Solutions Notes for DNY-NTCONSTRUCT Evan Chen January 17, 018 1 Solution Notes to TSTST 015/5 Let ϕ(n) denote the number of ositive integers less than n that are relatively rime to n. Prove that there exists a ositive integer m for which the equation ϕ(n) = m has at least 015 solutions in n. I consider the following ELEVEN PRIME NUMBERS: S = {11, 13, 17, 19, 9, 31, 37, 41, 43, 61, 71}. It has the roerty that for any S, all rime factors of 1 are one digit. Let N = (10) billion, and consider M = φ (N). For any subset T S, we have M = φ N T ( 1). Since T > 015 we re done. This solution was motivated by the dee fact that ϕ(11 1000) = ϕ(10 1000), for examle. T Solution Notes to IMO 003/6 Let be a rime number. Prove that there exists a rime number q such that for every integer n, the number n is not divisible by q. By orders, we must have q = k + 1 for this to be ossible. So we just need n k 1 (mod q). So we need a rime q 1 (mod ) such that k 1 (mod q). Wishfully we hoe the order of is and k. One way to do this is extract a rime factor from the cyclotomic olynomial 1 1 which does not haen to be 1 (mod ). Internal use: Olymiad Training for Individual Study (OTIS). Last udate January 17, 018. 1
Evan Chen (January 17, 018) 3 Solution Notes to December TST 015/ Prove that for every ositive integer n, there exists a set S of n ositive integers such that for any two distinct a, b S, a b divides a and b but none of the other elements of S. The idea is to look for a sequence d 1,..., d n 1 of differences such that the following two conditions hold. Let s i = d 1 + + d i 1, and t i,j = d i + + d j 1 for i j. (i) No two of the t i,j divide each other. (ii) There exists an integer a satisfying the CRT equivalences a s i (mod t i,j ) i j Then the sequence a + s 1, a + s,..., a + s n will work. For examle, when n = 3 we can take (d 1, d ) = (, 3) giving 5 {}}{ 10 }{{} 1 }{{} 15 because the only conditions we need satisfy are 3 a 0 (mod ) a 0 (mod 5) a (mod 3). But with this setu we can just construct the d i inductively. To go from n to n+1, take a d 1,..., d n 1 and let be a rime not dividing any of the d i. Moreover, let M = n 1 i=1 d i. Then we claim that d 1 M, d M,..., d n 1 M, is such a difference sequence. For examle, the revious examle extends as follows. 1507 {}}{ 907 {}}{ a }{{} b }{{} c }{{} d 600 900 The new numbers, + Md n 1, + Md n,... are all relatively rime to everything else. Hence (i) still holds. To see that (ii) still holds, just note that we can still get a family of solutions for the first n terms, and then the last n + 1th term can be made to work by Chinese Remainder Theorem since all the new + Md k are corime to everything. 7 4 Solution Notes to USAMO 017/1 Prove that there exist infinitely many airs of relatively rime ositive integers a, b > 1 for which a + b divides a b + b a. One construction: let d 1 (mod 4), d > 1. Let x = dd + d d+. Then set a = x + d, b = x d.
