Problem: [BRITAIN 1984/4] x^2 - |_x^2_| = ( x - |_x_| )^2. Count the number of solutions 1<=x<=n. Solution: Let x = m + d, where m = |_x_| and d = x - |_x_| which is 0<=d<1. Plugging x = m + d into x^2 - |_x^2_| = ( x - |_x_| )^2 we get (m + d)^2 - |_(m + d)^2_| = d^2 and after simplification 2md = |_2md + d^2_|. Therefore 2md is an integer k. Case m = 0: Then every 0<=d<1 is possible. Therefore all 0 <= x < 1 are valid. Case m > 0: Then 2md = k. That is d = k/(2m). As 0<=d<1 we have 0 <= k/(2m) < 1. So possible values are k in {0, 1, 2, ..., 2m-1 }. The corresponding x values are m, m + 1/(2m), m + 2/(2m), ..., m + (2m-1)/(2m). Let L(a,b) = { a<=x | n/a - n/b | < 1 => | 1/a - 1/b | < 1/n Since this is true for all n, 1/a = 1/b. Problem: [Jan87, EdM P964] Find all real pairs (a,b) such that for all positive integers n a |_b n_| = b |_a n_|. Solution: [BrF88, EdM P964] It is clear that a |_b n_| = b |_a n_| for all natural numbers n if either ab = 0, or if a=b, or if a and b are both integers. We show that this condition is also necessary. Thus we suppose a |_b n_| = b |_a n_| for all n, ab != 0, and a != b. Then, taking n = 1, we have bm = ak, where m = |_a_| and k = |_b_|. Thus 2m <= 2a < 2m + 2, so that either 2m <= 2a < 2m + 1 or 2m + 1 <= 2a < 2m + 2. Similarly, either 2k <= 2b < 2k + 1 or 2k + 1 <= 2b < 2k + 2. Taking n = 2 we conclude that in fact |_2a_| = 2m and |_2b_| = 2k. (Each of the other possibilities contradicts one of our hypotheses. E. g. assume |_2a_| = 2m + 1 and |_2b_| = 2k + 1 then b(2m+1) = a(2k+1) and as bm = ak we have the contradiction b = a.) Repeating this argument we inductively establish that |_2^r a_| = 2^r m and |_2^r b_| = 2^r k, so that m <= a < m + 1/2^r and k <= b < k + 1/2^r for all natural numbers r. Thus a = m and b = k, and our asertion is proven. Problem: (Sillke) Find all real pairs (x,y) such that x |_y_| = y |_x_|. Problem: Komal F3232 Let b(n) denote the minimum value of expression k + n/k, where k is a positive integer. Prove that for any natural number n, |_b(n)_| = |_sqrt(4n+1)_| Problem: Komal Gy2047 Solve the equation |_sqrt(|_x_|)_| = |_sqrt(sqrt(x))_| on the set of real numbers. Solution: First observe that |_sqrt(|_x_|)_| = |_sqrt(x)_|. Second set x = z^4 and get the nicer looking equation |_z^2_| = |_z_|. Case z < 0: |_z_| <= z < 0 <= |_z^2_|. No solution. Case z >= tau = (sqrt(5) + 1)/2: As z^2 >= tau*z >= z + 1 we get |_z^2_| >= |_z + 1_| = |_z_| + 1 > |_z_|. No solution. Case 0 <= z < tau: Analyze the range of |_z_| which is {0, 1}. So there are only two cases left. Case |_z^2_| = 0 = |_z_|: solutions for 0 <= z < 1. Case |_z^2_| = 1 = |_z_|: solutions for 1 <= z < sqrt(2). Collecting the results we get the solution 0 <= x < 4. Problem: Ouardini Problem 2-10 Solve the equation |_x^(1/2)_| = |_x^(1/3)_| on the set of real numbers. Solution: First set x = z^6 and get the nicer looking equation |_z^3_| = |_z^2_|. Case z < 0: |_z^3_| <= z^3 < 0 <= |_z^2_|. No solution. Case z >= 3/2: As z^3 >= 3/2 z^2 >= z^2 + 1 we get |_z^3_| >= |_z^2 + 1_| = |_z^2_| + 1 > |_z^2_|. No solution. Case 0 <= z < 3/2: Analyze the range of |_z^2_| which is {0, 1, 2}. So there are only three cases left. Case |_z^3_| = 0 = |_z^2_|: solutions for 0 <= z < 1. Case |_z^3_| = 1 = |_z^2_|: solutions for 1 <= z < 2^(1/3). Case |_z^3_| = 2 = |_z^2_|: solutions for 2^(1/2) <= z < 3^(1/3). Collecting the results we get the solution 0 <= x < 2 and 8 <= x < 9. Problem: SSM 3696 Solve the equation |_sqrt(x)_| = |_x/k_| on the set of real