Applied Mathematicsematics

Mathematical Gems II (Dolciani Mathematical Expositions No by Ross Honsberger

By Ross Honsberger

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This makesf(x,y) = 2, a prime. (b) B 2 = 0: For B 2 f(x,y) = = = 0 the value y - 1 2 y - 1 2 of the function is [1- 11- (-1)] + 2 [I+I]+2=y-l+2=y+1. In this case, however, B itself must also be 0, implying that the numbers substituted for x and y make x (y + 1) - (y! + 1) = 0, or x (y + 1) = y! + 1. Accordingly, the number y + 1 divides y! + 1. By Wilson's theorem, then, the number y + 1 is a prime. Consequently,f(x,y) yields prime numbers exclusively. We note that f(l, 1) = 2. Let p denote an odd prime number.

Since f(x - 1) = f( - x), we have f(O) = f( -1), f(l) = f( -2), f(2) = f( -3), .. " showing that the trinomial yields these primes again for x = -1, - 2, ... , - 40. Thus x 2 + x + 41 yields primes for the eighty consecutive integers x = - 40. - 39, ... , 38, 39. Equivalently, the function f(x - 40) = x 2 - 79x + 1601 gives these eighty prime values for x = 0, I. • 79. At present this shares the record for the longest string of consecutive integers for which a quadratic yields prime numbers exclusively.

W. Golomb, Combinatonal proof of Fermat's "Little" Theorem, Amer. Math. Monthly. 63 (1956) 718. CHAPTER 6 BICENTRIC POLYGONS, STEINER CHAINS, AND THE HEXLET 1. Bicentric Polygons. , center I and radius r) and a circumcircle O(R). Conversely, it is natural to ask when a given pair of nested circles are the incircle and circumcircle of a triangle. ) This old problem was solved by the great Euler (1707-1783). Denoting by s the distance between the centers I and 0 of the circles, he found that R2 = S2 + 2Rr if and only if I (r) and 0 (R) are the incircle and circumcircle of a triangle.

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