Evan Chen (January 17, 018) To see this works, first check that b is odd and a is even. Let d = a b be odd. Then: a + b a b + b a ( b) b + b a 0 (mod a + b) b a b 1 (mod a + b) b d 1 (mod d + b) ( ) d d d (mod d + b) d + b d d + d. So it would be enough that d + b = dd + d d + = b = 1 ( d d + d d + ) d which is what we constructed. Also, since gcd(x, d) = 1 it follows gcd(a, b) = gcd(d, b) = 1. Remark. Ryan Kim oints out that in fact, (a, b) = (n 1, n + 1) is always a solution. 5 Solution Notes to JMO 016/ Prove that there exists a ositive integer n < 10 6 such that 5 n has six consecutive zeros in its decimal reresentation. One answer is n = 0 + 19 = 54308. First, observe that 5 n 5 0 (mod 5 0 ) 5 n 5 0 (mod 0 ) the former being immediate and the latter since ϕ( 0 ) = 19. Hence 5 n 5 0 (mod 10 0 ). Moreover, we have 5 0 = 1 0 100 < 1 1000 100 = 10 6 10 0. Thus the last 0 digits of 5 n will begin with six zeros. 6 Solution Notes to Shortlist 007 N Let b, n > 1 be integers. Suose that for each k > 1 there exists an integer a k such that b a n k is divisible by k. Prove that b = An for some integer A. Just let k = b, so b C n (mod b ). Hence C n = b(bx + 1), but gcd(b, bx + 1) = 1 so b = A n for some A. 7 Solution Notes to IMO 000/5 Does there exist a ositive integer n such that n has exactly 000 rime divisors and n divides n + 1? 3
Evan Chen (January 17, 018) Answer: Yes. We say that n is Korean if n n + 1. First, observe that n = 9 is Korean. Now, the roblem is solved uon the following claim: Claim. If n > 3 is Korean, there exists a rime not dividing n such that n is Korean too. Proof. I claim that one can take any rimitive rime divisor of n 1, which exists by Zsigmondy theorem. Obviously. Then: Since ϕ(n) 1 it follows then that n. Moreover, n + 1 since n 1. Hence n n + 1 n + 1 by Chinese Theorem, since gcd(n, ) = 1. 8 Solution Notes to BAMO 011/5 Decide whether there exists a row of Pascal s triangle containing four airwise distinct numbers a, b, c, d such that a = b and c = d. An examle is ( ) ( 03 68 = 03 ) ( 67 and 03 ) ( 85 = 03 ) 83. To get this, the idea is to look for two adjacent entries and two entries off by one, and solving the corresonding equations. The first one is simle: ( ) ( ) n n = = n = 3j 1. j j 1 The second one is more involved: ( ) ( ) n n = k k = (n k + 1)(n k + ) = k(k 1) = 4(n k + 1)(n k + ) = 8k(k 1) = (n k + 3) 1 = ( (k 1) 1 ) = (n k + 3) (k 1) = 1 Using standard methods for the Pell equation: (7 + 5 )(3 + ) = 41 + 9. So k = 15, n = 34, doesn t work. (41 + 9 )(3 + ) = 39 + 169. Then k = 85, n = 03. 9 Solution Notes to TSTST 01/5 A rational number x is given. Prove that there exists a sequence x 0, x 1, x,... of rational numbers with the following roerties: (a) x 0 = x; (b) for every n 1, either x n = x n 1 or x n = x n 1 + 1 n ; (c) x n is an integer for some n. 4
Evan Chen (January 17, 018) Think of the sequence as a rocess over time. We ll show that: Claim. At any given time t, if the denominator of x t is some odd rime ower q = e, then we can delete a factor of from the denominator, while only adding owers of two to the denominator. (Thus we can just delete off all the odd rimes one by one and then double aroriately many times.) Proof. The idea is to add only fractions of the form ( k q) 1. Indeed, let n be large, and suose t < r+1 q < r+ q < < r+m q < n. For some binary variables ε i {0, 1} we can have x n = n t x t + c 1 ε1 q + c ε q + c s εm q where c i is some ower of (to be exact, c i = n r+i q, but the exact value doesn t r+1 matter). If m is large enough the set {0, c 1 } + {0, c } + + {0, c m } sans everything modulo. (Actually, Cauchy-Davenort imlies m = is enough, but one can also just use Pigeonhole to notice some residue aears more than times, for m = O( ).) Thus we can eliminate one factor of from the denominator, as desired. 10 Solution Notes to Shortlist 014 N4 Let n > 1 be an integer. Prove that there are infinitely many integers k 1 such that n k is odd. k If n is odd, then we can ick any rime dividing n, and select k = m for sufficiently large integers m. Suose n is even now. Then by Kobayashi s Theorem, there exist infinitely many rimes dividing some number of the form n nr 1 1. for some integer r. Let > n be such a rime, with corresonding integer r. It then follows that n nr n r (mod n r ) since this is clearly correct mod n r, and also correct modulo. If we select k = n r, we have n k = nr n r which is odd. k n r 5
