Nonexistence of Consecutive Powerful Triplets Around Cubes with Prime-Square Factors

#A103 INTEGERS 25 (2025) NONEXISTENCE OF CONSECUTIVE POWERFUL TRIPLETS AROUND CUBES WITH PRIME-SQUARE FACTORS Jialai She Phillips Academy, Andover, Massachusetts [email protected] Received: 7/9/25, Revised: 8/31/25, Accepted: 10/22/25, Published: 11/25/25 Abstract The Erd˝os–Mollin–Walsh conjecture, asserting the nonexistence of three consecutive powerful integers, remains a celebrated open problem in number theory. A natural line of inquiry, following recent work by Chan (2025), is to investigate potential counterexamples centered around perfect cubes, which are themselves powerful. This paper establishes a new non-existence result for a family of such integer triplets with distinct structural constraints, combining techniques from modular arithmetic, p-adic valuation, Thue equations, and the theory of elliptic curves. 1. Introduction A positive integer is called a powerful number if each of its prime factors appears with an exponent of at least two. Every powerful number nadmits a unique representation as n=a2b3,(1) where a, b ∈Zand bis square-free (meaning that it is not divisible by any perfect square other than 1) (see [5] for example). The Erd˝os–Mollin–Walsh conjecture [3, 9] asserts that no three consecutive integers are all powerful. A natural and interesting case arises when the triplet is centered at a perfect cube x3, itself always powerful. That is, does the triplet (x3−1, x3, x3+ 1) ever consist entirely of powerful numbers? Recently, Chan [2] resolved a subcase by proving no such triplets exist when x3−1=p3y2, x3+1=q3z2,(2) for primes p, q and integers x, y, z > 0. DOI: 10.5281/zenodo.17711516 INTEGERS: 25 (2025) 2 In this paper, we prove the non-existence of a new family of triplets in this setting, extending Chan’s result while addressing distinct constraints. Our main result is as follows. Theorem 1. There exist no consecutive powerful numbers of the form x3−1=p2a3, x3, x3+1=q2b3. where p, q are primes and a, b, x are integers. Notice that the powers in Theorem 1 differ from those in (2). Furthermore, unlike [2], we do not need to impose the restriction that variables x,a, and bare positive. This result establishes the non-existence of a class of consecutive powerful triplets not covered by previous literature. Corollary 1. For any primes p, q and any integers x, a with a= 0, the equation x6−1=p2q2a3has no solution. 2. Proof of the Main Result First, let us introduce some lemmas. Lemma 1. Let pbe a prime and R, S, C be integers that satisfy R S =p2C3, and set g= gcd(R, S). If g= 1, then one of R, S is a perfect cube and the other is p2times a perfect cube. If gis a prime, then there exist C1, C2∈Zsuch that either (R, S)=(gC3 1, gC3 2),{R, S}={gC3 1, g2p2C3 2},or{R, S}={gp2C3 1, g2C3 2}. Proof. For g= 1, if p∤C, any prime factor r=pof Cdivides exactly one of R, S, and if p|C, then all pfactors in p2C3must be contained entirely within either R or C, but not both. Assume gis a prime. If g=p, then both R/p and S/p are perfect cubes. Otherwise, the two factors can be written as gC3 1and g2p2C3 2, or gp2C3 1and g2C3 2, up to ordering. The conclusion follows. Lemma 2. The Diophantine equation u2+u+ 1 = 3v3has exactly two integer solutions (u, v)∈ {(−2,1),(1,1)}. The Diophantine equation u2−u+ 1 = 3v3has exactly two integer solutions (u, v)∈ {(2,1),(−1,1)}. Proof. The