Expressing whole numbers as sums of figurate numbers, including tetrahedral numbers, is a longstanding problem in number theory. Pollock's tetrahedral number conjecture states that every positive integer can be expressed as the sum of at most five tetrahedral numbers. Here we explore a generalization of this conjecture to negative indices. We provide a method for computing sums of two generalized tetrahedral numbers up to a given bound, and explore which families of perfect powers can be expressed as sums of two generalized tetrahedral numbers

Lycan, Ron

Ponomarenko, Vadim

English

California State University (CSU): Open Journal Systems

Sums of Two Generalized Tetrahedral NumbersR. Lycan and V. PonomarenkoAbstract - Expressing whole numbers as sums of figurate numbers, including tetrahedralnumbers, is a longstanding problem in number theory. Pollock’s tetrahedral number conjec-ture states that every positive integer can be expressed as the sum of at most five tetrahedralnumbers. Here we explore a generalization of this conjecture to negative indices. We providea method for computing sums of two generalized tetrahedral numbers up to a given bound,and explore which families of perfect powers can be expressed as sums of two generalizedtetrahedral numbers.Keywords : tetrahedral numbers; generalized tetrahedral numbers; figurate numbers;sums of figurate numbers; Waring’s problemMathematics Subject Classification (2020) : 11P051 IntroductionA historically important problem in number theory, known as Waring’s problem, is torepresent whole numbers as sums of perfect powers, and in particular to find the minimumamount of such powers required to sum to any given whole number. For instance, it isknown that every positive integer is the sum of at most 9 positive cubes ([4]). However,if we allow some of the cubes to be negative, this bound is considerably decreased, and itis conjectured that every integer is the sum of at most 4 integer cubes ([5]).Similar problems to Waring’s problem have been posed for other types of figuratenumbers, including tetrahedral numbers. In 1850, Frederick Pollock conjectured thatevery positive integer can be expressed as the sum of at most five tetrahedral numbers([6]), and the problem is still open. Recently, progress has been made on a generalizationof this problem which allows negative indices for the tetrahedral numbers. If we let Tenrepresent the nth tetrahedral number, Ten =n(n+1)(n+2)6, then for integer n, we call Ten a“generalized tetrahedral number”, or GTN. Note that Te0 = Te−1 = Te−2 = 0, and thatfor all n ∈ Z, Te−n = −Ten−2. So we see that the nonzero GTNs with negative indicesare the opposites of the positive GTNs. Thus an equivalent generalization of the problemwould be one allowing signed sums of tetrahedral numbers. Here we will instead use thenegative index representation.We say that an integer n is a sum of k GTNs if there exist integers t1, t2 · · · tk suchthat∑ki=1 Teti = n. Note that if n is a sum of k GTNs with indices t1, · · · tk, then −n isalso a sum of k GTNs with indices si = −ti−2. In [1] and [2] it was found that 2 < k ≤ 4for all n ∈ N. In this paper, we will prove two central theorems about sums of two GTNs.the pump journal of undergraduate research 5 (2022), 206–217 206The first allows us to readily compute sums of two GTNs below a given