aa r X i v : . [ m a t h . N T ] M a r EXPLICIT BOUNDS FOR THE DIOPHANTINE EQUATION A ! B ! = C ! LAURENT HABSIEGER
Abstract.
A nontrivial solution of the equation A ! B ! = C ! is a triple of positive integers( A, B, C ) with A ≤ B ≤ C −
2. It is conjectured that the only nontrivial solution is(6 , , C = 10 . Several estimates onthe relative size of the parameters are known, such as the one given by Erd¨os C − B ≤ C , or the one given by Bhat and Ramachandra C − B ≤ (1 / log 2+ o (1)) log log C .We check the conjecture for B ≤ and give better explicit bounds such as C − B ≤ log log( B +1)log 2 − . Introduction
Many authors [6] considered the diophantine equation(1) n ! = r Y i =1 a i !in the integers r, a , . . . , a r , with r ≥ a ≥ · · · ≥ a r ≥
2. A trivial solution is givenby a = n − n = Q ri =2 a i !. Hickerson conjectured that the only non-trivial solutionsare 9! = 7!3!3!2!, 10! = 7!6! = 7!5!3! and 16! = 14!5!2!. He checked it for n ≤ r = 2 (see [4]) and this was verified up to n = 10 by Caldwell [2].Luca [8] proved there are finitely many non-trivial solutions to (1), assuming the abc -conjecture. Erd¨os [4] showed that, if the largest prime number of n ( n + 1) is greater than4 log n for any positive integer n , then there are only finitely many nontrivial solutions to(1).From now on, we shall focus on the case r = 2, i. e. the equation(2) A ! B ! = C ! , which has been studied by Caldwell [2] for C ≤ . Erd¨os [5] proved that C − B ≤ C for C sufficiently large, and noted that it would be nice to obtain a bound ofthe form C − B = o (log log C ). His result was improved by Bhat and Ramachandra [1],who showed that C − B ≤ (1 / log 2 + o (1)) log log C . Hajdu, Papp and Szak´acs [7] recentlyproved that non-trivial solutions different from 10! = 7!6! satisfy to C < B − A ) and B − A ≥ . The aim of this paper is to get better explicit inequalities. Let a ≥ s a denote the sum of the digits of an integer written inthe basis a . When p is a prime, Legendre’s formula gives the exponent of p in n !: v p ( n !) = n − s p ( n ) p − . When we apply this formula to (2), we find A − v p ( A ) + B − v p ( B ) = C − v p ( C ). Since v p ( C ) ≥ v p ( n ) ≤ ( p −
1) log( n +1)log p (see Lemma 1 below), we obtain(3) C ≥ A + B + 1 − log( A + 1)log 2 − log( B + 1)log 2 . Since log C ! = log A ! + log B !, the condition (3) implies that A is much smaller that B .We shall make this assertion explicit by proving the following theorem. Theorem 1.
Let ( A, B, C ) = (6 , , be a nontrivial solutions triple of (2) . For anyreal number t > − − = − . . . . we have A ≤ log( B + 1)log 2 + 2 log log( B + 1)log 2 + t when B is sufficiently large. Moreover we have A ≤ log( B + 1)log 2 + 2 log log( B + 1)log 2 + 2 . . We can slightly improve on Bhat and Ramachandra’s result [1].
Theorem 2.
Let ( A, B, C ) = (6 , , be a nontrivial solution triple of (2) . For any realnumber u > − = − . . . . , we have C − B ≤ log log( B + 1)log 2 + u when B is sufficiently large. Moreover we have C − B ≤ log log( B + 1)log 2 + 1 . . We also deduce a better explicit estimate than B − A > C/
Theorem 3.
Let ( A, B, C ) = (6 , , be a nontrivial solution triple of (2) . For any realnumber v < = 2 . . . . , we have B − A > C − log( C + 1)log 2 − C + 1)log 2 + v when B is sufficiently large. Moreover we have B − A > C − log( C + 1)log 2 − C + 1)log 2 − . . QUATION A ! B ! = C ! 3 All these general estimates used the fact that B ≥ for nontrivial solutions tripledistinct from (6 , , Theorem 4.
Let ( A, B, C ) = (6 , , be a nontrivial solution triple of (2) . Then wehave B ≥ and A ≤ log( B + 1)log 2 + 2 log log( B + 1)log 2 − . ,C − B ≤ log log( B + 1)log 2 − . ,B − A > C − log( C + 1)log 2 − C + 1)log 2 + 2 . . Remark 5.
