On a Question of Mordell Downloaded by Guest on October 2, 2021 Downloaded by Guest on October 2, 2021 .Slto Density

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MATHEMATICS This leads to an algorithm that searches for solutions with |z| ≤ B by enumerating positive integers d ≤ αB, and, for each such d, determining the residue classes of all cube roots of k modulo d and searching the corresponding arithmetic progressions for values of z ∈ [−B, B] that make ∆(d, z) a square. With suitable optimizations, including sieving arithmetic progressions to quickly rule out integers that are not squares modulo primes in a suitably chosen set, this leads to an algorithm that requires only O (B(log log B)(log log log B))operations on integers in [0, B] for any fixed value of k. An attractive feature of this algorithm is that it finds all solutions with min{|x|, |y|, |z|} ≤ B, even those for which max{|x|, |y|, |z|} may be much larger than B (note that this is the case in our solution for k = 3). This algorithm was used in ref. 4 to find solutions for k = 33 and k = 795, leaving only the following 11 k ≤ 1,000 unresolved:

42, 114, 165, 390, 579, 627, 633, 732, 906, 921, 975. [1.3]

The search in ref. 4 also ruled out any solutions for these k with min{|x|, |y|, |z|} ≤ 1016. Here we make several improvements to this method in ref. 4 that allow us to find a new solution for k = 3 as well as four of the outstanding k listed above. • Instead of a single parameter B bounding |z| ≤ B and 0 < d ≤ αB, we use independent bounds dmax on d and zmax on |z|, whose ratio we optimize via an analysis of the expected distribution of |z|/d; this typically leads to a zmax/dmax ratio that is 10 to 20 times larger than the ratio 1/α ≈ 3.847332 used in ref. 4. • Rather than explicitly representing a potentially large set of sieved arithmetic progressions containing candidate values of z for a given d, we implicitly represent them as intersections of arithmetic progressions modulo the prime power factors of d and auxiliary primes. This both improves the running time and reduces the memory footprint of the algorithm, allowing for much larger values of |z|. • We dynamically optimize the choice of auxiliary primes used for sieving based on the values of k and d; when d is much smaller than zmax, this can reduce the number of candidate values of z by several orders of magnitude. • We exploit 3-adic and cubic reciprocity constraints for all k ≡ ±3 (mod 9); for the values of k listed in Eq. 1.3, this reduces the average number of z we need to check for a given value of d by a factor of between 2 and 4 compared to the congruence constraints used in ref. 4, which did not use cubic reciprocity for k 6= 3. Along the way, we compute, to high precision, the expected density of solutions to Eq. 1.1 conjectured by Heath-Brown (1), and compare it with the numerical data compiled by Huisman (6) for k ∈ [3, 1,000] and max{|x|, |y|, |z|} ≤ 1015. The data strongly support Heath-Brown’s conjecture that Eq. 1.1 has infinitely many solutions for all k 6≡ ±4 (mod 9). 2. Density Computations In this section, we study Heath-Brown’s conjecture in detail. In particular, we explain how to compute the conjectured density of solutions to high precision and compare the results with available numerical data. We further study the densities of divisors d | z 3 − k and arithmetic progressions z (mod d) that occur in our algorithm, which informs the choice of parameters used in our computations. √ √ k k ≥ 3 k 6≡ ±4 (mod 9) K = ( 3 k) F = ( −3) o o Let be a cube-free integer with and .√ Define Q and Q , and let K and F be their 1+ −3 × respective rings of integers. We have oF = Z[ζ6], where ζ6 = 2 is a generator of the unit group oF . Also, Disc(F ) = −3 and Disc(K ) = −3f 2, where, by ref. 7, lemma 2.1, ( Y 1 if k ≡ ±1 (mod 9), f = ( p) · 3 otherwise. p|k

We deﬁne two modular forms related to F and K . First, let f1 be the modular form of weight 1 and level | Disc(K )| such that ζK (s) = ζ(s)L(s, f1). It follows, from the ramiﬁcation description in ref. 7, section 2.1, that rational primes p decompose into prime ideals of oK as follows (subscripts denote inertia degrees):  p1p2 if p ≡ 2 (mod 3) and p - k,  00 p p0 p p ≡ 1 (mod 3) p k k p,  1 1 1 if and - and is a cube modulo poK = p3 if p ≡ 1 (mod 3) and p - k and k is not a cube modulo p,  2 0 p1p1 if p = 3 and k ≡ ±1 (mod 9),  3 p1 otherwise.

From this data, we ﬁnd that the local Euler factor of L(s, f1) at p is

 (p−1)/3 2 if p - k, p ≡ 1 (mod 3) and k ≡ 1 (mod p),  (p−1)/3 1 −1 if p - k, p ≡ 1 (mod 3) and k 6≡ 1 (mod p), Lp (s, f1) = , where cp (k) := −s Disc(K ) −2s 1 if p = 3 and k ≡ ±1 (mod 9), 1 − cp (k)p + p p  0 otherwise.