bound, while thesecond involves representing perfect powers as sums of two GTNs.To motivate our first result, let us imagine that we are trying to determine if a givenwhole number n is a sum of two GTNs. Our first step may be to determine whether n isa sum of two positive tetrahedral numbers. To do this, we can sum all pairs of positivetetrahedral numbers less than or equal to n and check if any of the sums are equal ton. We do not need to include tetrahedral numbers greater than n in this computation,because adding positive tetrahedral numbers to them will only increase the sum furtherbeyond n. In other words, we have an upper bound on which tetrahedral numbers toinclude in our sums. The following theorem provides a similar bound on the sizes of twogeneralized tetrahedral numbers required to sum to a given whole number n.Theorem 1.1 Let n ∈ N such that n = Tea + Te−b with a, b > 0. Then a and b− 1 areat most⌊√1+8n−12⌋.Proof.Let n = Tea + Te−b with a, b > 0. Note that b ≤ a + 1, or else n would be zero ornegative. Hence for a given a, the smallest possible positive value of Tea +Te−b isa(a+1)2,which occurs when b has its maximum value a + 1.Let m be the largest positive integer such that Tem + Te−m−1 ≤ n. Then we havem(m + 1)2≤ n < (m + 1)(m + 2)2.The left-hand side becomesm2 + m− 2n ≤ 0m ≤√1 + 8n− 12,while the right-hand side becomesm2 + 3m + 2− 2n > 0m >√1 + 8n− 32.Let s =√1 + 8n. Then m ∈ ( s−12− 1, s−12]. Since m is an integer, we getm =⌊s− 12⌋.Hence a ≤⌊s−12⌋. Since b ≤ a + 1, we also have b− 1 ≤ a ≤⌊s−12⌋.the pump journal of undergraduate research 5 (2022), 206–217 207Corollary 1.2 LetAm = {k ∈ N : ∃a, b ∈ N, (k = Tea + Te−b) ∧ (b− 1 ≤ a ≤ m)}.Then n ∈ N is not a sum of two GTNs if both n /∈ Am and n < (m+1)(m+2)2 .The above theorem allows us to easily conduct computer searches for sums of twoGTNs, because it gives a limit on how far we need to search. For instance, if n ≤ 100 is asum of two GTNs, we see that the largest possible index for the GTNs is⌊√801−12⌋= 13.By summing all pairs of GTNs with indices between -14 and 13 and choosing those sumswhich are in the interval [1, 100], we get all the sums of two GTNs below 100, shownbelow. The numbers which are sums of two positive tetrahedral numbers ([3]) form asubsequence, and are shown in bold.1 2 3 4 5 6 8 9 10 11 14 1516 19 20 21 24 25 28 30 31 34 35 3639 40 45 46 49 52 55 56 57 60 64 6670 74 76 78 80 81 83 84 85 88 91 94Of these 48 sums of two GTNs, 29 of them, or 60.4%, are also sums of two positivetetrahedral numbers. The following table shows the values of this proportion when thesums of two GTNs have indices in the interval [−m − 1,m] for various values of m. Forinstance, we have just seen that when m = 13, this value is 60.4%. The value seems todecrease as m increases.m Proportion100 44.4%200 41.6%300 40.0%400 39.3%500 38.7%One may wonder whether there is a result analogous to Theorem 1.1 which wouldprovide a bound on the indices when n is a sum of three GTNs. While such a resultwould undoubtedly be useful in future analysis, it is possible that no such bound exists.In the problem of sums of three integer cubes, it is often possible to add two large positivecubes and one large negative cube, or two large negative cubes and one large positive cube(where the indices are on the order of 1020), to achieve sums as small as 3 ([7]). A similarsituation may occur with tetrahedral numbers, so we will not attempt to find a bound onsums of three GTNs here.2 Modular AnalysisIt was shown in [2] that no prime numbers congruent to 2 (mod 5) apart from 2 are sumsof two GTNs. Here we will analyze whether this result has an effect on the distributionthe pump journal of undergraduate research 5 (2022), 206–217 208of sums of two GTNs within each equivalence class modulo 5. We first notice that theindex of a tetrahedral number modulo 5 determines its modulo 5 value according to thefollowing table:a mod 5 Tea mod 50 01 12 43 04 0We now let n be a sum of two GTNs, n = Tea + Teb. We wish to determine theprobability of n to be in each equivalence class modulo 5. To that end, we assume thateach of the indices a and b are chosen independently and uniformly from the modulo 5equivalence classes. Under this assumption, there are 25 equally likely choices for thepair (a, b). We can then compute the values of n modulo 5 in each of these 25 cases andcalculate the proportions which fall into each equivalence class. The resulting distribution,which we will refer to as the “theoretical distribution”, is shown below. We see that weshould expect 4% of sums of two GTNs to be congruent to 2 (mod 5).n mod 5 Proportion0 44%1 24%2 4%3 4%4 24%The following tables show the counts and proportions of sums of two GTNs between0 and m which are in each equivalence class modulo 5, for various values of m. They alsoshow the difference (in percentage points) between the theoretical distribution and theactual distributions up to m. We see that 2 (mod 5) is indeed slightly less common than3 (mod 5), possibly due to the lack of primes in this equivalence class. Also of note is thefact that multiples of 5 are considerably less common than expected.m = 10, 000:n mod 5 Count Proportion Difference0 599 39.18% −4.82%1 394 25.77% +1.77%2 69 4.51% +0.51%3 92 6.02% +2.02%4 375 24.53% +0.53%m = 100, 000:the pump journal of undergraduate research 5 (2022), 206–217 209n mod 5 Count Proportion Difference0 3239 41.17% −2.83%1 1982 25.19% +1.19%2 336 4.27% +0.27%3 401 5.1% +1.1%4 1910 24.28% +0.28%The accompanying Figure 1 shows the relative proportion of sums of two GTNs ineach equivalence class with respect to m. Together with the above data tables, thissuggests that the proportion of sums of two GTNs in each equivalence class will eventuallyapproach a limit as m increases. Assuming the limiting values are close to the values whenm = 100, 000, they are likely within 1.5 percentage points of the theoretical distribution,except in the 0 (mod 5) equivalence class.0%25%50%75%100%20000 40000 60000 800004 mod 5 3 mod 5 2 mod 5 1 mod 5 0 mod 5Figure 1: Stacked area chart showing the distribution of sums of two GTNs among themodulo 5 equivalence classes.We will now conduct a similar analysis modulo 7. The theoretical distribution modulo7 is as follows:n mod 7 Proportion0 26.53%1 14.29%2 6.12%3 16.33%4 16.33%5 6.12%6 14.29%the pump journal of undergraduate research 5 (2022), 206–217 210The following tables show the counts, proportions, and differences for sums of twoGTNs between 0 and m which are in each equivalence class modulo 7, for various valuesof m, similar to the modulo 5 tables above.m = 10, 000:n mod 7 Count Proportion Difference0 354 23.15% −3.38%1 242 15.83% +1.54%2 114 