core of the proof is to convert the original equations into the standard form of a Mordell curve through proper transformations. Taking the second equation as an example, we first multiply the equation by 32and substitute u′= 3uand v′= 3v, which yields (u′)2−3u′+ 9 = (v′)3.Multiplying by 4 and completing the INTEGERS: 25 (2025) 3 square on the left side allows for the substitution u′′ = 2u′−3 and v′′ =v′, leading to (u′′)2+ 27 = 4(v′′ )3.Finally, multiplying both sides by 24and letting x= 4v′′ and y= 4u′′ reduces the equation to the canonical Mordell form: y2=x3−432. The integer solutions to this well-known curve are (x, y) = (12,±36) (see [4] for example). Working backwards gives the two solutions for (u, v). The other equation can be handled by substituting u7→ −u. Lemma 3. The integer solutions for uto the equation u2+u+ 1 = v3are given by u=−19,−1,0,18. Similarly, the integer solutions for uto u2−u+ 1 = v3are −18,0,1,19. Proof. The first equation can be handled using the transformations and Mordell curve techniques in Lemma 2, or directly by citing the corollary of [10]. The second equation then follows by the substitution u7→ −u. Lemma 4. For any integer x, we have gcdx−1, x2+x+ 1= gcd(x−1,3) and gcdx+ 1, x2−x+ 1= gcd(x+ 1,3). Proof. Writing x2+x+ 1 = (x−1)(x+ 2) + 3 and x2−x+ 1 = (x+ 1)(x−2) + 3, the Euclidean algorithm on polynomials yields the conclusion. Lemma 5. The Diophantine equation u3−v3= 2 has the unique integer pair solution (1,−1),and the Diophantine equation u3−v3= 1 has the integer solutions (1,0) and (0,−1). Proof. We may assume uand vare both nonzero. Consider u3−v3= 2. Since u > v, both factors u−vand u2+uv +v2are positive integers. Therefore, we have two possibilities: u−v= 2, u2+uv +v2= 1, or u−v= 1, u2+uv +v2= 2. Simple calculation yields the unique solution (1,−1). The other equation can be treated similarly. We are now prepared to establish the main result. Suppose for contradiction that x3−1=p2a3, x3+1=q2b3,(3) with integers x, a, b and primes p, q. Set g−= gcd(x−1, x2+x+ 1) and g+= gcd(x+ 1, x2−x+ 1).By Lemma 4, g−=(3x≡1 (mod 3), 1 otherwise, g+=(3x≡2 (mod 3), 1 otherwise. INTEGERS: 25 (2025) 4 We proceed by casework since the pair (g−, g+) can be (1,1), (1,3), or (3,1) only. Case 1: (g−, g+) = (1,1),x≡0 (mod 3).By Lemma 1, we have the following possibilities (i) x−1=u3, x2+x+ 1 = p2v3, (ii) x−1=p2u3, x2+x+ 1 = v3,and (a) x+1=s3, x2−x+ 1 = q2t3, (b) x+1=q2s3, x2−x+ 1 = t3, where u, v, s, and tare integers. We explore each subcase below. •(i)+(a).We obtain x−1 = u3, x+1 = s3, or s3−u3= 2.By Lemma 5, s= 1 and u=−1.But then x= 0 and there exists no prime psatisfying (3). •(i)+(b).Applying Lemma 3 to x2−x+1 = t3yields x=−18,0,1,19. However, none of these values satisfies x+1=q2s3under the given constraints (for example, taking x= 19 leads to q2s3= 20 and thus q= 2, but no integer s exists). •(ii) + (a).This subcase can be treated similarly by applying Lemma 3 to the equation x2+x+ 1 = v3. •(ii)+(b).In this subcase, x2−x+ 1 and x2+x+1 are perfect cubes. Lemma 3 forces x= 0, but then no prime qexists satisfying (3). Case 2: (g−, g+) = (1,3),x≡2 (mod 3).Let vp(n) denote the p-adic valuation of n. First, applying the Lifting-the-Exponent lemma (see, e.g., Theorem 1.37 of [7]) to v3(x3+ 1) gives v3(x+ 1) + 1 = v3(x3+ 1) = v3(x+1)+v3(x2−x+ 1),or v3(x2−x+ 1) = 1.