√ × × Now let σ : F → C be the unique embedding for which we have =σ( −3) > 0. Let χf :(oF /3oF ) → C be the character defined by −1 χf (ζ6 + 3oF ) = σ(ζ6 ), and define χ∞(z) := σ(z)/|σ(z)|. Let χ be the Grossencharakter¨ of F defined by ( √ χ∞(α)χf (α + 3oF ) if α ∈ oF \ −3oF , χ(αoF ) := √ 0 if α ∈ −3oF .

2 of 11 | PNAS Booker and Sutherland https://doi.org/10.1073/pnas.2022377118 On a question of Mordell Downloaded by guest on October 2, 2021 Downloaded by guest on October 2, 2021 .Slto Density. Solution A. so Let L(s naqeto fMordell of question a On Sutherland and Booker to asymptotic have we (1), Heath-Brown by calculated as Then, Finally, prime a For is normalization) newform holomorphic a is there induction, automorphic By ssonaoe h eldniyde o eedo h rcs hieo h egtfunction height the of choice precise the satisfying on solutions depend the not to does applies density density real asymptotic the above, shown As where that so let Now ie prime a Given et-rw ojcue httenumber the that conjectures Heath-Brown of definition the extend we if summary, In • h eldniyo ouin to For solutions of density real The • χ α − h h f (x (λx = = (α) > ε 1 2 σ vol( , , χ( ∞ h a χ). y , p : , define 0, λy √ +3b 4 = S z R χ p )=0 = −3) ) 2 (ε))/ , 3 √ f > λz ≡ R → )= (−1) −3 1 ∞ 0 md3) (mod 2 = ) R p (t when vol( and ≥0 Define ≡ 3 oteElrfco at factor Euler the so , |λ 1) + md3) (mod 1 eahih ucin ywihw enafnto hti otnos ymti nisipt,adsatisfies and inputs, its in symmetric continuous, is that function a mean we which by function, height a be T Thus, −1. |h p x (ε)) = (x 3 ε→0 −2/3 lim T + S σ αo , ehave we , y p := (ε) := (ε) → y + , F = 3 z (2ε) dt min{|x othat so , 12 ) + let ,      o any for = σ x + 1 + 1 z 3 1 as −1 = χ(p) p 3 e 3 # 3 2 =lim := (B (x (x a + χ(p 0 = ε Γ( {(x vol(S p Γ( |, ( 3c α (1 → y , , p := ) p eoeteuiu nee o which for integer unique the denote e 3 1 |y y y −1) p 3 3 2 = 1 + p →∞ o λ o ) and ( − , , , ) 0 + 2 F k |, − p F y z z ∈ + ) 2 . p a )=1 lim 12 = (ε)) = ) χ(p ρ ) ) , |z = z σ p emyte compute then may We . − a R z sol √ 3 = ∈ ∈ # −1 3 xyz ( |}, p md3 : 3) (mod ) α) p × p n 0 = a p χ a R R )p 1 log {(x p othat so 12 = 4 = 4 = L . p 2 p (B ∞ ehave We . 3 3 − s1. is 0 6= p 3 + 2 + 2 1 and : : (s (p B , ihhih nteinterval the in height with −s |xyz ) χ(p x h Z Z y h , (x , Z ε→0 fsltost Eq. to solutions of )χ 1 0 ≥ , )(1 f (x 1 ∞ ∞ 0 z 2 2 a o as , | ∞ χ( f y = ) (mod ) F p , y 3 1 (p Z Z + − b ≥ y , , )N B 0 = Z p x 1 H (2ε) 1 √ , z = ) H ∞ = ) z 1 3 χ( z + 1 z ) ∞, → ∞ /h 2 −3 (p + ) and ≥ H ∈ /h 1 p) for ≤ ( −1 f χ zh − 1 y o 2 ( √ a 0, 3 p h f 3 = F 3 ( √ B p 3 σ ( )= (−1) fwih n level and 2 weight of e , vol( p t √ √ 2 1 3 ) p + √ ( h 3 3 : ) t α −s |x 2 1 p d o n iia hieof choice similar any for 6≡ −s +1,−t 3 t (x y )

z −s 3 +1,−t x 3 min = 3 oteElrfco of factor Euler the so , +1,−t ) T hnteElrfco of factor Euler the then 3) , (mod 1 3 p , +z 3 + + 1 a 2e ≡ = −y where (ε)) = + p 3 3 ,−1) +2 otecrepnigElrfco is factor Euler corresponding the so −1, y pt emtto,satisfying permutation, to up 1.1, k ,−y ,−1) + 1 y 3 + 1 p 9 , ,−1)∈H 3 {|x −z md9)} (mod + + ,−z + 2 p a H a ρ z p b dz 1 p 1−2s ) p z + z sol 3 √ 1−2s ) =[H := H ∈ p ≡ ≤ | 3 ∈H −s y −3 (t := (t ≡ md3) (mod 1 1 |, . 3 ε 3 3(y , k . + |x

! 6 1 1) + 1 1) + |x dt σ , dt , (mod p ∈ + H 3 ∞ 3 if if if 1−2s | 3o + Disc(F − 2 z 2/3 dy p p p 2/3 Y ] |, p F z y - | 3 = a hnb optda follows. as computed be then can h |y 3 3k k , . 3 p = nldn,frexample, for including, , σ dz and ) z e − and 2/3 + p . , )} σ h . )|N z ∞ etu ojcueta h same the that conjecture thus We . z . 3 4p L(s dz |}. p ≤ | log https://doi.org/10.1073/pnas.2022377118 (3o 3, 6= = , ε f H H a

F 2 p ) 2 , 1 2 27 = ) L(s 27b + , at max{|x , f p 2 ) uhthat such 2 at i t arithmetic its (in o some for |, p |y is PNAS |, | z L(s }≤ |} b | ∈ , f11 of 3 f [2.1] Z B 2 = ) >0 is .