7.46% +1.34%3 259 16.94% +0.61%4 248 16.22% −0.11%5 102 6.67% +0.55%6 210 13.73% −0.56%m = 100, 000:n mod 7 Count Proportion Difference0 1920 24.40% −2.13%1 1195 15.19% +0.9%2 545 6.93% +0.81%3 1317 16.74% +0.41%4 1263 16.05% −0.28%5 504 6.41% +0.29%6 1124 14.29% ±0.0%As in the modulo 5 case, we see that multiples of 7 are less common than expected.We conjecture that this will occur for any prime modulus p.3 Perfect PowersWe now turn our attention to perfect powers which are sums of two GTNs. To begin,we prove that all squares are representable in this way. Since all even powers are alsosquares, this result shows that any even power is a sum of two GTNs.Theorem 3.1 For all n ∈ N, n2 is a sum of two GTNs.Proof. For any n ∈ N, we haveTen + Te−n =16(n(n + 1)(n + 2) + (−n)(−n + 1)(−n + 2))=16(n3 + 3n2 + 2n− n3 + 3n2 − 2n) = n2Hence n2 is a sum of two GTNs. the pump journal of undergraduate research 5 (2022), 206–217 211We now move on to perfect cubes. We first conduct a computer search over all perfectcubes below 1203 and find that the cubes of the following integers are sums of two GTNs:1, 2, 4, 9, 11, 16, 25, 36, 49, 64, 81, 100, 109, 110The majority of these cubes are also squares, i.e. they are sixth powers. Since we alreadyknow that all squares are sums of two GTNs, we wish instead to focus on non-squarecubes. To this end, we introduce the following definition:Definition 3.2 We say that an integer p is a pure nth power iff n ∈ N \ {1} is minimalsuch that p = an for some a ∈ Z.We see that only 4 of the 14 cubes in the above list are pure, namely:• 23 = Te2 + Te2• 113 = Te19 + Te1• 1093 = Te197 + Te19• 1103 = Te199 + Te−25Based on the data above, it seems that it is relatively rare for a pure perfect cube tobe a sum of two GTNs. In fact, we will now show that there are infinitely many pureperfect cubes which are not a sum of two GTNs. The proof references the following twolemmas, which will be proven shortly.Lemma 3.3 Let n ∈ R and a ∈ {1, 2, 3, 6}. Let p ≥ 3. Then the equation3n2 + 6n + 2 =6pa− a2p4 + 3ap2(n + 1) (1)does not hold.Lemma 3.4 Let n ∈ N and a ∈ {1, 2, 3, 6}. Let p ∈ N be congruent to 3 or 18 (mod 35).Then the equation3n2 + 6n + 2 =6p2a− a2p2 + 3ap(n + 1) (2)does not hold.We are now ready to prove the main theorem.Theorem 3.5 If p is prime and p ≡ 3 or p ≡ 18 (mod 35), then p3 is not a sum of twogeneralized tetrahedral numbers.the pump journal of undergraduate research 5 (2022), 206–217 212Proof. Suppose that p is prime and p ≡ 3 or p ≡ 18 (mod 35). By way of contradiction,suppose also that p3 is a sum of two generalized tetrahedral numbers. Then either bothGTNs have positive indices, or exactly one GTN has a negative index. We consider eachof these cases in turn.Suppose p3 can be written as a sum of two GTNs with positive indices. Then thereexist integers n ≥ k ≥ 0 such thatp3 = Ten + Ten−k.From here we get6p3 = (2n− k + 2)(k2 − kn− k + n2 + 2n).Let A = 2n− k + 2, B = k2 − kn− k + n2 + 2n. Since p is prime, one of the followingcases holds:1. p3 | A2. p3 | B3. p2 | A and p | B4. p | A and p2 | BCase 1: p3 | A and B | 6. Note that B = n(n− k) + k(k− 1) + 2n ≥ 2n, so n ≤ 3. Wethen get A = 2n− k + 2 ≤ 8, so we must have p = 2. Since 2 is not congruent to 3 or 18(mod 35), we may disregard it here.Case 2: p3 | B and A | 6. Then (n, k) ∈ {(0, 0), (1, 1), (2, 0), (3, 2), (4, 4)}. The