A newly-generalized problem from a problem for the Mathematical Olympiad and the methods to solve it
AA newly-generalized problem from a problem for theMathematical Olympiad and the methods to solve it.
Yasushi Ieno
Abstract. A newly-generalized problem from a problem initially thought for the MathematicalOlympiad and the methods to solve it.Key Words and Phrases. diophantine equations, Cauchy problem0. IntroductionRecently I read [1], in which Dvornicich, Veneziano and Zannier show a problem initiallythought for the Romanian Mathematical Olympiad 2010 (see [2]). They present its two solutions,with the aid of its several interpretations. The one is an orthodox method, which is essentiallythe same as what the author of this problem (see [1]) has shown. And the other is a method inwhich they generalize the recurrence sequences in this problem to recurrence sequences relatedto a much longer set of diophantine equations, and use a linear differential equation of the secondorder.I have generalized this problem into a new problem and searched for its solutions by the refer-ence of [1].1. An original problem shortlisted for the Mathematical Olympiad and its newly-generalizedproblem.1-1 An original problem shortlisted for the Mathematical Olympiad.
Problem 1
Let x , x , x , . . . be the sequence defined byx = n + = + nx n ∀ n ≥ . (1) What are the values of n for which x n is an integer?1-2 A generalized problem from a problem for the Mathematical Olympiad. Problem 2
Let x , x , x , . . . be the sequence defined byx = n + = c + nx n ∀ n ≥ , where c is a constant positive integer . (2) What are the values of n for which x n is an integer?1 a r X i v : . [ m a t h . G M ] D ec s mentioned above, we have generalized (1) into (2).From now on we will research how to solve Problem 2 by the reference of [1].2.
2, because the solutions for c=1 arealready said in [1].We can prove by induction that
Lemma 1
For every n ≥ n − = c + (cid:112) ( n − )+ c < x n < c +√ + c = y n (3) Proof.
By a direct computation, we have c +√ + c
4. From the lemma we havec + (cid:112) ( n − )+ c < x n < c +√ + c (cid:112) ( n − )+ c < n − c < √ + c ( n − )+ c < ( n − c ) < + c However the last inequalities are inconsistent modulo 4, whether c is even or odd.Therefore we conclude that the only integral values of the sequence are x , x .Essentially the same solution may be reached by a slightly different approach.2he same conclusion as before can be reached if we show that n − − cx n 2, since1< ( c + ) − c ( c + ) = + <2.Now let an = x − cx n , and assume that the inequalities hold up to n. We may write a n − asa n + = x + − cx n + = x n + ( x n + − c ) = ( c + nx n ) nx n = n ( cx n + n ) x By the induction hypothesis we have:x < cx n + n ⇒ a n + > n cx n + ncx n + n = nand x > cx n + n − ⇒ a n + < n cx n + ncx n + n − < n + n > n ) from an arithmetic point of view we define an integer sequences(a n ) by the recurrences a = = ca n + = ca n + +( n + ) a n ∀ n ≥ , (4)with x n = a n a n − .Let us define d n = gcd ( a n , a n − ) ; d n tells us how much the reduced denominator of x n differsfrom a n − . So, to obtain a lower bound for said denominator, we need a lower bound for a n − andan upper bound for d n .Remark 2. By the recurrence (4) we see that d n + |a n + , and so d n + |d n + ; this will be helpfulin establishing an upper bound for d n .A lower bound for a n is easily obtained as in the following lemma. Lemma 2 For every n ≥ n ≥√ 𝑛 !. Proof. We argue by induction on n ≥ 0. We check that a =1= √ =c ≥√ n and a n + we get a n + = ca n + +( n + ) a n ≥ (cid:112) c ( n + ) ! +( n + )√ n! = (cid:112) ( n + ) ! c +√ n + √ n + = (cid:112) ( n + ) ! (cid:115) n + + c n + + √ n + + ≥ (cid:112) ( n + ) ! . (cid:3) 3o get an upper bound for d n we introduce the exponential generating