MATHEMATICS Let us now deﬁne

σp 1 · 1 rp := , where ζp (s) = and Lp (s,( )) = . L (1, f )3L (2, f )L (2, f )−6ζ (2)−6L (2, ( · ))−3 1 − p−s 3 p −s p 1 p 2 p 1 p p 3 1 − 3 p

A straightforward calculation shows that 3a c (k) + O(1) r = 1 − p p . p p3

−s Since −ap cp (k) is the coefﬁcient of p in the Rankin–Selberg L-function L(s, f1 f2), we expect square-root cancellation in the Q product p rp . Under the generalized Riemann hypothesis (GRH), for large X , we have

Y −2 3 −6 −6 · −3 Y σp = 1 + O(X log X ) L(1, f1) L(2, f2)L(2, f1) ζ(2) L(2, ( 3 )) rp . [2.2] p p≤X

9 Applying Eq. 2.2 with X = 10 allows us to compute the solution densities ρsol to roughly 18 digits of precision for all cube-free k ≤ 1,000. To evaluate the L-functions, we used the extensive functionality available for that purpose in PARI/GP (8). Since our goal is merely to gather some statistics, we content ourselves with a heuristic estimate of the error in this computation, although it could be rigorously certiﬁed with more work. Some examples are shown in Table 1. We compared Huisman’s dataset to an average form of Heath-Brown’s conjecture as follows. For an integer K ≥ 3, deﬁne

4 3 3 3 X NK (B) := #{(k, x, y, z) ∈ Z : x + y + z = k cube-free, 3 ≤ k ≤ K , |z| ≤ |y| ≤ |x| ≤ B} and ρK := ρsol(k). k∈Z∩[3,K ] k cube-free

Then Heath-Brown’s conjecture implies that, for ﬁxed K , we have NK (B) ∼ ρK log B as B → ∞. The plot in Fig. 1 compares 7.5 15 N1000(B) for B ∈ [10 , 10 ], computed from Huisman’s (6) data, with ρ1000 log B + C , where ρ1000 ≈ 363.869 and C ≈ −679.4 was chosen to minimize the mean square difference. Out of 6,256 points, the two plots never differ by more than 42, which gives strong evidence for Heath-Brown’s conjecture, at least on average over k.

B. Divisor and Arithmetic Progression Densities. In this section, we assume that k ≡ ±3 (mod 9) and derive estimates for the density of arithmetic progressions arising from cube roots of k modulo d. Deﬁne

( 3 ∞ 1 if ∃z ∈ Z s.t. z ≡ k (mod d) and ordp (d) ∈ {0, ordp (k/3)} ∀p | k, X δd δd := and F (s) := . 0 otherwise, d s d=1

As shown in ref. 4, any d arising from a solution to Eq. 1.1 must satisfy δd = 1, and we only consider such d in our algorithm.

15 Table 1. Selected ρsol and dexp(1/ρsol)e = min{B ∈ Z : e(B) ≥ 1} for k ≤ 1,000, including 10 smallest ρsol and all k with n(10 ) = 0 B = 105 B = 1010 B = 1015

k ρsol dexp(1/ρsol)e e(B) n(B) e(B) n(B) e(B) n(B) 858 0.028504 1,723,846,985,902,459 0.328 1 0.656 2 0.984 2 276 0.031854 43,031,002,119,138 0.367 1 0.733 1 1.100 2 390 0.032935 15,358,736,844,736 0.379 0 0.758 0 1.138 0 516 0.033062 13,665,771,588,173 0.381 0 0.761 1 1.142 1 663 0.033196 12,097,471,969,974 0.382 0 0.764 1 1.147 1 975 0.038722 164,297,126,902 0.446 0 0.892 0 1.337 0 165 0.039636 90,602,378,809 0.456 0 0.913 0 1.369 0 555 0.042706 14,770,444,441 0.492 1 0.983 2 1.475 2 921 0.044142 6,895,540,744 0.508 0 1.016 0 1.525 0 348 0.044632 5,378,175,303 0.514 2 1.028 2 1.542 3 906 0.049745 537,442,063 0.573 0 1.145 0 1.718 0 579 0.050838 348,939,959 0.585 0 1.171 0 1.756 0 114 0.058459 26,853,609 0.673 0 1.346 0 2.019 0 3 0.061052 12,985,612 0.703 2 1.406 2 2.109 2 732 0.063137 7,561,540 0.727 0 1.454 0 2.181 0 633 0.079660 283,059 0.917 0 1.834 0 2.751 0 33 0.088833 77,422 1.023 0 2.045 0 3.068 0 795 0.089491 71,273 1.030 0 2.061 0 3.091 0 42 0.113449 6,732 1.306 0 2.612 0 3.918 0 627 0.129565 2,249 1.492 0 2.983 0 4.475 0

