Baltic Way 1999 - Georg Mohr-Konkurrencen

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must be even as well. With n = 2n ′ and m = 2m ′ the above equation canbe restated as (19 n′ + 5 m′ )(19 n′ − 5 m′ ) = 6 which evidently has no solutionin positive integers.Case 2: Assume that n is odd. Then the last digit of k is 4. Consequently,we have 19 n − 5 m = 4. On the other hand, the remainder of 19 n − 5 mmodulo 3 is never 1, a contradiction.17. Answer: yes.Let n = 5 and consider the integers 0, 2, 8, 14, 26. Adding a = 3 or a = 5to all of these integers we get primes. Since the numbers 0, 2, 8, 14 and26 have pairwise different remainders modulo 5 then for any integer a thenumbers a + 0, a + 2, a + 8, a + 14 and a + 26 have also pairwise differentremainders modulo 5; therefore one of them is divisible by 5. Hence if thenumbers a + 0, a + 2, a + 8, a + 14 and a + 26 are all primes then one ofthem must be equal to 5, which is only true for a = 3 and a = 5.18. Squaring the second inequality gives (a − b) 2 < 5 + 4 √ 4m + 1. Sincem = ab, we have(a + b) 2 < 5 + 4 √ 4m + 1 + 4m = ( √ 4m + 1 + 2) 2 ,implyinga + b < √ 4m + 1 + 2 .Since a > b, different factorizations m = ab will give different values forthe sum a + b (ab = m, a + b = k, a > b has at most one solution in (a, b)).Since m ≡ 2 (mod 4), we see that a and b must have different parity, anda + b must be odd. Also note thata + b 2 √ ab = √ 4m .Since 4m cannot be a square we havea + b √ 4m + 1 .Since a + b is odd and the interval [ √ 4m + 1, √ 4m + 1 + 2 ) contains exactlyone odd integer, then there can be at most one pair (a, b) such thata + b < √ √4m + 1 + 2, or equivalently a − b < 5 + 4 √ 4m + 1.19. Note that the square of any prime p ≠ 3 is congruent to 1 modulo 3. Hencethe numbers k = 6m + 2 will have the required property for any p ≠ 3, as16
p 2 + k will be divisible by 3 and hence composite.In order to have 3 2 +k also composite, we look for such values of m for whichk = 6m + 2 is congruent to 1 modulo 5 — then 3 2 + k will be divisibleby 5 and hence composite. Taking m = 5t + 4, we have k = 30t + 26,which is congruent to 2 modulo 3 and congruent to 1 modulo 5. Hencep 2 + (30t + 26) is composite for any positive integer t and prime p.20. Answer: the only possible value is <strong>1999</strong>.Since a 2 − b 2 + c 2 − d 2 is odd, one of the primes a, b, c and d must be 2,and in view of a > 3b > 6c > 12d we must have d = 2. Now1749 = a 2 − b 2 + c 2 − d 2 > 9b 2 − b 2 + 4d 2 − d 2 = 8b 2 − 12 ,implying b 13. From 4 < c < b 2we now have c = 5 and b must be either11 or 13. It remains to check that 1749 + 2 2 − 5 2 + 13 2 = 1897 is not asquare of an integer, and 1749 + 2 2 − 5 2 + 11 2 = 1849 = 43 2 . Hence b = 11,a = 43 anda 2 + b 2 + c 2 + d 2 = 43 2 + 11 2 + 5 2 + 2 2 = <strong>1999</strong> .17
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Baltic Way 1999 - Georg Mohr-Konkurrencen

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