Caldwell’s result C ≥ concerning Sur´anyi’s conjecture is extended to themuch larger region C ≥ . We first establish useful general properties for the sum of digits and for the Γ function inthe next section. In section 3, we prove a key lemma that studies the asymptotic behaviourof log C ! − log A ! − log B ! under the condition (3), for A = log( B +1)log 2 + B +1)log 2 + t . Wededuce Theorems 1-3 in section 4. In section 5 we use these results to prove Theorem4, hence also to check Sur´anyi’s conjecture further, and to improve on the results of thepreceeding section. We end this paper with a few remarks on possible ways to get betterresults. 2. General properties of s a and ΓWe first give a tight upper bound for the sum of the digits function.
Lemma 1.
Let a ≥ be an integer. For any nonnegative integer n , we have the upperbound s a ( n ) ≤ ( a −
1) log( n + 1)log a . Proof.
Let n be a nonnegative integer. Write s a ( n ) = ( a − b + r , where b is a nonnegativeinteger and 0 ≤ r ≤ a −
2. We have n ≥ b − X i =0 ( a − a i + ra b = ( r + 1) a b − . The function x → x − ( a − log( x +1)log a is convex and vanishes at x = 0 and x = a − , a − s a ( n ) = ( a − b + r ≤ ( a −
1) log( a b )log a + ( a −
1) log( r + 1)log a ≤ ( a −
1) log( n + 1)log a . (cid:3) LAURENT HABSIEGER
Put Ψ( z ) = Γ ′ ( z ) / Γ( z ). Let γ denote Euler’s constant. We recall the formulas (see [3],p. 15)(4) Ψ( z ) = − γ + ∞ X k =0 (cid:18) k + 1 − z + k (cid:19) Ψ ′ ( z ) = ∞ X k =0 z + k ) , and Binet’s second expression for log Γ (see [3], p. 22)(5) log Γ( x ) = (cid:18) x − (cid:19) log x − x + log(2 π )2 + 2 Z ∞ arctan( t/x ) e πt − dt . From the bounds 0 ≤ arctan( t/x ) ≤ t/x and from (5), we get the well-known explicitSirtling’s formula(6) 0 ≤ log Γ( x ) − x (log x − − log(2 π/x )2 ≤ x . Derivating (5) also leads to the formulaΨ( x ) = log x − x − Z ∞ t ( t + x )( e πt − dt and the bounds 0 ≤ / ( t + x ) ≤ /x give the estimates(7) − x ≤ Ψ( x ) − log x + 12 x ≤ . The key lemma
Let us define R ( A, B ) = log Γ (cid:18) A + B + 2 − log( A + 1)log 2 − log( B + 1)log 2 (cid:19) − log Γ ( A + 1) − log Γ ( B + 1) , and let us put A t = log( B + 1)log 2 + 2 log log( B + 1)log 2 + t for any real number t . Lemma 2.
Let t be a real number, t > − − = − . . . . . There exists afunction C ( t, B + 1) such that R ( A t , B ) ≥ C ( t, B + 1) log( B + 1) , with lim B → + ∞ C ( t, B + 1) = t + 1 + 1 + 2 log log 2log 2 > . Moreover we can have C (2 . , B + 1) > for B ≥ . QUATION A ! B ! = C ! 5 Proof.
For B ≥
2, we can havelog ( A t + 1) = log (cid:18) log( B + 1)log 2 + 2 log log( B + 1)log 2 + t + 1 (cid:19) ≤ log log( B + 1) − log log 2 + 2 log log( B + 1) + ( t + 1) log 2log( B + 1)and therefore A t + B + 2 − log( A t + 1)log 2 − log( B + 1)log 2 = B + t + 2 + 2 log log( B + 1)log 2 − log( A t + 1)log 2 ≥ B + t + 2 + log log 2log 2 + log log( B + 1)log 2 − B + 1) + ( t + 1) log 2log 2 log( B + 1) > B + 1 , for B ≥
35. We thus get from (4) and (7), for B ≥ (cid:18) A t + B + 2 − log( A t + 1)log 2 − log( B + 1)log 2 (cid:19) − log Γ ( B + 1) ≥ (cid:18) log log( B + 1)log 2 + t + 1 + log log 2log 2 − B + 1) + ( t + 1) log 2log 2 log( B + 1) (cid:19) Ψ( B + 1) ≥ (cid:18) log log( B + 1)log 2 + t + 1 + log log 2log 2 − B + 1) + ( t + 1) log 2log 2 log( B + 1) (cid:19) × (cid:18) log( B + 1) − B + 1) − B + 1) (cid:19) . = (cid:18) log log( B + 1)log 2 + t + 1 + log log 2log 2 + ϕ ( t, B + 1) (cid:19) log( B + 1)with ϕ ( t, x ) = − x + ( t + 1) log 2log 2 log x − x (cid:18) x + 112 x (cid:19) (cid:18) log log x log 2 + t + 1 + log log 2log 2 − x + ( t + 1) log 2log 2 log x (cid:19) . Stirling’s formula (6) giveslog Γ( x ) ≤ x (log x −
1) + log(2 π/x )2 + 112 x ≤ x (log x − x ≥ . A t + 1) ≤ (cid:18) log( B + 1)log 2 + 2 log log( B + 1)log 2 + t + 1 (cid:19) × (cid:18) log log( B + 1) − − log log 2 + 2 log log( B + 1) + ( t + 1) log 2log( B + 1) (cid:19) , LAURENT HABSIEGER when A t ≥ . A t > A − − ≥ .