4 of 11 | PNAS Booker and Sutherland https://doi.org/10.1073/pnas.2022377118 On a question of Mordell Downloaded by guest on October 2, 2021 Downloaded by guest on October 2, 2021 ρ For naqeto fMordell of question a On details. Sutherland for and Booker 5 section of see sample computations; suitable distributed a listed large on the in algorithm of the allocation any running resource for by manage 3.5 estimated efficiently above be to never can and proportionality 3 of around constant is The average the range, 2.2: feasible Remark the beyond well is which 2.1: Remark estimates lists 2 Table For Then define now us Let have we turn, In at 2 order ntr,w have we turn, In div For d p p max G - - p k k / (s - h oa factor local the , ehave we , √ 3 k i.1. Fig. = ) s log and 1 = h vrg ubro rtmtcporsin modulo progressions arithmetic of number average The o n xdcoc fteratio the of choice fixed any For Q d max e p n oohrplsi h region the in poles other no and cte ltof plot Scatter > G G o h number the for p ehave we 0, p (s ( s L ) where ), d p L ( ≤d X p s (2s ρ F ,f ζ max ap p 1 p ( ) ,f ( s d ζ d s 1 N ) δ p G ≤d max X ) 3 ) d ( 1000 δ 2 ζ L s s lim (s L p p ∼ ) →1 max p e p (2s ( safnto of function a as (B) := ) ( o h number the for s = ρ s π ,( δ F ) ,f div d div ρ 3 1 · c + 1 = r (s ap ) p )) √ d (d d X (k 3 ∞ ) (k =1 = is log 3 )+2− max d (s ) 3 max + 1 δ O ∼ + 1 ) d d F − R r fadmissible of max G ( (p d p ρ d O 1) 3 p = (s ap s p O (k −3s ) (s (p 2 d othat so , := ) {s z (X ) max max = := ) B −3s as , Therefore, ). ∈ ∈ π −2 /d d + 1 ap        C [10 Therefore, ). where max )      as (d max 1 + 1 : 7.5 <(s F 1 + 1 + 1 O L(2, max 1 d ∞, → h oa unn ieo u loih sruhypootoa to proportional roughly is algorithm our of time running total the , 10 , d max (s − (X L(1, p ) ) ≤ − r p = ) 1+c 15 c d ≥ f ∞, → (1−s −2 faihei rgesosmodulo progressions arithmetic of p p 1 ord ] d (k ( s )ζ ae nHimnsdtst() oprdt h line the to compared (6), dataset Huisman’s on based f max k yrf ,term31 tflosthat follows it 3.1, theorem 9, ref. By 1}. −1 p Q 1 ) )+2− # = ) p ( 3p ) (2) k ord ) ( p ln ihterto fteequantities. these of ratios the with along , k ) L F s d ) L(1, F where s (s p ≤ ( p Y p 1, ≤X {z 3 p (s ( ) k d ) 3 f ) where max where ), 1 −1 3 a eoopi otnainto continuation meromorphic has · (mod G ) −1 ρ p itdi al ssrknl ml.Ee for Even small. strikingly is 2 Table in listed div p (1)L L Y ≤X if if if p := ρ if if if (1, d p p p ap p p p : ) p F 3 = - | =lim := (2, ζ f lim k 3 = | - p k 3 1 p z k , (1) k 3 , )ζ (1) 3 . , , s f s . 1 ≡ p →1 →1 3 )ζ (1) 2 L k L p G p F p (2) (mod (1, Γ (s (s 1, )(s . ) f 2 3 3 1 3 (s · ) d d − )}. − ≤ . https://doi.org/10.1073/pnas.2022377118 d 1). 1) d ≤ max 2 k d max . hs siae lo us allow estimates These . 3 1 <(s . swl sestimates as well as , ρ 1000 ) > log 1 3 B ihapl of pole a with , + PNAS d C max . ρ | 10 = ap f11 of 5 d max 24 , .

MATHEMATICS 12 Table 2. Comparison of estimated and actual counts of arithmetic progressions modulo d ≤ dmax = 10 for various k of interest √ ρ 3 log d π (d ) ρ π √ρdiv dmax π ap max ap max k apdmax ap(dmax) 3 div(dmax) ρ π (d ) log dmax div div max 3 476,709,085,641 476,709,082,386 221,480,415,360 222,316,170,600 2.152 2.144 42 439,262,042,312 439,262,055,314 194,525,166,395 195,043,114,314 2.258 2.252 114 346,031,225,026 346,031,232,985 169,944,552,313 169,697,769,695 2.036 2.039 165 398,768,628,911 398,768,635,237 201,820,401,130 201,648,107,384 1.976 1.978 390 361,424,697,190 361,424,750,258 170,411,108,873 170,119,932,464 2.121 2.125 579 467,532,879,762 467,532,936,236 220,746,986,113 221,627,128,720 2.118 2.110 627 544,308,148,137 544,308,117,802 238,234,806,279 240,026,258,762 2.285 2.268 633 510,771,397,972 510,771,391,669 227,368,579,096 228,697,959,163 2.246 2.233 732 396,862,883,895 396,862,943,789 145,013,347,786 145,167,910,326 2.737 2.734 906 353,110,285,004 353,110,236,539 166,128,603,588 165,813,813,631 2.126 2.130 921 420,143,131,383 420,143,101,621 212,693,499,876 212,924,474,063 1.975 1.973 975 461,977,372,770 461,977,396,756 194,140,103,965 194,481,735,572 2.380 2.375

