥1} and (S_{n}^{(k)})_{n≥1}. The numbers (a^{k})^{j} + (b^{k})^{j} = (a^{j})^{k} + (b^{j})^{k}, for positive integers a and b, appear trivially in both sequences. We are more interested in nontrivial common elements.

Equivalently, we study the equation a^{j} + b^{j} = c^{k} + d^{k} for positive integers a, b, c, d. We are interested mostly in the case 2 < j < k, but we begin by reviewing the case j = k, which has been studied extensively.

Many solutions in integers are known for the equation a^{j} + b^{j} = c^{j} + d^{j} = N when j = 2, 3, and 4. We exclude trivial solutions with {a, b} = {c, d}. Only primitive solutions, those with gcd(a, b, c, d) = 1, are considered because all other solutions may be derived from the primitive ones. For example, the solution 399^{4}+402^{4} = 177^{4}+474^{4} would be ignored because it is just 3^{4} times the solution 133^{4} + 134^{4} = 59^{4} + 158^{4}.

Theorem 412 in Hardy and Wright [10] asserts that some integers have arbitrarily many representations as the sum of two cubes. Parametric solutions are known for j = 2, 3, and 4. Equation (13.7.11) in Hardy and Wright [10] gives a parametric solution for the case j = 4 and Choudhry [3] offers another one. Wroblewski [14] lists all primitive solutions to a^{4} + b^{4} = c^{4} + d^{4} with a, b, c, d ≤ 10^{14}. No primitive solution to a^{j} + b^{j} = N = c^{j} + d^{j} is known with j > 4. According to Guy [8, Sec. D1], people have searched at least up to N < 10^{25} for solutions in the case j = 5. Fermat’s Last Theorem says there are no solutions with d = 0. Browning [2] used algebraic geometry to show that primitive solutions with j > 4 are rare if they exist at all.

We have searched for solutions to the equation a^{j} + b^{j} = c^{k} + d^{k} in integers a, b, c, d with 2 < j < k < 11. There are many solutions with j = 2 and all k > 2. We found primitive solutions in the three cases j = 3, k = 4, 5, and 6, but none for other values of 2 < j < k < 11.

When one studies the equation a^{j} + b^{j} = c^{k} + d^{k} with j < k, one ignores trivial identities such as (a^{k})^{j} + (b^{k})^{j} = (a^{j})^{k} + (b^{j})^{k}, even when gcd(a, b) = 1.

In case j < k and j | k some solutions with gcd(a, b, c, d) > 1 might be considered if one wants to see all solutions, because sometimes a common factor cannot be canceled. For example, one solution to a^{3} + b^{3} = c^{6} + d^{6} is

102^{3} + 330^{3} = 12^{6} + 18^{6}  (1)

and the common factor 6 of the four numbers cannot be canceled since it appears to different powers on the two sides of the equation. However, this solution comes from a solution to w^{3} + x^{3} = y^{3} + z^{3}. If k = gcd(a, b, c, d) > 1 for a solution to a^{3} + b^{3} = c^{6} + d^{6}, then this equation may be rewritten as (kw)^{3}+(kx)^{3} = (ke)^{6}+(kf)^{6} or k^{3}w^{3}+k^{3}x^{3} = (k^{2}e^{2})^{3}+(k^{2}f^{2})^{3} from which k^{3} may be canceled to give w^{3} + x^{3} = (ke^{2})^{3} + (kf^{2})^{3}, which is primitive.

A solution (k^{2}a)^{3} + (k^{2}b)^{3} = (kc)^{6} + (kd)^{6} for some integer k > 1 is easily derived from the solution a^{3} + b^{3} = c^{6} + d^{6} by multiplying by k^{6}, so it is ignored. For example, 918^{3} + 2970^{3} = 36^{6} + 54^{6} is just the solution 102^{3} + 330^{3} = 12^{6} + 18^{6} multiplied by 3^{6}. Likewise, we ignore solutions to a^{3} + b^{3} = c^{6} + d^{6} with a = c^{2} and b = d^{2}.

Similar reductions apply also whenever j < k and j | k. We tested the 11089 solutions to w^{4} + x^{4} = y^{4} + z^{4} found by Wroblewski [14] to see whether one would give a solution to a^{4} + b^{4} = c^{8} + d^{8}. None of them worked (because none had square values for both w and x).