numbers, where k is an integer. Problem: 20th MMO 9.1.2 Solve the equation x^3 - |_x_| = 3 on the set of real numbers. Solution: Rearranging the equation gives x^3 = 3 + |_x_|. Therefore the right hand side is an integer. Case x>=2: x^2 >= 4 => x^3 >= 4x >= 3 + x >= 3 + |_x_|. No solution. Case 2>x>=1: x^3 = 3 + |_x_| = 4 => x = 4^(1/3) = 1.587401 Case 1>x>=0: x^3 = 3 + |_x_| = 3 => x > 1. No solution. Case 0>x>=-1: x^3 = 3 + |_x_| = 2 => x > 1. No solution. Case x < -1: x^2 >= 1 => x^3 < x < 2 + x < 3 + |_x_|. No solution. Problem: Ouardini Problem 2-12 Solve the equation 1 + sin^2(x) + sin^2(x - |_sqrt(x)_|) = cos(x) on the set of real numbers. Solution: This equation looks difficult by the obeservation 1 + sin^2(x) + sin^2(x - |_sqrt(x)_|) >= 1 >= cos(x) makes it rather simple. So we are looking for the common solution of the three equations: cos(x) = 1, sin(x) = 0, and sin(x - |_sqrt(x)_|) = 0. The first has the solutions 2Pi*Z, the second Pi*Z. So we have only to check x = x_n = 2Pi*n with n in Z. But x_m = x_n - k for an integer k has only one solution k = 0 as Pi is irrational. As k=0 means x=0 we have only one solution for the original equation. Problem: Komal C596 Solve the equation |_1/(1-x)_| = |_1/(1.5-x)_| on the set of real numbers. Solution: case x < 1: 0 < 1-x < 1.5-x k <= 1/(1.5-x) < 1/(1-x) < k+1 k=0: x<0 case x > 1.5: 1-x < 1.5-x < 0 k=-1: x>=2.5 Problem: Komal C605 (Dec 2000) Solve the equation 1 1 ----- + --------- = x |_x_| x - |_x_| Solution: Multiply the equation by |_x_| and x - |_x_|. This gives x = x |_x_| (x - |_x_|) Case x=0: This don't solve the original equation. Case x!=0: Cancel x 1 = |_x_| (x - |_x_|) => x = |_x_| + 1/|_x_|. So for each integer n>=2 we get a valid solution x = n + 1/n. Problem: Determine the number of real solutions a of the equation |_a/2_| + |_a/3_| + |_a/5_| = a. Solution: There are 30 solutions. Since |_a/2_|, |_a/3_|, and |_a/5_| are integers, so is a. Now write a = 30p + q for integers p and q, 0 <= q < 30. Then |_a/2_| + |_a/3_| + |_a/5_| = a <=> 31p + |_q/2_| + |_q/3_| + |_q/5_| = 30p + q <=> p = q - |_q/2_| - |_q/3_| - |_q/5_|. Thus, for each value of q, there is exactly one value of p (and one value of a) satisfying the equation. Since q can equal any of thirty values, there are exactly 30 solutions as claimed. reference: - 37th International Mathematical Olympiad (July 1996) Mumbai India Problems proposed by not used. reprint: Crux Mathematicorum 23:8 (Dec 1997) 452 problem 18: Find all positive integers a and b for which |_a^2/b_| + |_b^2/a_| = |_(a^2+b^2)/(ab)_| + ab. - 20th Moskau Mathematical Olympiad (1957) Class 9 Round 1 Problem 2. Solve the equation x^3 - |_x_| = 3 on the set of real numbers. - Britain 1984/4: British Mathematical Olympiad - Komal Gy2006 - Komal Gy2047 - Komal F3232 May 1998 - Komal C596 October 2000 - SSM 3696 Douglas E. Scott - W. Janous; Problem 964 Elemente der Mathematik 42:3 (1987) 80 proposal by W. Janous Elemente der Mathematik 43:? (1988) 93 solved by J. L. Brenner, L. L. Foster - Abderrahim Ouardini; Mathématiques de Compétition, 112 problèmes corrigés, Ed. Ellipse, 2000, Paris, ISDN 2-7298-0125-1 - Arthur Engel; Mathematische Olympiadeaufgaben aus der UDSSR, Klett Verlag, Stuttgart, 1965, ISBN 3-12-710300-X - Arthur Engel; Problem-solving strategies, Problem Books in Mathematics. New York, NY: Springer. x, 403 p. (1998) ISBN 0-387-98219-1/hbk Zbl 887.00002 Chap 14.6 Integer Function -- http://www.mathematik.uni-bielefeld.de/~sillke/ mailto:Torsten.Sillke@uni-bielefeld.de