onlypair that gives a perfect-cube value for p3 is (2,0), giving p = 2, which we may disregard.Case 3: p2 | A and p | B. Then A = 2n− k + 2 = ap2 for some a | 6. Substituting intothe expression for B gives3n2 − 3ap2(n + 1) + 6n + a2p4 + 2 = 6pa3n2 + 6n + 2 =6pa− a2p4 + 3ap2(n + 1)By Lemma 3.3, p < 3, but then p cannot be congruent to 3 or 18 (mod 35).Case 4: p | A and p2 | B. Then A = 2n− k + 2 = ap for some a | 6. Substituting intothe expression for B gives3n2 − 3anp + 6n + a2p2 − 3ap + 2 = 6p2a3n2 + 6n + 2 =6p2a− a2p2 + 3ap(n + 1)By Lemma 3.4, this case is impossible. Hence p3 is not a sum of two positive GTNs.the pump journal of undergraduate research 5 (2022), 206–217 213Now suppose that p3 is a sum of two GTNs with exactly one of the indices negative.Then there exist integers n ≥ 0 and −1 ≤ k ≤ n such thatp3 = Ten + Te−(n−k).From here we get6p3 = (k + 2)(3n2 − 3kn + k2 + k).Let A = k + 2, B = 3n2 − 3kn + k2 + k. Since p is prime, one of the following casesholds:1. p3 | A2. p3 | B3. p2 | A and p | B4. p | A and p2 | BCase 1: p3 | A and B | 6. Note that B = 3n(n− k) + k(k + 1) ≥ k(k + 1). So k ≤ 2,and A = k + 2 ≤ 4. Since p ≥ 2, this contradicts p3 | A.Case 2: p3 | B and A | 6. Here we follow an argument similar to the one used in [2].Note that because p ≡ 3 (mod 5), we have p3 ≡ 2 (mod 5). Now A = k + 2 ∈ {1, 2, 3, 6}.If k = 0 then we get p3 = n2, impossible because p is prime. If k = 1 then we get3n2 − 3n + 2 = 2p3 ≡5 43n2 − 3n + 3 = 3(n2 − n + 1) ≡5 0n2 + 4n + 4 = (n + 2)2 ≡5 3,which is impossible. If k = 4 then we get3n2 − 12n + 20 = p3 ≡5 23n2 − 12n + 12 = 3(n2 − 4n + 4) ≡5 4n2 − 4n + 4 = (n− 2)2 ≡5 3,which is also impossible. If k = −1 then we getp3 = Ten + Te−n−1 =16(n3 + 3n2 + 2n− n3 + n) = n2 + n28p3 = (2p)3 = 4n2 + 4n = (2n + 1)2 − 1(2n + 1)2 − (2p)3 = 1.the pump journal of undergraduate research 5 (2022), 206–217 214By Catalan’s conjecture (proven in [8]), the only solution to the above occurs whenp = 1, which we are not considering here.Case 3: p2 | A and p | B. Hence A = k + 2 = ap2 for some a | 6. Substituting intothe expression for B givesB = 3n2 − 3n(ap2 − 2) + (ap2 − 2)2 + ap2 − 2 = 6pa3n2 + 6n + a2p4 + 2− 3ap2(n + 1) = 6pa3n2 + 6n + 2 = p(6a− a2p3 + 3ap(n + 1))By Lemma 3.3, p < 3, but then p cannot be congruent to 3 or 18 (mod 35).Case 4: p | A and p2 | B. Hence k = ap − 2 for some a | 6. Substituting into theexpression for B gives3n2 − 3anp + 6n + a2p2 − 3ap + 2 = 6p2a3n2 + 6n + 2 =6p2a− a2p2 + 3ap(n + 1)By Lemma 3.4, this case is impossible. Hence p3 is not a sum of two GTNs.Proof. [Proof of Lemma 3.3]Suppose that equation (1) holds under these conditions. We first rearrange it asfollows:3n2 + (6− 3ap2)n + 2 = 6pa− a2p4 + 3ap2The left-hand side is a quadratic in n; its minimum value is obtained when n = 12ap2−1.So we get3(12ap2 − 1)2+ (6− 3ap2)(12ap2 − 1)+ 2 ≤ 6pa− a2p4 + 3ap2−34a2p4 + 3ap2 − 1 ≤ 6pa− a2p4 + 3ap26pa+ 1 ≥ 14a2p4Since p ≥ 3 and a ≥ 1, we get6p + 1 ≥ 6pa+ 1 ≥ 14a2p4 ≥ 14p4 ≥ 274pSo 1 ≥ 34p or p ≤ 43. By contradiction, equation (1) does not hold.the pump journal of undergraduate research 5 (2022), 206–217 215Proof. [Proof of Lemma 3.4]Note that p ≡ 3 (mod 5). Reducing equation (2) modulo 5 gives3n2 + n + 2 ≡5 3(3a− 3a2 + 3an + 3a) ≡5−1a+ a2 − an− aIf a ≡5 1, then this becomes 3n2 +n+ 2 ≡5 −1−n⇒ 3n2−3n+ 3 ≡5 n2−n+ 1 ≡5 0,which is impossible. If a ≡5 