Lander [11] gave (1) as a solution when j = 3 and k = 6. He found parametric solutions to a^{j} + b^{j} = c^{k} + d^{k} (and to similar equations with any number of terms on each side) for every 2 ≤ j ≤ k, but none of his solutions are primitive. His solution to a^{j} + b^{j} = c^{k} + d^{k} is

(pw^{B})^{j} + (qw^{B})^{j} = (w^{A})^{k} + (w^{A})^{k},

where A, B and w are positive integers that depend on j, k, and the parameters p and q. For j = 5, k = 8, he gave a typical numerical solution (3 · 122^{3})^{5} + (122^{3})^{5} = (122^{2})^{8} + (122^{2})^{8}. For j = 3, k = 5, another numerical solution would be (4 · 36^{3})^{3} + (2 · 36^{3})^{3} = (36^{2})^{5} + (36^{2})^{5}.

The tables in the next section list only primitive solutions—those with gcd(a, b, c, d) = 1 (other than a = b = c = d = 1).

2 The results

We used the methods described in Bernstein [1] in the search for solutions. One heap (data structure) held the sums a^{j} + b^{j} and another one held the sums c^{k} + d^{k}. The program compared their least elements as they were removed. We used Algorithm 6 of Eisermann [4] to reduce the memory requirement of the program. The computation was done on the Brown cluster at Purdue’s RCAC.

We used local constraints to accelerate the algorithm. For instance, only five of the eleven residue classes modulo 11 are sums c^{5} + d^{5}, namely, −2, −1, 0, 1, 2 modulo 11. The map f(x) = x^{3} mod 11 is a permutation of the eleven classes. Thus for each a (mod 11), if a^{3} +b^{3} equals a sum of two fifth powers, then there are only five possible values of b (mod 11). For example, when a ≡ 3 (mod 11), we have a^{3} ≡ 5 (mod 11). Thus b^{3} mod 11 must be one of 4, 5, 6, 7, 8 (mod 11), so b mod 11 must be one of 5, 3, 8, 6, 2 (mod 11). To search these cases we ran five jobs, one for each possible value of b mod 11 (and all possible values for a, c, and d). We used the moduli 16, 11, 27, 29, 32, 19, 25, for k = 4, 5, 6, 7, 8, 9, 10, respectively, and most values of j < k.

If c ≤ d ≤ M, then a^{j} + b^{j} = c^{k} + d^{k} ≤ 2M^{k}, so a ≤ b ≤ L = (2M^{k})^{1/j}. The number of pairs a, b is about L^{2}/2. The search for solutions examines more a, b pairs than c, d pairs since j < k, and the naive running time would be O(L^{2}) or O(M^{2k/j}). However, we rewrote the equation as a^{j} − c^{k} = d^{k} − b^{j} and computed these differences instead of the sums. This reduced the work to O(M^{1+k/j}).

For j = 3, we used M = 10000, 5000, 1400, 700, 350, 200, 100 for k = 4, 5, 6, 7, 8, 9, 10, respectively. For j = 4, we used M = 18100, 3500, 1100, 460, 230, 135 for k = 5, 6, 7, 8, 9, 10, respectively. We made similar effort in the other cases. These limits kept the sums in range of 64-bit integer arithmetic in the three cases where we found solutions and in range of 128-bit integer arithmetic in the other cases. They also kept the search times reasonable.

Tables 1 and 2 show the primitive solutions to 0 ≤ a^{3} + b^{3} = c^{4} + d^{4} = N with 0 ≤ c ≤ d ≤ 10000 in order by the size of N.

Note the repeated values of c or d in these solutions.

