ADVANCED PROBLEMS AND SOLUTIONS. Edited by Raymond E. Whitney

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1 Edited by Raymond E. Whitney Please send all communications concerning ADVANCED PROBLEMS AND SOLUTIONS to RAYMOND E. WHITNEY, MATHEMATICS DEPARTMENT, LOCK HAVEN UNIVERSITY, LOCK HAVEN, PA This department especially welcomes problems believed to be new or extending old results. Proposers should submit solutions or other information that will assist the editor. To facilitate their consideration, all solutions should be submitted on separate signed sheets within two months after publication of the problems. PROBLEMS PROPOSED IN THIS ISSUE H-459 Proposed by Stanley Rabinowitz, Westford, MA Prove that for all n > 3, 3/5-9 r ( n? 2 0 L 2n+l + 4.4(-l) is very close to the square of an integer. H-460 Proposed by H.-J. Seiffert, Berlin, Germany Define the Fibonacci polynomials by F Q (x) = 0 5 F l (x) =, F n + 2 (x) = xf n+ (x) + F n (x). Show that, for all positive reals x s V // 2-2 &M - i2u " D ^ + l U * ) + (2tt + l)f 2n. l (2x) ( a ) ^ lfy x + s l n ^j = 4x(^ + DF 2n (2x) 2x 2? (b) n ~ l ' fcir\ l/(^2 + sin 2 ) ~ n/(x/x 2 + ), as n ^ w, / < = (c) ^ l / s i n 2 ^ = 2(n 2 - l ) / 3. H-46 Proposed by Lawrence Somer, Washington, D. C. Let {^n} = w(a s Z?) denote the Lucas sequence of the first kind satisfying the recursion relation U n + 2 ~ ^ ^ n + ~^~ bli n 9 where a and b are nonzero integers and the initial terms are UQ = 0 and u\ =. The prime p is a primitive divisor of u n if p u n but p\u m for < m < n -. It is known (see [], p. 200) for the Fibonacci sequence {F n } = w(l, ) that, if p is an odd prime divisor of ^2n+l» where n >, then p E (mod 4). 99] 377

2 (i) Find an infinite number of recurrences u(a, b) such that every odd primitive prime divisor p of any term of the form ^2n + l o r u hn satisfies p = (mod 4), where n >. (ii) Find an infinite number of recurrences u(a s b) such that every odd primitive prime divisor p of any term of the form u^n or Ui+ n + 2 satisfies p = (mod 4), where n >. Reference. E. Lucas. "Theorie des Fonctions Numeriques Simplement Periodiques." Amer. J. Math. (878):84-240, SOLUTION Either Way H-44 Proposed by Albert A. Mullin, Huntsville, AL (Vol. 28, no. 2, May 990) By analogy with palindrome, a Validrome is a sentence, formula, relation, or verse that remains valid whether read forward or backward. For example, relative to prime factorization, 34 is a factorably validromic number since 34 = 3, and when backward gives 3 = 43, which is also correct. () What is the largest factorably validromic square you can find? (2) What is the largest factorably validromic square, avoiding palindromic numbers, you can find? Here are three examples of factorably validromic squares: 3 3, 0 0, 3 3. Solution by Paul S. Bruckman, Edmonds, WA Suppose n = 0x02...Q r is in denary notation; we write n' = Q r Q r..i...!. Given two natural numbers m and n, we say the product m x n is validromic if and only ±f mx n = m r x n r. A natural number n is said to be a validromic square root if and only if: () (n 2 ) ' = in') 2 -. Let V denote the set of validromic square roots; we also write n V if equation () holds. In this case, we also call n 2 - a validromic square. Some interesting properties of such numbers are derived by analyzing the familiar "long multiplication" process, somewhat modified. The multiplication for n z = n x n is indicated below: (2) a o ef s i a i h 6 r-l 0 i e 2 '.. e r _ e 2 0 l e r - l 6 l s 2. ' S r- a 2. a r-l u l u r e 2 6 r - l 6 r - l 0 2 M l S r a r u 2 u r 0 3 e r - l e r 0 2 S r+l * a r + l ' ' r- r 0 0, v r - S 2 r - 2 a 2 r - 2 S 2r- a 2r-l 378 [Nov.