(4) We split into two subcases based on q. •q= 3 : For r= 3 as an arbitrary prime factor of x2−x+ 1,since g+= 3, we have vr(x2−x+ 1) = vr((x+ 1)(x2−x+ 1)) = vr(9b3)=3vr(b). Combined with (4), x2−x+ 1 can be written as 3 Qn i=1 r3αi i,or 3s3for some s∈Z. Applying Lemma 2 gives x=−1 or 2.Yet neither yields a valid solution for (3) with prime p. •q= 3 : Here, we have v3(x3+ 1) = v3(q2b3)=3v3(b). Combined with (4), v3(x3+ 1) ≥3, from which it follows that v3(x+ 1) = v3((x3−1)/(x2−x+ 1)) ≥3−1 = 2, INTEGERS: 25 (2025) 5 and thus x≡ −1 (mod 9).(5) Next, applying Lemma 1 to (x−1)(x2+x+1) = p2a3, we have two possibilities: (i) x−1 = u3, x2+x+1=p2v3,and (ii) x−1=p2u3, x2+x+1=v3. Combining (i) and (5) yields u3≡ −2 (mod 9), which is impossible. For (ii), (5) and Lemma 3 force x=−19 or −1.Neither yields a valid solution for (3) with prime p. Case 3: (g−, g+) = (3,1),x≡1 (mod 3).This case is analogous to the previous one, with xreplaced by −x. The result follows from a combination of p-adic analysis and Lemmas 1, 2, and 3. 3. Proof of the Corollary The proof of Corollary 1 combines the arguments for our main theorem with that of Corollary 1 of [2]. We begin with a few preparatory lemmas. Lemma 6. If 3|x−1, then v3(x2+x+1) = 1; if 3|x+1, then v3(x2−x+1) = 1. Proof. The argument is analogous to the one used in the main theorem’s proof. Lemma 7. The only integer solutions of the Diophantine equation u3−2v3= 1 are (1,0) and (−1,−1). Proof. Both (1,0) and (−1,−1) satisfy u3−2v3= 1. By the Delone–Nagell theorem (see, e.g., [1, Theorem V, §72]), there is at most one solution in addition to (1,0). The conclusion follows. Although the following result is likely known, we were unable to locate a specific reference for these parameters. For the sake of completeness, we provide a brief, self-contained proof based on Lemma 2. Lemma 8. For d∈ {4,18,36}, the Diophantine equation u3−dv3= 1 has the unique integer solution (1,0). Proof. First, for d= 4, we consider the equation u3−4v3= 1. Reducing this modulo 9 implies that u3≡1 (mod 9) and v3≡0 (mod 9), and so u≡1 (mod 3) and v≡0 (mod 3). Let v= 3v0for some integer v0. The equation becomes (u−1)(u2+u+ 1) = 4 ·33·v3 0. INTEGERS: 25 (2025) 6 Lemma 6 gives v3(u2+u+1) = 1, from which it follows that gcd(u−1, u2+u+1) = 3. We can therefore set u−1=3aand u2+u+ 1 = 3bwith gcd(a, b) = 1, and obtain ab = 12v3 0. Since u2+u+ 1 is odd, bmust also be odd. As v3(3b)=1,3∤b. It follows that b=s3, and u2+u+ 1 = 3s3. By Lemma 2, u= 1,−2. Noting umust be odd, the conclusion follows. Similarly, for d= 18, the equation u3−18v3= 1 taken modulo 9 gives u3≡1 (mod 9), which implies u≡1 (mod 3). By Lemma 6, v3(u2+u+ 1) = 1, and thus gcd(u−1, u2+u+1) = 3. Setting u−1 = 3aand u2+u+1 = 3bwith gcd(a, b)=1 results in ab = 2v3. Since bmust be odd and is coprime to a, we have b=s3, which yields u2+u+ 1 = 3s3. The unique integer solution (1,0) again follows by Lemma 2. The case for d= 36 follows from an identical argument, applying a modulo 9 reduction along with Lemma 6 and Lemma 2. We now prove Corollary 1 by contradiction. The equation in the corollary can be written as x3−1x3+ 1=p2q2a3.(6) Since gcd(x3−1, x3+ 1) |x3+ 1 −(x3−1) = 2, we have gcd(x3−1, x3+ 1) = 1 or 2. For gcd(x3−1, x3+ 1) = 1, there are two possibilities for the factors on the left-hand side of (6). Assume p2q2divides one of the factors. This implies that the other factor must be a perfect cube. If x3+1=a3 1or x3−1=a3 1, then xhas no valid solution by Lemma 5. Therefore, p2divides one factor and q2divides the other; without loss of generality, assume x3−1 = p2a3 1and x3+ 1 = q2a3 2. But this contradicts our main theorem. Next, consider gcd(x3−1, x3+ 1) = 2. If one of the primes is 2, say p= 2, then we have two possible systems of equations: (x3−1=2q2u3 x3+ 1 = 2v3,or (x3−1=2u3 x3+ 1 = 2q2v3.