45^{3} + 133^{3} = 39^{4} + 19^{4}
161^{3} + 176^{3} = 39^{4} + 52^{4}
887^{3} + 6457^{3} = 456^{4} + 690^{4}
2941^{3} + 14771^{3} = 456^{4} + 1338^{4}
6377^{3} + 7977^{3} = 227^{4} + 935^{4}
5528^{3} + 8529^{3} = 398^{4} + 935^{4}

\begin{matrix}a & b & c & d & a & b & c & d \\ 17 & 24 & 8 & 11 & 3658 & 29849 & 825 & 2262 \\ 45 & 133 & 19 & 39 & 7202 & 29977 & 225 & 2286 \\ 161 & 176 & 39 & 52 & 26224 & 27217 & 167 & 2486 \\ 641 & 960 & 133 & 170 & 18226 & 33641 & 396 & 2577 \\ 993 & 1400 & 32 & 247 & 16101 & 35149 & 1225 & 2595 \\ 840 & 1681 & 73 & 270 & 4401 & 37096 & 1644 & 2573 \\ 1417 & 1634 & 78 & 291 & 11986 & 39881 & 288 & 2841 \\ 41 & 3801 & 193 & 481 & 23097 & 42217 & 195 & 3059 \\ 4272 & 4337 & 35 & 632 & 14521 & 46850 & 1959 & 3090 \\ 887 & 6457 & 456 & 690 & 31249 & 44928 & 215 & 3318 \\ 6377 & 7977 & 227 & 935 & 39893 & 42669 & 2261 & 3275 \\ 5528 & 8529 & 398 & 935 & 32321 & 47936 & 981 & 3458 \\ 8669 & 9043 & 60 & 1086 & 23645 & 56133 & 401 & 3713 \\ 2941 & 14771 & 456 & 1338 & 35653 & 61979 & 1470 & 4086 \\ 9928 & 16849 & 1145 & 1418 & 17432 & 65969 & 1806 & 4097 \\ 2201 & 19721 & 1239 & 1519 & 39777 & 69704 & 2131 & 4418 \\ 6885 & 22253 & 571 & 1831 & 23281 & 74176 & 266 & 4529 \\ 5381 & 23901 & 913 & 1903 & 217 & 76679 & 2418 & 4518 \\ 20368 & 20449 & 371 & 2030 & 51072 & 79217 & 950 & 5009 \\ 10745 & 25113 & 1457 & 1883 & 69793 & 70440 & 3618 & 4771 \\ 17777 & 25104 & 1469 & 2024 & 12721 & 90136 & 4076 & 4627 \end{matrix}

Table 1: Primitive solutions to a^{3} + b^{3} = c^{4} + d^{4}, Part 1.

a  b  c  d  a  b  c  d
15248 91169 3207 5060 111875 133933 3606 7764
77169 77912 3964 5117 120089 136087 5496 7602
77393 77904 4633 4670 74384 157697 6811 6834
15333 100205 1781 5623 77375 157633 6252 7308
7849 101257 257 5677 104528 154625 1791 8336
72753 103520 3809 5986 89673 161689 675 8387
12269 116357 3021 6217 127992 143009 1757 8414
70181 109325 3177 6275 85529 172402 2691 8688
1129 118857 4087 6117 10320 183793 6732 8029
45176 129521 1222 6897 145373 183379 570 9804
109009 110017 1861 7151 64989 210277 2379 9883
17717 138189 557 7171 18403 216605 4092 9972
47233 137824 3388 7135 162754 180713 7365 9234
25176 142801 197 7356 94201 215481 8459 8697
41361 145361 407 7487

Table 2: Primitive solutions to a^{3} + b^{3} = c^{4} + d^{4}, Part 2.

Table 3 shows the primitive solutions to 0 ≤ a^{3} +b^{3} = c^{5} +d^{5} = N with 0 ≤ c ≤ d ≤ 5000 in order by the size of N. Note that c = 187 occurs in both the second and third solutions.

\begin{matrix}a & b & c & d & a & b & c & d \\ 2467 & 3071 & 115 & 119 & 315515 & 410576 & 96 & 2515 \\ 7755 & 15102 & 187 & 326 & 324657 & 450095 & 1458 & 2600 \\ 14475 & 29190 & 187 & 488 & 111359 & 526113 & 1286 & 2700 \\ 9301 & 40290 & 80 & 581 & 74484 & 597517 & 293 & 2924 \\ 41144 & 144677 & 653 & 1244 & 480397 & 480926 & 1690 & 2909 \\ 39032 & 150265 & 1065 & 1158 & 50387 & 710804 & 123 & 3244 \\ 177898 & 451093 & 1765 & 2404 & 265936 & 1225217 & 3838 & 4001 \end{matrix}

Table 3: Primitive solutions to a^{3} + b^{3} = c^{5} + d^{5}.