3 In t h i s product, the terms Q^dj are not reduced (mod 0) as they would normally be 3 nor are the columnar sums s k. Therefore, Q v = Z* O^-,-) or more precisely, < i, j < r min(fe s v ) (3) s k = e A + l - i > fe ". 2,.... 2r -. i =max(a: -r +, ) Thus, the terms 0^6^ and the sums s^ are not necessarily denary d i g i t s. However., the a k s (indicated below the s k s) are denary d i g i t s, obtained by the process of "carrying forward and bringing down familiar to any schoolchild. We do not preclude the possibility a Q = 0. Next, we carry out the similar multiplication for (n*) 2 - = n? xn f t u r - l 8 r - l u r l 8 r - l 6 2 r - l l e r U 2 U (4) e, e r - r J 3 r - e 2 e r 2 u r - l 6 l e r D l u r - As in the first product, we allow b = 0. The observation that the columnar sums s k in (4) are identical to those in (2) (except in reverse order) is a consequence of their consisting of the same components 6^0j, albeit in permuted order. In fact, we see that if we reverse the order of the r "product-rows in (4), then reverse the order of the digits in each such row, we obtain the corresponding product-rows of (2). Using the notation introduced, we call the product n x n proper if and only if, for all i, j'e{l, 2,..., r}, ke{l, 2,..., 2r - }, the products 6 i Qj and the s? fe s are all denary digits. Otherwise, we say that the product n x n is improper* We now prove a useful characterization of validromic square roots. Theorem : n e V if and only if n x n is proper. Proof: First, suppose n x n is proper. Looking at (2) and (4), it is clear that a Q = b Q = 0, and moreover that a k = s k = b 2r _ k, /c =, 2,..., 2r -. Equivalently, (n 2 ) f = (n f ) 2, or nev. Conversely, suppose that s k * a k for some k. Let s k = a k5 for all k > h 9 but s h = a h + 0(3^ for some J^ >. Inspection of (4) yields: b 2r -h = a h* b u t blr-h-l = s h+l + ^h ( mod 0 ) 5 if 7z = 2r -, we define s 2 r = ^ 2 P = 0- If d h ^ 0 (mod 0), then blr-h- E a /z + l + ^7z ( mod 0 ) * If d h = 0 (mod 0), then s o ^2r-7z-l * y 2r-h - = a h+l> b u t h 2. r-h-2 3 h+2 + ^fr + l + J *h + 2 h+ (mod 0), where d h+l = d h /0«We apply the same argument until we find a nonzero remainder that is not a multiple of 0; eventually, there exists a value of k such that a h+l- 99] 379

4 b^v-k * a k' Thus, if nxn is improper, then n V. This completes the proof of Theorem. The theorem just proved greatly facilitates the search for validromic square roots (and validromic squares). A by-product of its proof is that if nev and n has v digits, then n 2 - has 2r - digits; to avoid trivial variants, we adopt the convention that, if n = 6 02.«.e p ef, then 0} * 0, d r * 0. Thus, n 2 < 0 2r ~, which implies the following Corollary: If n e V has v digits, then n < [I0 r ~^] = 362,... O digits) Let n v denote the largest r-digit validromic square root. Then, by the Corollary, n\ < 3, n 2 ^ 3, n$ < 36, etc. We readily find that n^ = 3 (trivially), and ri2 = 3. There are other useful observations that may be made to facilitate extension of these initial results. In what follows, we suppose that n r = 662 v E V (with the conventions as described previously). First, we surmise that 0! = 3 for all r; this is easily proved. Clearly, this is true for r = and r = 2. If r > 2, define then m T = ; m 2 = , and (jn' r ) 2 = , so m v E V. Since n v > m r, by definition of n r9 thus 0]_ > 3. But the Corollary implies Q\ < 3. Hence, 0]_ = 3. Clearly, if n > m and m x m is improper, so is nxn. This observation allows us to reject all candidates for n r which exceed a previously excluded candidate and differ from it in only one or more digits. However, a much more powerful result may be inferred, which greatly reduces our search for n r. Given that 0 = 3 and 0 P =, then the formula in (3) implies: s k - 20i0 fe = 60^, for k = 2, 3,..., r. However, s k must be a digit; this implies 0^ = 0 or. Therefore, n v must be composed of "binary" digits, except for Its leading digit, which is 3, and its last digit must equal. Proceeding largely by trial and error, with the tools developed thus far, we find n r, at least for the initial values of r. We begin from the left with 0 = 3, then affix as many consecutive l? s as possible to the right. When one or more 0's must be used, we try to minimize the number of such 0's, and to push them as far to the right as possible, subject only to the condition that 0 P =. As we proceed, we keep track of the rejected candidates, so as to reduce our search. Thus, if 0]_02.-.6r i s such a rejected value for n r, then we know that n r+ i < 0]_02- O^l. Proceeding in this fashion, we find the following values of n r, up to v = 5 (though we could have continued the table, by these same methods): (5) Inspection of the foregoing table leads to the conjecture that Q k is constant for all sufficiently large r; a rigorous proof of this premise seems possible but was not attempted. A related observation is that, for sufficiently large k, the values of Q k do not affect the leading digits of n [Nov