3. Cubic Reciprocity 3 3 3 In ref. 10, Cassels used cubic reciprocity to prove that, whenever x, y, z ∈ Z satisfy x + y + z = 3, we must have x ≡ y ≡ z (mod 9). For fixed d = |x + y|, it follows that z is determined modulo 81. Colliot-Thélène and Wittenberg (11) later recast this phenomenon in terms of Brauer–Manin obstructions, and showed that, for any k, the solutions to Eq. 1.1 are always forbidden for some residue classes globally but not locally.∗ In this section, we extend Cassels’ analysis to all cube-free k ≡ ±3 (mod 9), and derive constraints on the residue class of z (mod q) for a certain modulus q | 27k. We assume throughout that k ≡ 3 (mod 9) for a fixed ∈ {±1}. √ √ α −1+ −3 Given α, β ∈ oF with β∈ / −3oF , let β 3 be the cubic residue symbol, as defined in ref. 12, chapters 9 and 14. Put ζ3 = 2 ∈ oF . For integers x, y satisfying x ≡ y ≡ (mod 3), define

−1 (y−x)/3 ζ3x + ζ3 y χk (x, y) := ζ3 . k/3 3

Note that χk (x, y) depends only on the residue classes of x, y (mod 3k). 2 Definition 3.1: We say that a pair (d, z) ∈ Z is admissible if there exist x, y ∈ Z satisfying the following conditions: d 1) x + y ≡ − 3 d (mod 27k); 2) x 3 + y 3 + z 3 ≡ k (mod 81k); 3) {χk (x, y), χk (x, z), χk (y, z)} ⊆ {0, 1}. Note that this definition depends only on the residue classes of d, z (mod 27k). 3 Lemma 3.2. Let (x, y, z) ∈ Z be a solution to Eq. 1.1, and let d := |x + y|. Then (d, z) is admissible. Proof: Recall that k ≡ 3 (mod 9). Since every cube is congruent to 0 or ±1 (mod 9), we have x ≡ y ≡ z ≡ (mod 3), so that d d x + y ≡ − ≡ − 3 d (mod 3). As d = |x + y|, it follows that x + y = − 3 d, so condition 1 of the definition is satisfied. Condition 2 then follows directly from Eq. 1.1. Now let −1 γ := (ζ3x + ζ3 y) = −y + (x − y)ζ3. By ref. 12, chapter 9, example 19, we have

−1 (y−x)/3 ζ3x + ζ3 y 3 γ −3 γ χk (x, y) = ζ3 = = , k/3 3 γ 3 k/3 3 γ 3 −k/3 3

3 where the last equality follows from the fact that (α/β)3 depends only on the ideal βoF and (±/β)3 = ((±) /β)3 = 1. By cubic reciprocity (ref. 12, chapter 14, theorem 1), this equals

−3 −k/3 k = . γ 3 γ 3 γ 3

3 3 3 Noting that x + y = (x + y)γγ, we have k ≡ z (mod γoF ), whence

z 3 χk (x, y) = ∈ {0, 1}, γ 3

and, by symmetry, we also have χk (x, z), χk (y, z) ∈ {0, 1}; thus condition 3 holds as well.

*Thus strong approximation fails for Eq. 1.1, but this is never enough to forbid the existence of integer solutions outright, so there is no Brauer–Manin obstruction.