448 for B ≥
23, we getlog Γ ( A t + 1) ≤ (cid:18) log log( B + 1)log 2 − ϕ ( t, B + 1) (cid:19) log( B + 1)for B ≥
23, with ϕ ( t, x ) = 2 log log x + ( t + 1) log 2log 2 log x (cid:18) log log x − log log 2 + 2 log log x + ( t + 1) log 2log x (cid:19) . We deduce R ( A t , B ) ≥ (cid:18) t + 1 + 1 + 2 log log 2log 2 + ϕ ( t, B + 1) − ϕ ( t, B + 1) (cid:19) log( B + 1) , for B ≥
35, and we put C ( t, B + 1) = t + 1 + + ϕ ( t, B + 1) − ϕ ( t, B + 1). Notethat the functions ϕ ( t, x ) and ϕ ( t, x ) tend to 0 when x goes to infinity, which provesthe first part of the lemma.For t ≥ − x ≥ , we have − C ( t, x ) ≤ x + ( t + 1) log 2log 2 log x (cid:18) log log x + 1 − log log 2 + 2 log log x + ( t + 1) log 2log x (cid:19) + (cid:18) x + 112 x (cid:19) (0 . . t + 1)) + t + 1 + 1 + 2 log log 2log 2 , a decreasing function of x . We thus deduce C (2 . , ) > . (cid:3) Proof of the first three theorems
Proof of Theorem 1.Lemma 3. If ( A, B, C ) is a solution of (2) , then R ( A, B ) ≤ . The function R is anincreasing function of A for ≤ A ≤ B .Proof. The first claim follows directly from (3): R ( A, B ) ≤ log C ! − log A ! − log B ! = 0.We compute ∂ R∂A∂B ( A, B )= (cid:18) − A + 1) log 2 (cid:19) (cid:18) − B + 1) log 2 (cid:19) Ψ ′ (cid:18) A + B + 2 − log( A + 1)log 2 − log( B + 1)log 2 (cid:19) . QUATION A ! B ! = C ! 7 From (4) we get ∂ R∂A∂B ( A, B ) ≥ ≤ A ≤ B . We use (4) to deduce ∂R∂A ( A, B ) ≥ ∂R∂A ( A, A ) = (cid:18) − A + 1) log 2 (cid:19) Ψ (cid:18) A + 2 − A + 1)log 2 (cid:19) − Ψ( A + 1)= γ ( A + 1) log 2 + ∞ X k =0 k + A + 1 − − A +1) log 2 k + 2 A + 2 − log( A +1)log 2 ! = γ ( A + 1) log 2 + ∞ X k =0 k ( A +1) log 2 + A + 1 − log( A +1)log 2 + ( k + A + 1)( k + 2 A + 2 − log( A +1)log 2 ) > A + 1 ≥ max (cid:16) log( A +1)log 2 − , log( A +1)log 2 (cid:17) ≥
0, which is true for A ≥ (cid:3) Thus we only need to find ¯ A such that R ( ¯ A, B ) > A < ¯ A . For t > − − we have R ( A t , B ) > B large enough by Lemma 2, which givesthe first part of Theorem 1. Hajdu, Papp and Szak´acs [7] proved B − A ≥ , whichensures us that B ≥ . We can therefore deduce the second part of the theorem fromthe inequality C (2 . , B + 1) >
0, also given in Lemma 2.4.2.
Proof of Theorem 2.
Note thatlog A ! = log C ! B ! ≥ ( C − B ) log( B + 1) . For A ≤ A t , we have showed in the proof of Lemma 2 thatlog A ! ≤ log Γ( A t + 1) ≤ (cid:18) log log( B + 1)log 2 − ϕ ( t, B + 1) (cid:19) log( B + 1) . Therefore C − B ≤ log log( B + 1)log 2 − ϕ ( t, B + 1) , thus proving the first part of the theorem, since ϕ ( t, x ) tend to 0 when x goes to infinity.Each monomial term (log log x ) n (log x ) − m defining ϕ is a positive decreasing functionof x for t ≥ − x ≥ . We find − + ϕ (2 . , ) < .