Table 4 shows the primitive solutions to 0 ≤ a^{3} +b^{3} = c^{6} +d^{6} = N with 0 ≤ c ≤ d ≤ 1400 in order by the size of N.

a  b  c d  a  b  c  d
3441 7708 57 88 265008 500137 636 653
28105 28596 40 189 85656 620785 534 775
50145 350428 286 591 296305 1233132 238 1113
225681 458812 233 690
Table 4: Primitive solutions to a^{3} + b^{3} = c^{6} + d^{6}.

3 The first heuristic argument

We estimate the probability that a^{j} + b^{j} = c^{k} + d^{k} has a nontrivial solution using a form of the Birthday Paradox. This argument is old and well known. Weaver [13], page 135, told an amusing anecdote about the Birthday Paradox during World War II.

Our first heuristic assumption is that the four terms a^{j}, b^{j}, c^{k}, d^{k}, are random integers of about the same size B, say, between B/2 and 3B/2. Then a and b will be near B^{1/j} and there will be about that many possible values for each of them, so that there are about B^{2/j} pairs (a, b). Likewise, there are about B^{2/k} pairs (c, d) with c and d near B^{1/k}.

The sums a^{j} + b^{j}, c^{k} + d^{k} will be random integers in the interval [B, 2B] of length B, and we assume they are independent. Then the probability that one particular sum a^{j} +b^{j} differs from every one of the B^{2/k} sums c^{k} + d^{k} is (1 − 1/B)^{B^{2/k}}. By independence, the probability that every sum a^{j} + b^{j} differs from every sum c^{k} + d^{k} is

((1 − 1/B)^{B^{2/k}})^{B^{2/j}} = (1 − 1/B)^{B^{2/j+2/k}},

and the probability that the equation a^{j} +b^{j} = c^{k} +d^{k} has at least one solution with numbers of this size is 1 minus this probability.

Since B is large we may approximate 1 − 1/B by exp(−B^{−1}). The probability of at least one solution becomes 1 − exp(−B^{−1+2/j+2/k}). When 2/j + 2/k < 1 the probability is near 0, so we expect no solution. When 2/j + 2/k > 1 the probability is near 1, so we expect many solutions. When 2/j + 2/k = 1 we have a borderline case where there may be a few solutions or none at all.

This heuristic argument explains why there are many solutions to a^{j} + b^{j} = c^{k} + d^{k} when j = 2: 2/j + 2/k > 1 for every integer k > 1.

When j = 3 the argument predicts many solutions for k = 4 and 5, just a few solutions when k = 6, and no solution when k > 6. It also predicts no solution for 3 < j < k. This is just what we found.

The case j = k = 4 is also a borderline case and it has infinitely many solutions perhaps because there is a parametric solution. Wroblewski [14] found 11089 solutions with a, b, c, d ≤ 10^{14}.

Note that the comparison of number of solutions is a bit unfair because of different search limits M. Tables 1, 2, 3, and 4 give all solutions with roughly the same limit on the sum N.

4 The second heuristic argument

The argument in this section is modeled on that of Erdős and Ulam [6]. We will show that for most sequences having the growth rates of (S_{n}^{(j)})_{n≥1} and (S_{n}^{(k)})_{n≥1} there is a nontrivial intersection if and only if 2/j + 2/k ≤ 1. This theorem proves nothing about particular sequences like (S_{n}^{(j)})_{n≥1} or (S_{n}^{(k)})_{n≥1}, but it suggests that the same statement is likely to hold for them.

We define a probability measure on the space of sequences of positive integers. See Erdős and Rényi [5] or Halberstam and Roth [9] for more about this measure. Let γ > 1 be a real number and n be a positive integer. Let the measure of the set of all sequences containing n be c_{1}n^{−1+1/γ} and the measure of the complement be 1 − c_{1}n^{−1+1/γ}. Here c_{1} and other c_{i} used later are appropriate positive constants. In this case, c_{1} is chosen so that the measure of the set of all sequences is 1. Call this measure the γ measure. The phrase “almost all sequences A” will mean “for all A except for a set of sequences of γ measure 0.” Let P_{γ}(n) = c_{1}n^{−1+1/γ} be the probability that n ∈ A.

