44th International Mathematical Olympiad: Short-listed

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Extra resources for 44th International Mathematical Olympiad: Short-listed problems and solutions

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If s = 0, then n = 0, b has n digits, its square c has at most 2n digits, and so does d, a contradiction. Thus the last digit of a is not 0. Consider now, for example, the case s = 4. Then f must be 6, but this is impossible, since the squares of numbers that start in 4 can only start in 1 or 2, which is easily seen from 160 · · · 0 = (40 · · · 0)2 ≤ (4 ∗ · · · ∗)2 < (50 · · · 0)2 = 250 · · · 0. Thus s cannot be 4. The following table gives all possibilities: s f = last digit of (· · · s)2 f = first digit of (s · · · )2 1 1 1, 2, 3 2 4 4, 5, 6, 7, 8 3 9 9, 1 4 6 1, 2 5 5 2, 3 6 6 3, 4 7 9 4, 5, 6 8 4 6, 7, 8 9 1 8, 9 Thus s = 1, s = 2, or s = 3 and in each case f = s2 .

Note that the radius of Γr/2 is r/2. Now consider the inversion with respect to the circle with radius 1 and centre (0, 0). 50 Γe Γg Γf Γd Γr/2 1/r 2β −2α Let Γd , Γe , Γf , Γg , Γr/2 be the images of Γd , Γe , Γf , Γg , Γr/2 , respectively. Set α = 1/4e, β = 1/4f and R = α + β. The equations of the lines Γe , Γf and Γr/2 are x = −2α, x = 2β and y = 1/r, respectively. Both of the radii of the circles Γd and Γg are R, and their centres are (−α + β, 1/r) and (−α + β, 1/r + 2R), respectively. Let D be the distance between (0, 0) and the centre of Γg .

The discriminant D of the equation ( ) is the square of some integer d ≥ 0: D = (2b2 k − b)2 + 4k − b2 = d2 . If e = 2b2 k − b = d, we have 4k = b2 and a = 2b2 k − b/2, b/2. Otherwise, the clear estimation |d2 − e2 | ≥ 2e − 1 for d = e implies |4k − b2 | ≥ 4b2 k − 2b − 1. If 4k − b2 > 0, this implies b = 1. The other case yields no solutions. 2. Assume that b = 1 and let s = gcd(2a, b3 −1), 2a = su, b3 −1 = st , and 2ab2 −b3 +1 = st. Then t + t = ub2 and gcd(u, t) = 1. Together with st | a2 , we have t | s.

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