5 To s t r e s s dependence of r (as well as k), we use the expanded n o t a t i o n - 0 ( r ) - 0 s ( P ) = q If 2^ represents the minimum value of r such that 6^} = 0 fc5 a constant for all T > p fcj we can tabulate our apparent results as follows: k K. r k _Iii v k (6) etc. Of course, in order to form this table, we first need to compute n T for r» 8; even then? we cannot always be certain that the values in (6) are correct, at least for the higher values of k. However, if we can accept these values as gospels we may then extend the table of n f P s. The number of terms 6^0^ in s^ is maximized when k = r, and such number is T. A necessary (but not sufficient) test, therefore, is that s^ be a digit. Other values of s ^ also need to be tested., of course; since the ones most likely to fail are the ones whose terms contain 0 ]_ = 3, we test those first. We illustrate by finding n^, assuming that (6) is correct. We note that thus, (2 7) ^ fl(27) fl(2 7) _ 9 V f l fl (27) 4- ft 2 T H f h A ( 2 7 ) -» = = h ) - 2 ( e > + e 7 >e 2 + e ( 5 7) e ^ 6, + e ( 2 2 > 5 + e%h 7 + e ( 2 7 ) e 9 ) + e f, - 2 ( 3 + e ( ) + e ( ) + e ( 2\ 7) + e ( ) + e ( 2 27) + e (2 9 7) ) +. (2 7) (2 7) To maximize n we may attempt 0ig - ; however, since s 2 7 is to be a digit, this forces 0 ^ = Q ( fp = 0 (^7) = 0 ( 2 2 5?) = ^ = A t t h i s. P o i n t * nothing can (9 7) (? 7) be inferred about 0 2 Q or ; f o r t n i s * w e need to consider the following: 4 2 o? ) = e ( r ) e 2 2 i 7 - i - 2(e ( ) 9 + e ( 2 7 ) e e, + e 4 e 7 + e 2 e 9 ) i = =. 2(0 ( o 7) ), (27) assuming 0, g =. In order for this last expression to be a digits we must have 0 ( 2 2 o 7) = 0. Likewise, we find that 0 (2 g 7) = implies s ( ) = 2(0 ( ) (2 7) + ), which can only be a digit if = 0. Therefore, we surmise that n 27 is given by using the values of Q k shown by (6) for its first 8 digits, then, with 0 ( i 2 9 7)0 2 2 O 7) * ) = Testing this as a candidate for n 7, we find that it works; hence, we conclude that n 2 7 is as just described. Continuing in this fashion, we may extend (5) and (6) by alternating back and forth between tables. With considerable effort, the following additional values of n r were derived (manually) by these methods: 99] 38

6 r n r In theory, one could extend these results indefinitely, however, without the aid of a computer, human endurance wanes. It seems quite plausible that a program might be devised, enabling extension of the foregoing tables to an arbitrary degree. The aim of such extension would be to discover any lurking pattern in the sequence of "binary" digits among the 6^fs, as k increases. It may be surmised that, having discovered such a pattern, one might be able to prove its validity rigorously. This exercise is left for the interested reader. As for this particular solver, he gave up the effort at p = 30. Therefore, to "answer" both parts of the problem simultaneously (since neither n r nor n^9 clearly, are palindromes), the largest validromic square found was ^ 3 Q, where ^ Note: The proposer noted that 44 to squares is unnecessary. 2*2, so that the restriction of factors 382 [Nov.