6 of 11 | PNAS Booker and Sutherland https://doi.org/10.1073/pnas.2022377118 On a question of Mordell Downloaded by guest on October 2, 2021 Downloaded by guest on October 2, 2021 n let and let Finally, aifigEq. satisfying d that follows It suffices. est .5 .9 .7 .6 0.585 0.962 0.970 0.590 of roots cube of set the denote of factor a by density the reduces solutions 81 mod 0.250 to solutions 9 mod from A. passing particular, in 4; ref. in used not were Density k residues admissible of density average the is which have we Moreover, naqeto fMordell of question a On Sutherland and Booker Set 2) divisor positive each For 1) such if mod residue cubic a (mod If let and )Set 4) Let 3) primes desired. y that show to suffices it theorem, remainder − p n primes and p loih 3.5: Algorithm eusvl nmrt l oiieintegers positive all enumerate Recursively some for modest is improvement the Although If 3.3. Lemma Let through 3.4: factors Example admissibility of definition the Thus, : Proof Set Algorithm. admissible. order. ordering. the in prime next the is ≡ | bp p (x d md3) (mod 1 0 p a A - a m b o some for d 2 otefollowing: the do , p q (x ) (z = , n,if and, 1, := − easto ml uiir primes auxiliary small of set a be etepoutof product the be < sacbcrsdemod residue cubic a is ups that Suppose etedvsrof divisor the be := z 2 , c y 256 − z 1 d 2 0 0 ) 0 > p − y 1.1 Let ∈ Let qa and 2 h al eo hw h ratio the shows below table The - o dividing not c z hnw a choose may we then ), Z b dk Given and , 0 n let and , )/p with k ∈ > satisfy (x define , p χ ≡ 27kp p 1 c othat so , - k 0 ) n ems have must we and , 1 3 (x (x |x # and 3 qd (d k md9) (mod P (y + C | 0 + , z −2 , 0 > (p , Z d 27k 0 y d c c < z y max 34 1 633 114 42 33 3 ≡ Then Z. |y 0 2 (m 2 1 = ) ) k = ) hnw have we then ), 0 d z md27k (mod primes > ) z Let . | samsil.Let admissible. is z max 1 endin defined 3 > k , ) 0 (mod 1 z (z + of χ S p etesbe of subset the be = modulo eapstv nee,ad o ahpstv integer positive each for and, integer, positive a be max hnodrthe order then , |z otherwise. eoeitgr htw ilcos ootmz efrac (typically, performance optimize to choose will we that integers denote k But . d (x z k (p s |, /3 0 + := p √ ∈ , ) := ) ) q 3 y #{z ∈ ap k Z p ). with ), ≡ b >0 A < > em 3.3. Lemma P m Let . Then p χ x ( C d 3 p othat so |z o dividing not k 2 Let . 3 - (d d (m {k {z with x (x (x - ota n ouint Eq. to solution any that so , gcd(d C + ≤ | implies (mod md27k (mod k ≡ , (p 0 2 + := ) + z d (d y x , (x hs rdc exceeds product whose y 1 − 0 0 z e z P 3 , p Z/m k ) p d p 1 max , y 0 = 0 + (mod 1 ) = ) Z samsil nti case. this in admissible is y ) k z b ≡ eapiedvsrof divisor prime a be etesto primes of set the be z - 2 + {z , 3 q C × · · · × eitgr aifigtecniin in conditions the satisfying integers be ) 2 (x hs ae tl eetfo moiglclcntansmod constraints local imposing from benefit still cases those , p d q and , z 3d : k : ) (mod c 3 (y + ) 1 samsil iff admissible is e ) 3 Z =27k := /d + χ p in 2 1 Z/27k or Z} # 3 + 2 = (2) ∃x da k − · · · hti dnie with identified is that (4s nmrt l pairs all enumerate 9), (mod m 1 p A (x {z p 1 = ) 0 |x ), y hsnuigete h reigcmue ntepeiu tpo fixed a or step previous the in computed ordering the either using chosen , ) Z p by , p | 3 q 2 (z [az p (mod (mod z (x + n : e ) ) (z + Z (p p meaning −1, n and ), z 3 > o # log − ≡ =2 mn l oal emte eius o few a for residues, permitted locally all among set , y 3 × + ≤ n 2 − e ord | 2 n ≡ + or = Z/q y d 0 ) k and ) az p Y q q max k ). p c A × 3 ) b d d S ( p |k ( : ) χ ) k (x − ≡ 2 (2)=−1 d (d := ≤ )=2 Z. k s.t. where , (mod p (p (mod (y 2 d k d p d d d , - − 3 d (q )/ , z max max x 0 md81k (mod x - 1.1 ) 27k 0 , z 0 k 3 hc en that means which , y d ) ≡ z p ) log ) p z (d + 1 2 × 0 o which for sadmissible. is sfollows: as m ≡ with sadmissible} is max = ) p −1 )] /q n let and p Y hl,if while, ), p p ⊆ )} md3) (mod 1 ≡ 1 |a , (mod npatclcmuain,w a take may we computations, practical in ; n consider and , n,while and, , χ > − 3p d S k · · · (y d x ). = Z/m [az m (p ) , 3 A |x p # > let , z (d ) 2 + )} d yiseto.Thus inspection. by ), C + p Z + (q , z (p n p z and y 3 b ) ) | r rmsin primes are ) if if | c (x ≡ n let and 3.1, Definition ∈ (x etestof set the be > 0 has 2x p p qd k 2 Z 2 z for 0; + 2 = > ( − 3 2 0 0 md3q (mod p sgn ≡ + pa y −1)/3 o hc hr exist there which for 2, y then ), . 2 z z https://doi.org/10.1073/pnas.2022377118 < ](mod )] 3 z p md27k (mod ≡ = z ∈ 6≡ max (mod 0 c s P )} (mod 1 0 P n,frpstv integers positive for and, , z replace , p ete have then we , ≈ , + | and k p (d z 4, q 2 finterest. of Z /p n n hieof choice any and ). , c p z 1 e o which for p yteChinese the By ). 0 ) i ≈ ) q x ∈ o2cno be cannot 2 so ), a and oeo which of some , samsil,as admissible, is 0 and 50, = Z by PNAS (x >0 x 2 pa 4/9. A , ≡ #C + o each For . y where , ob the be to , (−z bp (d c | z (p 2 ) , , f11 of 7 ≈ 3 = ) ∈ z y /x ) 0 6), Z = ) is p b 3 3

MATHEMATICS via the Chinese remainder theorem. Let

Z(m, s, zmax) := {z ∈ Z : z + mZ ∈ Z(m), sgn z = s, and |z| ≤ zmax}.