(7) By Lemma 7, x=±1, which violates a= 0. For the remainder of the proof, assume p= 2, q= 2, and (6) becomes (x3− 1)(x3+ 1) = 8p2q2b3. One possibility is that p2q2divides one of the factors on the left. This leads to four subcases, each of which yields a contradiction by applying either Lemma 7 or Lemma 8: •x3−1=2p2q2u3, x3+ 1 = 4v3. By Lemma 8, x=−1, violating a= 0. •x3−1=4p2q2u3, x3+ 1 = 2v3. By Lemma 7, x=±1, violating a= 0. •x3−1=2u3, x3+ 1 = 4p2q2v3. By Lemma 7, x=±1, violating a= 0. •x3−1=4u3, x3+ 1 = 2p2q2v3. By Lemma 8, x= 1, violating a= 0. INTEGERS: 25 (2025) 7 Therefore, p2must divide one of x3−1 and x3+ 1, and q2must divide the other. It remains to study the system x3−1=2p2u3, x3+ 1 = 4q2v3.(8) Indeed, when the factors of 2 and 4 are swapped, x3−1 = 4p2u3and x3+1 = 2q2v3 reduce to the form in (8) under the change of variables x7→ −x, u 7→ −v, v 7→ −u, p 7→ q, and q7→ p. The equations in (8) resemble those in (3), but are distinct: for example, x3−1 here is not necessarily a powerful number. Nevertheless, the same proof strategy used in the previous section can be applied. Recall the definitions of g−and g+in the proof of the main theorem. We proceed in a similar manner. Case 1: (g−, g+) = (1,1).The factorization of the first equation in (8) yields four possibilities: (i) x−1 = 2t3 1, x2+x+ 1 = p2t3 2, (ii) x−1=2p2t3 1, x2+x+ 1 = t3 2, (iii) x−1 = t3 1, x2+x+ 1 = 2p2t3 2, and (iv) x−1 = p2t3 1, x2+x+ 1 = 2t3 2, and similarly, the second equation gives: (a) x+ 1 = 4t3 3, x2−x+ 1 = q2t3 4, (b) x+ 1 = 4q2t3 3, x2−x+1=t3 4, (c) x+ 1 = t3 3, x2−x+ 1 = 4q2t3 4, and (d) x+1=q2t3 3, x2−x+ 1 = 4t3 4, where tiare integers. Since x2±x+1 = x(x±1)+1 is always odd, (iii), (iv), (c), and (d) are impossible. Among the remaining combinations, those involving (ii) or (b) can be excluded by Lemma 3. For example, under (i) + (b),we have xodd, and Lemma 3 forces x∈ {1,19};x= 1 gives a= 0 (a contradiction), while x= 19 yields x+ 1 = 20, so no prime qexists satisfying (8). Thus only (i) + (a) remains. In this case, 2t3 3−t3 1= 1.By Lemma 7, t1=±1, hence x∈ {3,−1}, but neither value produces a prime psatisfying (8). Case 2: (g−, g+) = (3,1).If p= 3, Lemma 8 implies x= 1, violating a= 0. Assume p= 3. The case condition requires x≡1 (mod 3), and applying Lemma 6 gives v3(x2+x+ 1) = 1. Combining this with the parity argument from Case 1, it suffices to analyze the factorizations: (i) x−1 = 18t3 1, x2+x+ 1 = 3p2t3 2, or (ii) x−1 = 18p2t3 1, x2+x+ 1 = 3t3 2for the first equation in (8), and (a) x+ 1 = 4t3 3, x2−x+1=q2t3 4, or (b) x+ 1 = 4q2t3 3, x2−x+1=t3 4for the second equation. By a similar line of reasoning, applying Lemma 2 and Lemma 3 eliminates all but one nontrivial combination, (i) + (a), from which it follows that 2t3 3−9t3 1= 1. Reducing the equation modulo 9 gives 2t3 3≡1 (mod 9), a contradiction. Case 3: (g−, g+) = (1,3).If q= 3, Lemma 8 implies x=−1, violating a= 0. Assume q= 3. Applying Lemma 6 gives v3(x2−x+ 1) = 1. Combining the 3-adic valuation and the parity argument as in Case 2, the first equation in (8) leads to two possibilities: (i) x−1=2t3 1, x2+x+1 = p2t3 2, or (ii) x−1=2p2t3 1, x2+x+1 = t3 2, and the second equation leads to two possibilities: (a) x+ 1 = 36t3 3, x2−x+ 1 = 3q2t3 4, or (b) x+1 = 36q2t3 3, x2−x+1 = 3t3 4. Applying Lemma 2 and Lemma 3 leaves only one nontrivial case, x−1=2t3 1, x2+x+1 = p2t3 2, x+1 = 36t3 3, and x2−x+1 = 3qt3 4. This combination implies 18t3 3−t3 1= 1. Then Lemma 8 gives t1=−1, and thus INTEGERS: 25 (2025) 8 x=−1, violating a= 0. The proof is complete. 