Hungarian Problem Book II: Based on the Eötvös Competitions by Jozsef Kurschak, Elvira Rapaport

By Jozsef Kurschak, Elvira Rapaport

The Eötvös Contests in common arithmetic were open to Hungarian scholars of their final yr of highschool ever given that 1894. they're recognized for the simplicity of the strategies hired, the mathematical intensity reached, and the variety of common mathematical fields touched. yet might be their such a lot impressive function is the impact that they, including a arithmetic magazine for college students, appear to have had at the youngsters of that small state. one of the winners of the 1st 11 contests (i.e., these contained in quantity eleven) many changed into scientists of foreign reputation; e.g., L. Fejér, T. von Kármán, D. Kónig, M. Riesz. one of the winners of the following twenty contests (i.e., these contained during this quantity) are G. Szegö, T. Radó, E. Teller; all 3 are popular within the usa, the place they now live. This translation of the Eötvös Contests difficulties from 1894-1928 is predicated at the revised Hungarian variation of J. Kürschák's unique compilation. Kürschák mixed his excellence in arithmetic along with his curiosity in schooling while he provided the dependent options and illuminating factors.

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The theorem just proved is the simplest instance of the following statement made by Fermat: If n is an integer greater than 2, the equation x" + y" = zn . t In spite of the efforts of many distinguished mathematicians, no proof has been found for Fermat's conjecture to this day. Certain special cases are easily proved: the following, for example, includes the problem given above. If n is an odd integer greater than 1, thm three consecutive terms x, y, E of an arithmtk @ogression never satisfy x" Let x = y If we divide (1) + y" = P.

Let m and n be the lengths of the altitudes from Q of these triangles and observe that they form two sides of a rectangle (the other two sides of which lie along A C and BD) whose diagonal is the radius r/2 of the circle k, so that m2 n2 = (r/2)2. We now express the areas as follows: + (QA QC)2 + (QB QD)2 = ( A C m ) 2 + (BD n)2 and hence r p 3 = --#) s = 2r4 - 2 - 4 2 which depends only on the radius r of the given circle. p, Figure 56 Second Solution. Without loss of generality we assume that the radius of the given circle with center 0 is 1.

Product is even. Since n is odd, we may write n = 2k 1, where k is an integer, and observe that our product has 2k 1 factors. Moreover, of the 1, exactly k 1 are odd because numbers 1, 2 , 3 , * * - , 2k + + + + 1 = 2 . 1 - 1, 3=2*2-1, 5=2*3-1, . . . . +1 2k 2(k + 1) - 1. + + + Since the a’s also consist of the numbers from 1 to 2k 1, exactly k 1 of the a’s are odd. Therefore, there are exactly 2 ( k 1) = n 1 odd a,,; 1 , 2 , n appearing numbers among the 2n numbers al, aa, in the above product.

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