function of the sequence ( a n ) n ∈ N , namely F ( x ) = ∞ (cid:213) n = a n n! x n We consider F(x) merely as a formal power series, although one could prove that it convergesfor every complex x. From the recurrence on (a n ) we can obtain a differential equation for F; in fact,we can multiply (6) by a n n! and sum it for n ≥ 0; since clearly F’(x)= (cid:205) ∞ n = n + n! x n and F”(x)= (cid:205) ∞ n = n + n! x n ,we obtain that F satisfies the conditions F ( ) = (cid:48) ( ) = cF (cid:48)(cid:48) ( x ) = ( c + x ) F (cid:48) ( x )+ F ( x ) (5) The Cauchy problem (5) may be solved (in the ring of formal power series) to getF ( x ) = e cx + x22 and we can use this explicit form to get a formula for a n . In fact ∞ (cid:213) n = a n n! x n = e cx + x22 = e cx e x22 = ∞ (cid:213) m = c m x m m! ∞ (cid:213) s = x s s!a n n! = (cid:213) + m = n c m s m!s!a n = (cid:213) ≤ n n!c − n s ( n − ) !s! = (cid:213) ≤ n c n − (cid:18) n2s (cid:19) ( − ) !!where the semifactorial (2s − − · (2s − · · · · Lemma 3. Let p be an odd prime. If p | n, then a n ≡ c n (mod p). Proof. Let p be an odd prime dividing n.If p<2s, we have that p | (2s − − (cid:0) n2s (cid:1) is divisible by p, as the p factor in n is not cancelled by (2s)!.So we have that in formula (8) only the term with s = 0 is not divisible by p, whencea n = (cid:213) ≤ n c n − (cid:18) n2s (cid:19) ( − ) !! ≡ c n ( mod p ) (cid:3) Applying this lemma we get the following property of d n , Corollary 4. For every n ≥ 1, d n is a power of 2 and all of the odd prime factors of c.4 roof. If an odd prime p, that is not a prime factor of c, divides d m for some m ≥ 1. Then, by Remark2, p divides d n (and hence a n ) for all n ≥ m, so also for n=pm. But this is not possible becausea p m ≡ (cid:3) We are now ready to prove an upper bound for d n , which will follow by using again the expo-nential generating function F(x). Proposition 5. For every n ≥ n ≤ n − ( p − p − ... p − j ) n − ( − ) j / where p ,p ,. . .,p j are all of the distinct odd prime factors of c. Proof. We have the following identities concerning the above generating function F(x): ( ∞ (cid:213) m = a m x m m! )( ∞ (cid:213) r = (− ) r a r x r r! ) = F ( x ) F (− x ) = e ( x ) = ∞ (cid:213) n = n n!Comparing the coefficients of x for any n ≥ 1, we obtain (cid:213) m + r = (− ) r (cid:18) (cid:19) a m a r = ( ) !n! = n ( − ) !!(6) Now, as observed in Remark 3 above, d n + divides any a m with m ≥ n, so it divides the left-handside of (6), and we know from the Corollary 4 that it is a power of 2 and all of the odd prime factorsof c.Therefore at first, the exponent of 2 in d n + is less than or equal to n. Next, as mentioned above,we assume that c has as many as j distinct odd prime factors, p ,p ,. . .,p j .Now we define m i as the upper bound of the exponent of p i in the prime factorization of d n + ,such that 1 ≤ i ≤ j, then because ( − ) !! is odd,m i = t i (cid:213) k = (cid:34) ( − )+ p ki ki (cid:35) (7)where t i =max t (∈ R ) such that (cid:20) ( − )+ p ti ti (cid:21) ≥ . (8) 5t follows by (7) and (8) thatm i = t i (cid:213) k = (cid:34) − ki + (cid:35) ≤ t i (cid:213) k = (cid:32) − ki + (cid:33) = − ( p i − ) (cid:18) − it i (cid:19) + t i ≤ − ( p i − ) (cid:18) − − (cid:19) + t i = n − i − + t i ∴ d n + ≤ n p m p m ... p m j j ≤ n p n − − + t12 p n − − + t22 ... p n − − + tj2 j = n ( p − p − ... p − j ) n − ( p t12 p t22 ... p tj2 j )≤ n ( p − p − ... p − j ) n − ( − ) j / ( see ( ) ) Therefore for every n ≥ n ≤ n − ( p − p − ... p − j ) n − ( − ) j / (cid:3) We denote by D n the reduced denominator of x n . We have:D n ≥ a n − d n ≥ (cid:112) ( n − ) !2 n − ( p − p − ... p − j ) n − ( − ) j / ∀ n ≥ ( n ) as follows:E ( n ) = (cid:112) ( n − ) !2 n − ( p − p − ... p − j ) n − ( − ) j / ∀ n ≥ n →∞ E ( n ) = lim n →∞ (cid:112) ( n − ) !2 n − ( p − p − ... p − j ) n − ( − ) j / = ∞ (11)It follows (11) that ∃ h (∈ N ) D n ≥ ∀ n ≥ h(12)This means that if n ≥ h then x n is never an integer.At 2, then D n ≥ ( n )≥ ≥ 2. 