For each z ∈ Z(m, s, zmax), if z + pZ lies in Sd (p) for all p | b, check whether ∆(d, z) is square, and, if so, output the pair (d, z). Remark 3.6: The following remarks apply to the implementation of Algorithm 3.5. • The algorithm can be easily parallelized by restricting the range of p1 and, for very small values of p1, fixing p1 and restricting the range of p2. en en en • The recursive enumeration of d0 = p1 ··· pn ensures that, typically, only the value of pn changes from one d0 to the next, e1 en allowing the product C(p1 ) × · · · × C(pn ) to be updated incrementally rather than recomputed for each d0. e √ • The sets C(p ) are precomputed for p ≤ dmax, as are the sets Ad (q) for each d ∈ {1, ... q − 1} not divisible by 3, and the sets Sd (p) for each p ∈ A and d ∈ {1, ... , p − 1}. This allows the sets Z(m) to be efficiently enumerated using an explicit form of the Chinese remainder theorem that requires very little space. We shall refer to this procedure as CRT enumeration. • For p ∈ A, the precomputed sets Sd (p) for d ∈ {1, ... , p − 1} are also stored as bitmaps, as are Cartesian products of pairs of these sets and certain triples; this facilitates testing whether z + pZ lies in Sd (p) for p | b. 16 15 Example 3.7: For k = 33 and d = 5, we have C(d) = {2} and sgn z = +1. For zmax = 10 , this leaves 2 × 10 candidate pairs (5, z) to 13 check. We have #Ad (q) = 14 with q = 891, which reduces this to approximately 3.143 × 10 candidate pairs. The table below shows the benefit of including additional primes p | a.

p | a #Sd (p) #Z(m) m #Z(m, s, zmax) — — 14 4,455 3.143 × 1013 2 1 14 8,910 1.571 × 1013 7 1 14 62,370 2.245 × 1012 13 3 42 810,810 5.180 × 1011 17 9 378 13,783,770 2.742 × 1011 23 12 4,536 317,026,710 1.431 × 1011 29 15 68,040 9,193,774,590 7.401 × 1010 43 19 1,292,760 395,332,307,370 3.270 × 1010 67 27 34,904,520 26,487,264,593,790 1.318 × 1010 103 43 1,500,894,360 2,728,188,253,160,370 5.501 × 109

The net gain is a factor of more than 363,541 over the na¨ıve approach; we gain a factor of about 63 from cubic reciprocity and local constraints mod q, and a factor of about 5,712 from the p | a. In general, including auxiliary p | a ensures that the number of (d, z) we need to consider for small values of d is a negligible proportion of the total computation. Remark 3.8: With CRT enumeration, we avoid the need to store the sets Z(m), analogs of which were explicitly constructed in ref. 4. This greatly reduces the memory required when d is small. In this way, we no longer rely on computations of integral points on the elliptic curve defined by Eq. 1.2 to rule out very small values of d. Nevertheless, we note that one can improve the integral point search carried out in ref. 4, using a trick of Bremner (13) to pass to a 3-isogenous curve. Using this approach, we were able to unconditionally rule out any solutions to Eq. 1.1 with d ≤ 100 for the k listed in Eq. 1.3, and with d ≤ 20, 000 assuming the GRH. It is thus now possible to certify, under GRH, Bremner’s heuristic search of the same region in 1995. 4. Heuristics In this section, we present a heuristic analysis of the distribution of solutions to Eq. 1.1 for a fixed k. We then use this to optimize the choice of the ratio R := zmax/dmax. 3 3 3 3 From Eq. 2.1, we see that, on V = {(x, y, z) ∈ R : x + y + z = 0, |x| ≥ |y| ≥ |z|}, the proportion of the real density contributed by points satisfying y/z ∈ [t1, t2] is Z t2 −1 dt 4σ∞ 3 2/3 . [4.1] t1 (t + 1) 3 3 3 3 2 Given a large solution (x, y, z) ∈ Z to x + y + z = k, with |x| ≥ |y| ≥ |z|, the projective point [x : y : z] ∈ P (R) lies close to the Fermat curve x 3 + y 3 + z 3 = 0. We conjecture that, for fixed k, the ratios y/z are distributed as above: The proportion of points (ordered by any height function as in section A) with y/z ∈ [t1, t2] should converge to the quantity in Eq. 4.1. z Let us assume that this is the case and work out the distribution of r := − x+y for (x, y, z) ∈ V . We have √ √ y 2r 3 + 1 − 12r 3 − 3 z r( 12r 3 − 3 − 3) − = and − = , x 2(r 3 − 1) x 2(r 3 − 1)

so that √ y 12r 3 − 3 − 3 dt r 3 t := = and (t 3 + 1)−2/3 = . z 6r dr 4r 3 − 1 Hence, for any R ≥ α−1, we have

