aa r X i v : . [ m a t h . N T ] M a y Applications of the abc conjecture to powerful numbers
P.A. CrowdMathMay 18, 2020
Abstract
The abc conjecture is one of the most famous unsolved problems in number theory.The conjecture claims for each real ǫ > a + b = c with c > (rad( abc )) ǫ . Iftrue, the abc conjecture would imply many other famous theorems and conjectures ascorollaries. In this paper, we discuss the abc conjecture and find new applications topowerful numbers, which are integers n for which p | n for every prime p such that p | n .We answer several questions from an earlier paper on this topic, assuming the truth ofthe abc conjecture. For each positive integer n define the radical rad( n ) of n to be the product of the distinctprime divisors of n . In 1985, David Masser conjectured [4] that for every positive real number ǫ , there are only finitely many triples ( a, b, c ) of coprime positive integers with a + b = c such that c > (rad( abc )) ǫ . This conjecture has come to be called the abc conjecture , anda proof of the abc conjecture would affirm many other famous conjectures and theorems innumber theory. For example, both Fermat’s Last Theorem for sufficiently large powers andRoth’s Theorem on diophantine approximation of algebraic numbers are corollaries of theabc conjecture [5], as are Lang’s conjecture on the N´eron-Tate height and the Erd˝os-Woodsconjecture on sequences of consecutive integers.Call a positive integer x a powerful number if for every prime p such that p | x , we alsohave p | x . Call x a k -powerful number if p k | x for every prime p with p | x . In [2], Cushingand Pascoe found new ways to apply the abc conjecture to prove conditional results aboutpowerful numbers. In addition to proving new corollaries of the abc conjecture, they alsoasked several questions about powerful numbers. In this paper we answer four of the ques-tions from [2], assuming the truth of the abc conjecture. In particular, we prove that thereare only finitely many solutions to the equation x + y = z with ( x, y ) = 1 and x, y, z all4-powerful, solving Problem 2 from [2]. We also show for n ≥ x n + y n where ( x, y ) = 1, partially solving Problem 3from [2] for n ≥
5. For n ≤
3, there are infinitely many powerful numbers of the form x n + y n with ( x, y ) = 1. Problem 3 is not completely solved, since the n = 4 case is still open.1e solve Problem 4 from [2] by proving that there are only finitely many powerfulnumbers of the form k n + r for fixed positive integers k and r with ( k, r ) = 1. We also solveProblem 5 from [2] by showing that for any fixed positive integers r and k , the numbers( n !) r + k are powerful only finitely often. In addition, we present the solution for anotherproblem from the author of [2] about powerful numbers that was posted on the CrowdMath2018 forum. In the last section, we discuss several open problems. For the remainder of the paper, we assume that the abc conjecture is true. We start with asolution to Problem 2 in [2].
Theorem 2.1.
There are only finitely many solutions to the equation x + y = z with ( x, y ) =1 and x, y, z all -powerful.Proof. Let a = x , b = y , c = z , and ǫ = . Observe that(rad( xyz )) ≤ (rad( x ) rad( y ) rad( z )) ≤ ( x y z ) = x y z which is clearly less than z . Thus by the abc conjecture, there are only finitely many solutionsto the equation x + y = z with ( x, y ) = 1 and x, y, z all 4-powerful.The next theorem partially solves Problem 3 from [2], which was to determine when x n + y n is powerful, with ( x, y ) = 1. Note that for n = 2 and n = 3, there are infinitelymany powerful numbers of the form x n + y n with x, y coprime, since there are infinitely manycoprime solutions to x + y = z and x + y = z . The problem is still open for n = 4. Theorem 2.2.
