Problem of the Month (February 2004)

It is interesting to note that

11 + 21 + 61 = 41 + 51
12 + 22 + 62 = 42 + 52.

This is an example of a constellation, an equation involving sums of powers that is true for more than one power. In fact, this is the smallest (1,2) constellation with distinct bases, though there is a smaller one with repeated bases. Can you find it? Can you show that there are infinitely many (1,2) constellations? Can you find the smallest (1,3) constellation? Can you find the smallest (1,-1) constellation? What is the smallest constellation you can find for other pairs of integers? Are there constellations that correspond to any finite subset of integers? Can you find a (1,2,3) constellation smaller than this one found by Boris Kordemsky?

11 + 41 + 121 + 131 + 201 = 21 + 31 + 101 + 161 + 191
12 + 42 + 122 + 132 + 202 = 22 + 32 + 102 + 162 + 192
13 + 43 + 123 + 133 + 203 = 23 + 33 + 103 + 163 + 193

ANSWERS

Chen Shuwen considered this problem in the 1990's. L.J. Lander, T.R. Parkin and J.L. Selfridge considered the problem back in the 1960's. G. Tarry considered some of these problems in the 1910's! Bill Clagett sent me lots of small constellations. Joseph DeVincentis and Philippe Fondanaiche found various small (1,2) constellations, and proved that there are infinitely many. Philippe Fondanaiche found various families of constellations (for example, [1,2,x,3+x]=[1+x,2+x,3] is a (1,2) constellation for positive integers x). Joseph DeVincentis proved that a constellation exists for any subset of positive integers. Alex Rower sent me a large constellation.

We call a constellation ideal if there are n different powers and no more than n+1 powers on each side of the equation. Here are the ideal constellations with distinct terms and the smallest known sums:

(1,2): 1n + 2n + 6n = 4n + 5n (EF)
(1,3): 1n + 5n + 9n = 7n + 8n (EF, BC, PF)
(1,4): 3n + 25n + 38n = 7n + 20n + 39n (CS)
(1,5): 3n + 54n + 62n = 24n + 28n + 67n (LL)
(2,3): 37n + 62n = 21n + 26n + 64n (EF)
(2,4): 5n + 6n + 11n = 1n + 9n + 10n (BC)
(2,6): 3n + 19n + 22n = 10n + 15n + 23n (LL)
(1,2,3): 4n + 7n + 11n = 1n + 2n + 9n + 10n (BC, PF)
(1,2,4): 2n + 7n + 11n + 15n = 3n + 5n + 13n + 14n (CS)
(1,2,5): 1n + 28n + 39n + 58n = 8n + 14n + 51n + 53n (CS)
(1,2,6): 7n + 16n + 25n + 30n = 8n + 14n + 27n + 29n (CS)
(1,3,4): 3n + 140n + 149n + 252n = 50n + 54n + 201n + 239n (CS)
(1,3,5): 6n + 16n + 18n + 24n = 7n + 13n + 21n + 23n (CS)
(1,3,7): 184n + 443n + 556n + 698n = 230n + 353n + 625n + 673n (CS)
(1,2,3,4): 1n + 2n + 10n + 14n + 18n = 4n + 8n + 16n + 17n (PF)
(1,2,3,5): 1n + 8n + 13n + 24n + 27n = 3n + 4n + 17n + 21n + 28n (CS)
(1,2,3,6): 7n + 18n + 55n + 69n + 81n = 9n + 15n + 61n + 63n + 82n (CS)
(1,2,4,6): 1n + 19n + 22n + 37n + 42n = 2n + 14n + 29n + 33n + 43n (CS)
(1,2,3,4,5): 1n + 2n + 10n + 12n + 20n + 21n = 5n + 6n + 16n + 17n + 22n (PF)
(1,2,3,4,5,6): 14n + 16n + 45n + 54n + 73n + 83n = 3n + 5n + 28n + 34n + 65n + 66n + 84n (PF)
(1,2,3,4,5,6,7): 1n + 5n + 10n + 24n + 28n + 42n + 47n + 51n = 2n + 3n + 12n + 21n + 31n + 40n + 49n + 50n (AR)
(1,2,3,4,5,6,7,8):1n + 25n + 31n + 84n + 87n + 134n + 158n + 182n + 198n = 2n + 18n + 42n + 66n + 113n + 116n + 169n + 175n + 199n (AR)
(-1,1): 4n + 10n + 12n = 5n + 6n + 15n (CS)
(-1,2): 35n + 65n + 84n = 39n + 52n + 91n (CS)
(-2,1): 60n + 105n + 140n = 65n + 84n + 156n (CS)

Here are the ideal constellations with the smallest known sums, if different than above:

(1,2): 3n + 3n = 1n + 1n + 4n (EF, PF)
(2,4): 3n + 5n + 8n = 7n + 7n (EF)
(1,2,3): 1n + 1n + 6n + 6n = 3n + 4n + 7n (PF)
(1,2,4,6): 3n + 7n + 10n + 16n + 16n = 4n + 5n + 12n + 14n + 17n (CS)
(1,2,3,4,5): 3n + 5n + 11n + 13n + 16n= 1n + 1n + 8n + 8n + 15n + 15n (GT)

Boris Bukh tells me that the problem of estimating the number of (1,2,...,k) constellations, where the number of summands on the left and on the right is the same, is at the heart of the Vinogradov's mean-value method for bounding exponential sums. He suggests Titchmarsh's "Theory of Riemann zeta-function", chapter VI for discussion of the method, and known results.

If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 3/8/04.