Z ∞ r −1 3 Pr[r ≤ R] = 1 − Pr[r > R] = 1 − 4σ∞ 3 dr = 1 − cK (R), R 4r − 1

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MATHEMATICS We seek to maximize the expected solution count, which, to leading order, is Z ∞ dr ρ log dmax + C − ρc K (r) . R r Differentiating with respect to R, this gives ρ ∂d ρcK (R) max + = 0, dmax ∂R R ∂dmax so that ∂R = −cdmaxK (R)/R. Substituting this into the above, we obtain Td (dmax, Rdmax) 1 −1 3/2 1 = R − 1 ≈ c R 1 − 3 − R =: CR. Tz (dmax, Rdmax) cK (R) 56R

In Table 3, we show computed Td /Tz ratios for k = 3 and various values of R and dmax. For a given dmax, we wish to choose R so that Td /Tz ≈ CR. It is difficult to measure Td /Tz precisely; it is the ratio of two small numbers, and this ratio is easily influenced by small differences in timings when running computations on different hardware. To compute the values below, we used a single hardware platform and took medians of five runs to compute each row. 35 50 From Table 3, we can see that, for k = 3 and dmax ≥ 2 , the optimal choice of R is greater than 32, and, for dmax ≥ 2 , it is greater than 64. For other values of k, the pattern is similar, although the Td /Tz vary slightly; this is to be expected, given the varying benefit of cubic reciprocity constraints. 5. Computational Results A. Implementation. We implemented the algorithm described in section A using the gcc C compiler (14) and the primesieve library for fast prime enumeration (15). We parallelized by partitioning the set of primes p ≤ dmax into subintervals [pmin, pmax] of suitable size, with the work distributed across jobs that checked all of the (d, z) candidates with the largest prime factor p1 | d lying in the assigned interval. Each job was run on a separate machine, with local parallelism achieved by distributing the p1 across available cores (and, for small values of p1, also distributing the p2), as noted in Remark 3.6. When choosing the number of jobs and the sizes of the intervals [pmin, pmax], we use the ρap density estimates derived in section A, as noted in Remark 2.2. We used a standard Tonelli–Shanks approach to computing cube roots modulo primes; this involves computing a discrete logarithm × in the 3-Sylow subgroup of (Z/pZ) , using O(1) group operations on average, and O(1) exponentiations. Hensel lifting√ was used to compute cube roots modulo prime powers; these were precomputed and cached for all prime powers up to min{pmax, dmax}. For the values of dmax that we used, this precomputation typically takes just a few seconds, and the cache size is well under 1 gigabyte. We r × use Montgomery representation (16) for performing arithmetic in (Z/p Z) , but switch to standard integer representation and use Barrett reduction (17) during CRT enumeration of cube roots of k modulo d, and when sieving arithmetic progressions via auxiliary primes. For the k of interest, the sets Ad (q) giving constraints modulo the integer q defined in Lemma 3.3 for admissible pairs (d, z) were precomputed and cached; again, this takes only a few seconds for the largest values of k. In order to avoid using arithmetic progressions of modulus larger than zmax, we project these constraints to residue classes modulo a suitably chosen divisor of q when qd > zmax.

B. Computations. In September 2019, we ran computations for the 11 unresolved k ≤ 1,000 listed in Eq. 1.3 on Charity Engine’s 17 crowd-sourced compute grid consisting of approximately 500,000 personal computers. For this initial search, we used zmax = 10 and 17 dmax = αzmax to search for all solutions to Eq. 1.1 with min{|x|, |y|, |z|} ≤ 10 . This search yielded the solutions for k = 42, k = 165, 18 and k = 906 listed in the Introduction. We then ran a search for k = 3 using zmax = 10 and dmax = αzmax/9 and found the solution for k = 3 listed in the Introduction. These computations involved a total of several hundred core-years but were completed in just a few weeks (it is difﬁcult to give more precise estimates of the computational costs, due to variations in processor speeds and resource

Table 3. Td /Tz versus CR for various values of dmax and R = zmax/dmax for k = 3

R dmax zmax Td Tz Td/Tz CR 32 235 240 2.804 × 10−08 2.359 × 10−10 118.9 60.3 32 240 245 2.738 × 10−08 2.247 × 10−10 121.8 60.3 32 245 250 2.922 × 10−08 2.175 × 10−10 134.4 60.3 32 250 255 3.113 × 10−08 2.150 × 10−10 144.8 60.3 32 255 260 3.678 × 10−08 2.100 × 10−10 175.2 60.3 64 235 241 3.140 × 10−08 1.813 × 10−10 173.2 197.1 64 240 246 2.771 × 10−08 1.730 × 10−10 160.2 197.1 64 245 251 3.112 × 10−08 1.613 × 10−10 192.9 197.1 64 250 256 3.187 × 10−08 1.506 × 10−10 211.6 197.1 64 255 261 3.862 × 10−08 1.612 × 10−10 239.6 197.1 128 235 242 3.749 × 10−08 1.238 × 10−10 302.8 618.5 128 240 247 3.407 × 10−08 1.216 × 10−10 280.2 618.5 128 245 252 3.826 × 10−08 1.530 × 10−10 250.1 618.5 128 250 257 3.768 × 10−08 1.185 × 10−10 318.0 618.5 128 255 262 4.096 × 10−08 1.091 × 10−10 375.4 618.5

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