819 and thetheorem follows, as in the previous subsection.4.3.
Proof of Theorem 3.
We write B − A = C − A − ( C − B ) and we use Theorems1 and 2 to get B − A ≥ C − log( B + 1)log 2 − B + 1)log 2 − . . The second part of the theorem follows, and the first part is straightforward.
LAURENT HABSIEGER The proof of Theorem 4
Theorems 2 and 3 show that both A and C − B are small with respect to B . Let usput k = C − B to simplify the statements. Lemma 4.
Let ( A, B, C ) a be a nontrivial solutions triple of (2) . For k = C − B ∈{ , , . . . , } , we have B = B k ( A ) := ⌈ ( A !) /k − ( k + 1) / ⌉ .Proof. We have A ! = k Y i =1 ( B + i ) = k Y i =1 p ( B + i )( B + k + 1 − i ) < (cid:18) B + k + 12 (cid:19) k , which shows that B > ( A !) /k − ( k + 1) / Q ki =1 ( B + i ) − ( B + ( k − / k is apolynomial in B − B = 1, for2 ≤ k ≤
12. This implies that
B < ( A !) /k − ( k − /
2, and the lemma follows. (cid:3)
We checked that the inequality A ! = Q ki =1 ( B k ( A ) + i ) never occurred for A ≤ ≤ k ≤
12 using MAPLE; we asked for a 40000-digits precision (enough to write allthe digits of A !), and this required about twenty-eight hours of computations.For B ≤ , Theorems 2 and 3 give A ≤ k ≤
12, so that the equation (2)has no solution for 10 ≤ B ≤ . We can get better inequalities in these theorems,using B ≥ . Computing C ( − . , ) and ϕ (1 . , ) leads to A ≤ log( B + 1)log 2 + 2 log log( B + 1)log 2 − . ,C − B ≤ log log( B + 1)log 2 − . . For 10 ≤ B ≤ , we thus obtain A ≤ k ≤
11, and the equation (2) hasno solution on this interval. Computing C ( − . , ) and ϕ (1 . , ) givesthe inequalities from Theorem 4.6. Concluding remarks
Our method is based on two informations: an arithmetical information obtained byconsidering the dyadic valuation of the factorials, and an asymptotic information obtainedfrom Stirling’s formula. In order to improve on the orders of magnitude of our estimates,one should get more arithmetical information. First, we applied the estimate from Lemma1 both for A ! and for B !, and it is quite uncommon that this estimate can be sharp inboth cases. Second, we did not use any property of the p -adic valuations for p ≥
3, andany useful information could lead to improvements.The algorithm we used to check that A ! B k ( A )! = ( B k ( A )+ k )! is rather basic. A smarterone should lead to an even much larger bound than ours. QUATION A ! B ! = C ! 9 References [1] K. G. Bhat, K. Ramachandra,
A remark on factorials that are products of factorials , Math. Notes (2010), 317–320.[2] C. Caldwell, The diophantine equation A ! B ! = C !, J. Recreat. Math. (1994), 128–133.[3] A. Erd´elyi, W. Magnus, F. Oberhettinger, F. G. Tricomi, Higher transcendental functions
Vol 1,McGraw-Hill Book Company, Inc., New York-Toronto-London, 1953.[4] P. Erd¨os,
Problems and results on number theoretic properties of consecutive integers and relatedquestions , Proceedings of the Fifth Manitoba Conference on Numerical Mathematics (Univ. Mani-toba, Winnipeg, Man., 1975), pp. 2544. Congressus Numerantium, XVI, Utilitas Math., Winnipeg,Man., 1976.[5] P. Erd¨os,
A consequence of a factorial equation , Amer. Math. Monthly (1993), 407–408.[6] R. K. Guy,
Unsolved problems in number theory , third edition, Springer 2004, section B23.[7] L. Hajdu, ´A Papp, T. Szak´acs,
On the equation A ! B ! = C !, J. Number Theory (2018), 160–165.[8] F. Luca, On factorials which are products of factorials , Math. Proc. Cambridge Philos. Soc. (2007), 533–542.
MSC2010: 11D85 11B65 11D41 11N64
Universit´e de Lyon, CNRS UMR 5208, Universit´e Claude Bernard Lyon 1, InstitutCamille Jordan, 43 boulevard du 11 novembre 1918, 69622 Villeurbanne Cedex, France
E-mail address ::