For n ≥ , there are only finitely many powerful numbers of the form x n + y n where ( x, y ) = 1 .Proof. Let x n + y n = z where z is a powerful number and ( x, y ) = 1. Without loss ofgenerality, we let x ≤ y . Then, we have that( xy ) = ( p x y ) ≤ ( x + y ) < ( x + y ) < ( x n + y n ) = z . Now, we apply the abc conjecture. We let a = x n , b = y n , and c = z for relatively prime a , b , and c . Letting ǫ = , we see that(rad( x n y n z )) ≤ (rad( x n )) (rad( y n )) (rad( z )) ≤ (rad( x )) (rad( y )) ( z ) ≤ ( xy ) z < z. Thus for n ≥
5, there are a finite number of solutions.Next we solve Problem 4 from [2], which was to determine whether 2 n + 1 is powerful onlyfinitely often. We prove the more general result that there are a finite number of powerfulnumbers of the form k n + r for any fixed positive integers k and r with ( k, r ) = 1.2 heorem 2.3. If k are r are fixed positive integers with ( k, r ) = 1 , then there are a finitenumber of powerful numbers of the form k n + r .Proof. Suppose that k n + r = z for powerful z . Let a = k n , b = r , c = z , and ǫ = . Observethat (rad( k n rz )) ≤ (rad( k n )) (rad( r )) (rad( z )) ≤ (rad( k )) (rad( r )) ( z ) ≤ k r z which is less than z for n sufficiently large, since k and r are fixed. Therefore if the abcconjecture is true, then for any fixed k ≥ r ≥
1, there must be a finite number ofpowerful numbers of the form k n + r . For k = 1, we have only one possible value of k n + r .Therefore, for any fixed positive k and r with ( k, r ) = 1, there are a finite number of powerfulnumbers of the form k n + r .The next result solves Problem 5 from [2]. For the proof of this result, we define the primorial of x denoted x x . Theorem 2.4.
For any fixed positive integers r and k , the numbers ( n !) r + k are powerfulonly finitely often.Proof. Let ( n !) r + k = z for powerful z , and let n ≥ k , so that k | n ! and the equation becomes ( n !) r k + 1 = zk . Let a = ( n !) r k , b = 1, and c = zk . Observe that:rad( ( n !) r k ) ≤ rad(( n !) r ) = rad( n !) ≤ n zk ) ≤ rad( z ) ≤ z rad( abc ) = rad( ( n !) r k zk ) ≤ ( n z )So, letting ǫ = , (rad( abc )) ≤ ( n z < z = c for n sufficiently large, since n ≤ n and z > n !. Thus, ( n !) r + k is powerful only finitely often, if the abc conjecture is true.During CrowdMath 2018, Cushing proposed two more problems about powerful numbersand the abc conjecture on the CrowdMath forum [1]. We solved one of the problems andmade some progress on the other [1, 3]. The next lemma and theorem together solve thefirst problem by showing that liminf | a n +1 − a n | = ∞ , where a < a < a . . . denotes theset of 3-powerful numbers. We discuss the other problem in the next section. Lemma 2.5.
Let
M, k be fixed positive integers. Then there are finitely many x ∈ Z > suchthat p < M for every prime p dividing x ( x + k ) .Proof. It suffices to prove that there are finitely many x ∈ Z > with ( x, k ) = 1 such that p < M for every prime p dividing x ( x + k ). Let x be an integer with x > ( kM !) . Then wehave rad( kx ( x + k )) < k · Y p
4, there are only finitelymany y where both y and y + k are 3-powerful numbers. We note that in most of our results, the set of powerful numbers can be replaced with the setof numbers which are at least their radical squared, and the set of k -powerful numbers canbe replaced with the set of numbers which are at least their radical to the k th power. Thefollowing problem about powerful numbers was proposed during CrowdMath 2018. Therewas some progress on this problem for some specific polynomials [3], but no progress ingeneral. Question 3.1.
Let P be a polynomial with integer coefficients and at least simple roots.Is it true that P ( n ) is powerful only finitely often? Although we answered most of the questions from [2], Problem 1 in that paper is stillunsolved.
Question 3.2. [2] Does every coprime arithmetic progression contain infinitely many pow-erful pairs?
We also answered Problem 3 from [2] for all n except n = 4. Question 3.3. Is x + y powerful only finitely often when ( x, y ) = 1 ? Acknowledgments
References [1] DCushing, More Cushing-Pascoe problems. https://artofproblemsolving.com/polymath/mitprimes2018/f/c583627h1573657_more_cushingpascoe_problems [2] D. Cushing, J. Pascoe, Powerful numbers and the abc conjecture. https://arxiv.org/pdf/1611.01192.pdf [3] JGeneson, polynomial powerful numbers problem. https://artofproblemsolving.com/polymath/mitprimes2018/f/c583627h1577586_polynomial_powerful_numbers_problemhttps://artofproblemsolving.com/polymath/mitprimes2018/f/c583627h1577586_polynomial_powerful_numbers_problem