4. Conclusion We have proved that no three consecutive integers centered at a perfect cube can all be powerful under the structural constraints studied here. This result extends recent advances on the Erd˝os–Mollin–Walsh conjecture by eliminating a notable family of potential counterexamples through modular and elliptic curve methods. One natural extension is to examine whether similar non-existence holds when centered around higher powers or other special integers, and what further constraints might eliminate all consecutive powerful numbers entirely. A related and more fundamental question that arose during our research is the following: we conjecture that for every integer x > 1 and every integer n>2, the number xn−1 is not powerful. This is a stronger claim than Mih˘ailescu’s Theorem (formerly Catalan’s Conjecture) [8]. A proof of this general assertion would have significant implications; the specific case for n= 3 would resolve the question addressed in Theorem 1 and provide deeper insight into the structure of powerful numbers. Acknowledgement. The author is grateful to Tudor Popescu for bringing Chan’s work to his attention and for proposing the key conjecture addressed in this paper. In an earlier draft [6], the author established the result of the corollary with 2xin place of x; the author is grateful to Dr. Tsz Ho Chan for his insightful suggestion to investigate removing the factor of 2, which was accomplished in this revision. The author also thanks the anonymous referee for independently suggesting this same improvement, providing a helpful proof sketch, and offering many stylistic suggestions that improved the exposition of the paper. The conjecture stated in the final section was first raised by the author in the personal communication with Dr. Chan. References [1] B. N. Delone and D. K. Faddeev, The Theory of Irrationalities of the Third Degree, Translations of Mathematical Monographs, vol. 10, American Mathematical Society, Providence, RI, 1964. [2] T.H. Chan, A note on three consecutive powerful numbers, Integers 25 (2025), #A7, 7 pp. [3] P. Erd˝os, Problems and results on consecutive integers, Publ. Math. Debrecen 23 (1976), 271-282. INTEGERS: 25 (2025) 9 [4] J. Gebel, A. Peth˝o, and H.G. Zimmer, On Mordell’s equation, Compos. Math. 110 (1998), 335-367. [5] S.W. Golomb, Powerful numbers, Amer. Math. Monthly 77 (1970), 848-852. [6] J. She. Nonexistence of Consecutive Powerful Triplets Around Cubes with Prime-Square Factors, preprint, arXiv:2507.16828v2. [7] E. Kaya, Analytic Number Theory for Beginners, Springer Undergraduate Mathematics Series, Springer, 2023. [8] P. Mih˘ailescu, Primary cyclotomic units and a proof of Catalan’s conjecture, J. Reine Angew. Math. 572 (2004), 167–195. [9] R.A. Mollin and P.G. Walsh, On powerful numbers, Int. J. Math. Math. Sci. 9(1986), 801-806. [10] N. Tzanakis, The Diophantine equation x3 −3xy2 −y3= 1 and related equations, J. Number Theory 18 (1984), 192-205.