2, then we get 3n2 + n + 2 ≡5 4 − 2n ⇒ 3n2 + 3n + 3 ≡5n2 + n + 1 ≡5 0, which is also impossible. So we see that a /∈ {1, 2, 6}.If a = 3, then equation (2) becomes3n2 + 6n + 2 = p(−7p + 9n + 9)Reducing modulo 7 gives3n2 + 6n + 2 ≡7 p(2n + 2)5n2 + 3n + 1 ≡7 p(n + 1)Note that either p ≡ 3 (mod 7) or p ≡ 4 (mod 7). If p ≡ 3 (mod 7), then we get5n2 + 3n + 1 ≡7 3n + 35n2 + 5 ≡7 0n2 ≡7 6,which is impossible. If p ≡ 4 (mod 7), then we get5n2 + 3n + 1 ≡7 4n + 45n2 + 6n ≡7 3n2 + 4n + 4 = (n + 2)2 ≡7 6,which, again, is impossible. So we see that a /∈ {1, 2, 3, 6}, and equation (2) does nothold.Theorem 3.5 provides two infinite families of pure cubes which are never sums of twoGTNs. However, it does not cover all the pure cubes that are not sums of two GTNs. Forinstance, Theorem 3.5 alone does not prevent 53 from being a sum of two GTNs, but wehave seen that it is not. There are likely many more families of pure cubes that are neversums of two GTNs, which would help to explain the observed scarcity of such cubes.We conclude with a brief discussion of fifth powers. A computer search showed thatthere are no pure fifth powers less than or equal to 305 which are sums of two GTNs.However, as we have seen in the case of cubes, the pure perfect powers which are sumsof two GTNs are relatively sparse. As such, we are not prepared to conjecture that nopure fifth powers are sums of two GTNs. We have also not managed to prove a resultanalogous to Theorem 3.5 for fifth powers; it is still an open question whether there areinfinitely many fifth powers which are not sums of two GTNs.the pump journal of undergraduate research 5 (2022), 206–217 216AcknowledgmentsThe authors would like to thank an anonymous referee, whose suggestions and contribu-tions singificantly improved this work.References[1] V. Ponomarenko, Pollock’s Generalized Tetrahedral Numbers Conjecture, Amer. Math. Monthly,128 (2021), 921.[2] V. Ponomarenko, Sums of Generalized Tetrahedral Numbers, Amer. Math. Monthly, 129 (2022),474.[3] The On-Line Encyclopedia of Integer Sequences, available online at the URL: https://oeis.org/A102798[4] W.J. Ellison, Waring’s Problem, Amer. Math. Monthly, 78 (1971), 10–36.[5] P. Revoy, Sur les sommes de quatre cubes, Enseign. math., 29 (1983), 209–220.[6] F. Pollock, On the extension of the principle of Fermat’s theorem of the polygonal numbers tothe higher orders of series whose ultimate differences are constant. With a new theorem proposed,applicable to all the orders, Abstracts of the Papers Communicated to the Royal Society of London,5 (1843), 922–924.[7] A.R. Booker, A.V. Sutherland, On a question of Mordell, Proc. Natl. Acad. Sci. USA, 118 (2021).[8] P. Mihăilescu, Primary cyclotomic units and a proof of Catalan’s conjecture, J. Reine Angew. Math.,2004 (2004), 167–195.Ron LycanSan Diego State University5500 Campanile DriveSan Diego, CA 92182E-mail: rlycan0041@sdsu.eduVadim PonomarenkoSan Diego State University5500 Campanile DriveSan Diego, CA 92182E-mail: vponomarenko@sdsu.eduReceived: July 5, 2022 Accepted: December 17, 2022Communicated by Tamás Forgácsthe pump journal of undergraduate research 5 (2022), 206–217 217

Sums of Two Generalized Tetrahedral Numbers

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Sums of Two Generalized Tetrahedral Numbers

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