Evan Chen (January 17, 018) 11 Solution Notes to USAMO 006/3 For integral m, let (m) be the greatest rime divisor of m. By convention, we set (±1) = 1 and (0) =. Find all olynomials f with integer coefficients such that the sequence { ( f ( n )) n} n 0 is bounded above. (In articular, this requires f(n ) 0 for n 0.) If f is the (ossibly emty) roduct of linear factors of the form 4n a, then it satisfies the condition. We will rove no other olynomials work. In what follows, assume f is irreducible and nonconstant. It suffices to show for every ositive integer c, there exists a rime and a nonnegative integer n such that n 1 c and divides f(n ). Firstly, recall there are infinitely many odd rimes, with > c, such that divides some f(n ), by Schur s Theorem. Looking mod such a we can find n between 0 and 1 (since n ( n) (mod )). We claim that only finitely many from this set can fail now. For if a fails, then its n must be between 1 c and 1. That means for some 0 k c we have ( ( ) ) ( 1 ( 0 f k f k + 1 ) ) (mod ). There are only finitely many dividing ( c ( f k + 1 ) ) k=1 unless one of the terms in the roduct is zero; this means that 4n (k + 1) divides f(n). This establishes the claim and finishes the roblem. 1 Solution Notes to USAMO 013/5 Let m and n be ositive integers. Prove that there exists an integer c such that cm and cn have the same nonzero decimal digits. One-line soiler: 14857. To work out the details, there exist arbitrarily large rimes such that 10 e m n for some ositive integer e, say by Kobayashi theorem (or other more mundane means). In that case, the eriodic decimal exansions of m and n are cyclic shifts of each other. Thus if one looks at 1 the reeating block of decimals, one may take c to be that resulting integer. Remark. The official USAMO solutions roose using the fact that 10 is a rimitive root modulo 7 e for each e 1, by Hensel lifting lemma. This argument is incorrect, because it breaks if either m or n are divisible by 7. One may be temted to resort to using large rimes rather than owers of 7 to deal with this issue. However it is an oen conjecture (a secial case of Artin s rimitive root conjecture) whether or not 10 (mod ) is rimitive infinitely often, which is the condition necessary for this argument to work. 6
Evan Chen (January 17, 018) 13 Solution Notes to RMM 01/4 Prove there are infinitely many integers n such that n does not divide n + 1, but divides n +1 + 1. Zsig hammer! Define the sequence n 0, n 1,... as follows. Set n 0 = 3, and then for k 1 we let n k = n k 1 where is a rimitive rime divisor of n k 1 +1 + 1 (by Zsigmondy). For examle, n 1 = 57. This sequence of n k s works for k 1, by construction. It s very similar to IMO 000 Problem 5. 14 Solution Notes to Shortlist 013 N4 Determine whether there exists an infinite sequence of nonzero digits a 1, a, a 3,... such that the number a k a k 1... a 1 is a erfect square for all sufficiently large k. The answer is no. Assume for contradiction such a sequence exists, and let x k = a k a k 1... a 1 for k large enough. Difference of squares gives def A k B k = (x k+1 x k )(x k + x k+1 ) = a k 10 k with gcd(a k, B k ) = gcd(x k, x k 1 ) since x k and x k 1 have the same arity. Note that we have the inequalities A k B k < x k+1 < 10 k+1. The idea will be that divisibility issues will force one of A k and B k to be too large. We now slit the roof in two cases: First, assume ν 5 (x k ) k for all k. Then in articular a 1 = 5, so all x k are always odd. So one of A k and B k is divisible by k 1. Moreover, both divisible by at least 5 k/. So for each k, min(a k, B k ) k 1 5 k/ which is imossible for large enough k. Next assume ν 5 (x m) = e < m for some m. Then since x k+1 x k (mod 10k ), we obtain ν 5 (x k ) = e for all k > m. Now, min(a k, B k ) 5 k e which again is imossible for k large enough. 15 Solution Notes to EGMO 014/3 We denote the number of ositive divisors of a ositive integer m by d(m) and the number of distinct rime divisors of m by ω(m). Let k be a ositive integer. Prove that there exist infinitely many ositive integers n such that ω(n) = k and d(n) does not divide d(a + b ) for any ositive integers a, b satisfying a + b = n. 7