If A is a sequence of positive integers and x is a real number, let A(x) be the number of a ∈ A with a ≤ x. It is easy to see that A(x) = (1 + o(1))c_{1}γx^{1/γ} for almost all A. Hence the n-th term of A is (1 + o(1))(n/c_{1}γ)^{γ} for almost all A.

Let 2A denote the sequence of all sums a + a′ with a ∈ A and a′ ∈ A. Note that if γ = j ≥ 2, then, for almost all A, A and 2A have growth rates similar to those of (n^{j})_{n≥1} and (S_{n}^{(j)})_{n≥1}, respectively.

Now let 1 < α ≤ β. Write P_{α}(n) = c_{1}n^{−1+1/α} and P_{β}(n) = c_{2}n^{−1+1/β}.

Theorem 1. Let 1 < α ≤ β. If 2/α + 2/β < 1, then for almost all sequences A in α measure and almost all sequences B in β measure the intersection 2A ∩ 2B is finite. But if 2/α + 2/β ≥ 1, then for almost all sequences A in α measure and almost all sequences B in β measure the intersection 2A ∩ 2B is infinite.

The proof is based on that in Erdős and Ulam [6].

Proof. We will prove the first statement by showing that the expected number E of integers n in the intersection is finite. We have

E = ∑_{n=1}^{∞} ( ∑_{u+v=n} P_{α}(u)P_{α}(v)) ( ∑_{u+v=n} P_{β}(u)P_{β}(v))

= ∑_{n=1}^{∞} c_{1}^{2}c_{2}^{2} ( ∑_{u+v=n}(uv)^{−1+1/α}) ( ∑_{u+v=n}(uv)^{−1+1/β})

< c_{3} ∑_{n=1}^{∞} n^{−1+2/α}n^{−1+2/β}

= c_{3} ∑_{n=1}^{∞} n^{−2+2/α+2/β},

where the inner sums were estimated by integrals. For example, the first inner sum is estimated by ∫_{0}^{n}(u(n − u))^{−1+1/α}du = n^{−2+2/α}nc_{4} = c_{4}n^{−1+2/α}. Since 2/α + 2/β < 1, E < ∞ and the intersection is finite by the Borel-Cantelli lemma. See Feller [7] for the Borel-Cantelli lemma.

Now suppose 2/α + 2/β ≥ 1. We will give the proof for the case 2/α + 2/β = 1. The case 2/α + 2/β > 1 is similar. Let E(x) denote the expected number of integers n ≤ x in the intersection. We have

E(x) = ∑_{n=1}^{x} ( ∑_{u+v=n} P_{α}(u)P_{α}(v)) ( ∑_{u+v=n} P_{β}(u)P_{β}(v))

= ∑_{n=1}^{x} c_{1}^{2}c_{2}^{2} ( ∑_{u+v=n}(uv)^{−1+1/α}) ( ∑_{u+v=n}(uv)^{−1+1/β})

= (1 + o(1))c_{5} ∑_{n=1}^{x} n^{−2+2/α+2/β}

= (1 + o(1))c_{5} ∑_{n=1}^{x} n^{−1} = (1 + o(1))c_{5} log x.

Now we use a second moment argument to show that for almost all A and almost all B the size f(A, B, x) of 2A ∩ 2B ∩ [1, x] satisfies f(A, B, x) = (1 + o(1))c_{5} log x so that lim_{x→∞} f(A, B, x)/E(x) = 1. The expected value of f(A, B, x) is E(x), which we just computed. Let E^{2}(x) be the expected value of f(A, B, x)^{2}. Then E^{2}(x) =

∑_{n_{1}=1}^{x} ∑_{n_{2}=1}^{x} ∑_{u_{1}+v_{1}=n_{1}} ∑_{u_{2}+v_{2}=n_{1}} ∑_{u_{3}+v_{3}=n_{2}} ∑_{u_{4}+v_{4}=n_{2}} P (u_{1}, u_{2}, u_{3}, u_{4}, v_{1}, v_{2}, v_{3}, v_{4}),

where P (u_{1}, u_{2}, u_{3}, u_{4}, v_{1}, v_{2}, v_{3}, v_{4}) is the probability that u_{1}, v_{1}, u_{2}, and v_{2} are in A and that u_{3}, v_{3}, u_{4}, and v_{4} are in B. If these eight integers were distinct, we would have P (u_{1}, u_{2}, u_{3}, u_{4}, v_{1}, v_{2}, v_{3}, v_{4}) =

P_{α}(u_{1})P_{α}(v_{1})P_{α}(u_{2})P_{α}(v_{2})P_{β}(u_{3})P_{β}(v_{3})P_{β}(u_{4})P_{β}(v_{4}),

and the sum would be (E(x))^{2}, but if some integers are repeated, the probability is larger. Hence, E^{2}(x) > (E(x))^{2}. To get the opposite inequality we add terms to E^{2}(x) to account for possible repeated values.