6or example on Problem 1, c=1 and the inequality below holds.E ( n ) = (cid:112) ( n − ) !2 n − ≥ ∀ n ≥ , (13)It is evident that x n is not integral for n ≥ 10. But we are still left to inspect the value of x n with0 ≤ n ≤ 9, as follows;x = 1, x = 1, x = 2, x = 2, x = , x = , x = , x = , x = and x = .Thus by inspecting the change of E(n) according to n and computing its values for small n’s, itcan be known that the integral values of the sequence (x n ) are x , x , x and x .Similarly on Problem 2, for c ≥ 2, if h on (12) is found, then we are to get the solution only bycomputing x , x . . .x h − .But for now, it is far from easy to find h as a function of c. Especially for the cases that j is greator p − p − ... p − j is great, it is hard to compute h.Owing to (4), we can obtain another expression as follows;D n ≥ a n − d n > c n − n − ( p − p − ... p − j ) n − ( − ) j / ∀ n ≥ In this case c ≥ p . . .p j , soc n − n − ( p − p − ... p − j ) n − ( − ) j / = c n − n2 (√ − p − ... p − j ) n − ( − ) j / ≥ c n − n2 c n2 − ( − ) j / = (cid:16) c2 (cid:17) n2 ( − ) j / (15)We denote the rightmost side of (15) by E ( n ) as follows:E ( n ) : = (cid:16) c2 (cid:17) n2 ( − ) j / Then E ( ) = (cid:0) c2 (cid:1) = c2 ≥ ( n + ) E ( n ) = (cid:16) c2 (cid:17) (cid:18) − − (cid:19) j / ≥ ( p p ... p j ) (cid:18) (cid:19) j / ≥ ( n ) ≥ ∀ n ≥ 2, which results in D n > ≥ ≥ 5, so c n − n − ( p − p − ... p − j ) n − ( − ) j / ≥ c n − n − ( p p ... p j ) n − ( − ) j / ≥ c n − n − ( c ) n − ( − ) j / = c n − n − ( − ) j / (16)We denote the rightmost side of (16) by E ( n ) as follows:E ( n ) : = c n − n − ( − ) j / Then E ( ) = c2 > 1, andE ( ) = c / ( j / ) ≥ ( p p ... p j ) (√ ) j > ( n + ) E ( n ) = c / (cid:18) − − (cid:19) j / ≥ ( p p ... p j ) (cid:18) (cid:19) j / > ∀ n ≥ ( n ) ≥ ∀ n ≥ 2, which results in D n > ≥ =3 and the other factors are all greater than 3, if they exist.We assume the exponent of 3 in the prime factorization of c is more than 1, thenc ≥ p ... p j ≥ ( / p / ... p / ) ≥ ( p − p − ... p − j ) , thereforec n − n − ( p − p − ... p − j ) n − ( − ) j / ≥ c n − n − ( p p ... p j ) n − ( − ) j / ≥ c n − n − ( c ) n − ( − ) j / = c n − n − ( − ) j / (17)Now we use E ( n ) again, then similarly as (15) E ( ) > 1, E ( ) > E ( n + ) E ( n ) > ∀ n ≥ 3, whichresults in D n > ≥ =3.Besides we assume that j>1, so p ≥ ≥ n − n − (cid:18) p − p − ... p − j (cid:19) n − ( − ) j / = c n − n − (cid:18) √ − ... p − j (cid:19) n − ( − ) j / ≥ c n − n − (cid:32) (cid:18) . p . − ... p . − j (cid:19) . (cid:33) n − ( − ) j / ≥ c n − n − (cid:16) (cid:0) p ... p j (cid:1) . (cid:17) n − ( − ) j / ≥ c n − n − (cid:16) c (cid:17) n − ( − ) j / = c n − n − ( − ) j / (18) 8e denote the rightmost side of (18) by E ( n ) as follows:E ( n ) : = c n − n − ( − ) j / Then E ( ) = c2 > 1, andE ( ) = c / ( j / ) ≥ ( p p ... p j ) (√ ) j > ( n + ) E ( n ) = c / (cid:18) − − (cid:19) j / ≥ c / (cid:18) (cid:19) j / > ∀ n ≥ ( n ) ≥ ∀ n ≥ 2, which results in D n > ≥ ( n ) = (cid:112) ( n − ) !2 n − (√ ) n − ( − ) / ∀ n ≥ ( n + ) E ( n ) = √ n ( − ) / √ ( − ) / ∀ n ≥ ( E ( )) = √ ≥ . × . × ≥ . (cid:18) E ( ) E ( ) (cid:19) = × × ≥ . ( E ( )) > 1, so E ( ) > 1. And on (20) if n>30, then E ( n + ) E ( n ) > 1, which results inE ( n ) > ≥ 31. Sufficiently it holds that D n > ≥ n where n ≤ ≤∀ n ≤ 15, a n and d n = gcd ( a n , a n − ) can be obtained as follows.9 n is evidently much more than 1 as above.Next for 16 ≤∀ n ≤ 30, it holds thatD n ≥ a n − d n > a × n − n − (√ ) n − √ − = a (√ ) √ − (cid:18) √ (cid:19) n − > a (√ ) √ (cid:18) √ (cid:19) = a √ > , , , , , × . > . × . × > . n > ∀ n ≥ , x1