Evan Chen (January 17, 018) Weird roblem. The condition is very artificial, although the construction is kind of fun. I m guessing the low scores during the actual contest were actually due to an unusually tricky P. Let n = 1 t, where t 5 (mod 6), ω(t) = k 1, and t is a sufficiently large rime. Let a + b = n and a + b = c. We claim that d(c), which solves the roblem since (n). First, note that 3 a + b, since 3 n. Next, note that c < n < 5 1 (since t) so no exonent of an odd rime in c exceeds. Moreover, c < 3 1. So, it remains to check that ν (c) / { 1, 1}. On the one hand, if ν (a) < ν (b), then ν (a) = 1 and ν (c) = ν (a) =. On the other hand, if ν (a) = ν (b) then ν (a), and ν (c) = ν (a) + 1 is odd and less than 1. 16 Solution Notes to USAMO 01/3 Determine which integers n > 1 have the roerty that there exists an infinite sequence a 1, a, a 3,... of nonzero integers such that the equality holds for every ositive integer k. a k + a k + + na nk = 0 Answer: all n >. For n =, we have a k +a k = 0, which is clearly not ossible, since it imlies a k = a 1 k for all k. For n 3 we will construct a comletely multilicative sequence (meaning a ij = a i a j for all i and j). Thus (a i ) is determined by its value on rimes, and satisfies the condition as long as a 1 + a + + na n = 0. The idea is to take two large rimes and use Bezout s theorem, but the details require significant care. We start by solving the case where n 9. In that case, by Bertrand ostulate there exists rimes and q such that n/ < q < n/. and 1 (q 1) < < q 1 Clearly q, and q 7, so > 3. Also, < q < n but q > n, and 4 4 ( 1 (q + 1)) > n. We now stiulate that a r = 1 for any rime r, q (in articular including r = and r = 3). There are now three cases, identical in substance. If,, 3 [1, n] then we would like to choose nonzero a and a q such that 6 a + q a q = 6 + q 1 n(n + 1) which is ossible by Bézout lemma, since gcd(6, q) = 1. Else if, [1, n] then we would like to choose nonzero a and a q such that 3 a + q a q = 3 + q 1 n(n + 1) which is ossible by Bézout lemma, since gcd(3, q) = 1. 8
Evan Chen (January 17, 018) Else if [1, n] then we would like to choose nonzero a and a q such that a + q a q = + q 1 n(n + 1) which is ossible by Bézout lemma, since gcd(, q) = 1. (This case is actually ossible in a few edge cases, for examle when n = 9, q = 7, = 5.) It remains to resolve the cases where 3 n 8. We enumerate these cases manually: For n = 3, let a n = ( 1) ν 3(n). For n = 4, let a n = ( 1) ν (n)+ν 3 (n). For n = 5, let a n = ( ) ν 5(n). For n = 6, let a n = 5 ν (n) 3 ν 3(n) ( 4) ν 5(n). For n = 7, let a n = ( 3) ν 7(n). For n = 8, we can choose (, q) = (5, 7) in the rior construction. This comletes the constructions for all n >. 17 Solution Notes to TSTST 016/3 Decide whether or not there exists a nonconstant olynomial Q(x) with integer coefficients with the following roerty: for every ositive integer n >, the numbers Q(0), Q(1), Q(),..., Q(n 1) roduce at most 0.499n distinct residues when taken modulo n. We claim that Q(x) = 40(x 1) works. Clearly, it suffices to rove the result when n = 4 and when n is an odd rime. The case n = 4 is trivial, so assume now n = is an odd rime. First, we rove the following easy claim. ( ) Claim. For any odd rime, there are at least 1 ( 3) values of a for which 1 a = +1. Proof. Note that if k 0, k ±1, k 1, then a = (k + k 1 ) 1 works. Also a = 0 works. Let F (x) = (x 1). The range of F modulo is contained within the 1 ( + 1) quadratic residues modulo. On the other hand, if for some t neither of 1 ± t is a quadratic residue, then t is omitted from the range of F as well. Call such a value of t useful, and let N be the number of useful residues. We aim to show N 1 4. 9
Evan Chen (January 17, 018) We comute a lower bound on the number N of useful t by writing ( N = 1 [( ( )) ( ( ))] ( 1 t 1 + t 1 1 1 4 t 1 [( ( )) ( ( ))] 1 t 1 + t 1 1 1 4 t ( ( ) ) 1 t 1 ( )) ( 1 ( )) ) = 1 4 + t 1 4 ( + (+1) 1 ( 3) + 0 + ( 1) (( ) 1 ( 3))) 1 1 ( 5). 4 Thus, the range of F has size at most This is less than 0.499 for any 11. 1 ( + 1) 1 N 3 ( + 3). 8 Remark. In fact, the comutation above is essentially an equality. There are only two oints where terms are droed: one, when 3 (mod 4) there are no k = 1 in the lemma, and secondly, the terms 1 (/) and 1 ( /) are droed in the initial estimate for N. With suitable modifications, one can show that in fact, the range of F is exactly equal to 1 8 (3 + 5) 1 (mod 8) 1 ( + 1) 1 1 N = 8 (3 + 7) 3 (mod 8) 1 8 (3 + 9) 5 (mod 8) 1(3 + 3) 7 (mod 8). 8 10