If the eight integers were distinct, we could pair the sums

( ∑_{u_{1}+v_{1}=n_{1}} P_{α}(u_{1})P_{α}(v_{1})) ( ∑_{u_{3}+v_{3}=n} P_{β}(u_{3})P_{β}(v_{3}))

and obtain (1 + o(1))c_{6} log x as in the calculation for E(x). But if say, u_{3} = u_{1}, then v_{3} = v_{1} and the two sums become

∑_{u_{1}+v_{1}=n_{1}} P_{α}(u_{1})P_{α}(v_{1})P_{β}(u_{1})P_{β}(v_{1}) = ∑_{u_{1}+v_{1}=n_{1}}(u_{1}v_{1})^{−1+1/α}(u_{1}v_{1})^{−1+1/β}
= ∑_{u_{1}+v_{1}=n_{1}}(u_{1}v_{1})^{−2+1/α+1/β}
= ∑_{u_{1}+v_{1}=n_{1}}(u_{1}v_{1})^{−3/2}.

When we approximate this sum by the integral ∫_{0}^{n_{1}}(u_{1}(n_{1} − u_{1}))^{−3/2}du we get a constant times n_{1}^{−2}, and the sum on n_{1} is finite. The sum on n_{2} pairs the other two sums and gives (1 + o(1))c_{7} log x.

Likewise, the other added terms are all less than constants times log x. Therefore, (E(x))^{2} < E^{2}(x) < (E(x))^{2} + c_{8} log x. By the Tchebycheff inequality the α measure of the set of A and the β measure of the set of B for which

|f(A, B, x) − E(x)| > ϵ log x

are less than c_{9}/ϵ^{2} log x. Let x_{k} = 2^{k(log k)^{2}}. By the Borel-Cantelli lemma we have

lim_{k→∞} f(A, B, x_{k})/E(x_{k}) = 1.

Therefore, since f(A, B, x_{k}) ≤ f(A, B, x) ≤ f(A, B, x_{k+1}) when x_{k} < x < x_{k+1}, for almost all A and almost all B we have lim_{x→∞} f(A, B, x)/E(x) = 1. In the same way, lim_{x→∞} f(A, B, x)/E(x) = 1 when 2/α + 2/β > 1. Since E(x) is unbounded as x → ∞, the intersection 2A ∩ 2B is infinite for almost all A and almost all B when 2/α + 2/β ≥ 1.

Now let α = j and β = k. If (n^{j})_{n≥1} and (n^{k})_{n≥1} were typical sequences, this theorem would predict infinitely many solutions to a^{j} + b^{j} = c^{k} + d^{k} when 2/j + 2/k ≥ 1 and finitely many solutions when 2/j + 2/k < 1. Of course, Lander [11] found infinitely many solutions to this equation for all j and k. Thus (n^{j})_{n≥1} and (n^{k})_{n≥1} are special sequences in the exceptional set of measure 0 when 2/j + 2/k < 1. But since the theorem predicts that there should be no solutions, perhaps Lander found all solutions in this case.

5 Acknowledgments

I thank Bjorn Poonen for suggesting an improvement to the first heuristic argument. I thank the anonymous referee for suggestions and references that improved the manuscript and speeded the program.

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[14] J. Wroblewski, 2006. Available at http://www.math.uni.wroc.pl/~jwr/422/422.zip.

2020 Mathematics Subject Classification: Primary 11D41. Secondary 11Y50. Keywords: equal sums like powers.

(Concerned with sequences \underline{A003336}, \underline{A003347}, \underline{A003358}, and \underline{A004999}.)

Received January 8 2022; revised version received February 24 2022; March 6 2022. Published in Journal of Integer Sequences, March 6 2022.

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