Algebra.Com's Answer #739126 by ikleyn(52816)![\"\"](\"/images/stars/stars-13.gif\")  
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document.write( " It is a Math Olimpiad level problem. So I will write the solution as if I talk with
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document.write( " an advanced level student, whose level corresponds to the problem's level.
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document.write( " In other words, I will point the major ideas accurately, but will not go in details.\r
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document.write( "1. The solution starts from a remarkable formula for the sum of the first n odd integer numbers\r\n" );
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document.write( "  = 1 + 3 + 5 + 7 + . . . + (2n-1) =  , (1)\r\n" );
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document.write( " which is valid for any natural \"n\".\r\n" );
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document.write( "2. The series under the question (the arifmetic progression) can start from a positive first term; \r\n" );
document.write( " in this case it lies entirely in the domain of positive integer numbers.\r\n" );
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document.write( " It also can partly lie in the domain of negative integer numbers and partly in the domain \r\n" );
document.write( " of positive integer numbers. The condition does not prohibit it . . . \r\n" );
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document.write( " But it can not lie entirely in the domain of negative integer numbers, since the sum must be \r\n" );
document.write( " positive number.\r\n" );
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document.write( "3. If the progression starts from some positive number, let say\r\n" );
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document.write( " S = (2k+1) + (2k+3) + . . . + (2n-1),\r\n" );
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document.write( " then its sum is \r\n" );
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document.write( " S = -  , (2)\r\n" );
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document.write( " according to the formula (1). (Indeed, in this case S = -  , and then the formula (1) works).\r\n" );
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document.write( " If the progression starts from some negative integer number and partly lies in the negative domain\r\n" );
document.write( " and partly in the positive domain, the same formula (2) works.\r\n" );
document.write( " In this case the positive addend corresponds to the part of the series, which is in positive\r\n" );
document.write( " domain, while the negative addend  corresponds to the part of the series, which is in negative domain.\r\n" );
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document.write( " So, the formula (2) works for any of the two admitted cases.\r\n" );
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document.write( "4. We want the sum of the series be the fourth degree of some integer \"d\". It means that the equation\r\n" );
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document.write( "  = - , (3)\r\n" );
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document.write( " must be hold.\r\n" );
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document.write( "5. Decompose - as (n+k)*(n-k). Then the equiation (3) becomes\r\n" );
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document.write( " = (n+k)*(n-k). (4)\r\n" );
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document.write( "6. Now notice that n-k must be equal to 384, the number of the terms in the series.\r\n" );
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document.write( " The number 384 has the decomposition into the product of prime numbers 2 and 3: \r\n" );
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document.write( " 384 = 3*128 =  .\r\n" );
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document.write( " So, it should be\r\n" );
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document.write( " =  . (5)\r\n" );
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document.write( " In order for the equality (5) was possible, the factor (n+k) must complement to the full fourth degree integer.\r\n" );
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document.write( " The minimal such complement is n+k =  = 27*2 = 54.\r\n" );
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document.write( " It gives us two equations for n and k:\r\n" );
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document.write( " n - k = 384 (6) (as was noticed above), and\r\n" );
document.write( " n + k = 54 (7)\r\n" );
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document.write( " From these equations, it is easy to get\r\n" );
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document.write( " n = 219, k = -165.\r\n" );
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document.write( "7. Thus we get that the progression starts from 2*(-165)+1 = -329 and lasts till 2*219-1 = 437.\r\n" );
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document.write( " So, the progression partly lies in the negative and partly in positive domain.\r\n" );
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document.write( "Answer. Arithmetic progression\r\n" );
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document.write( " -329, -327, - 325, . . . , 437\r\n" );
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document.write( "(first term is -329, the common difference is 2, the last term is 437, the number of terms is 384) has the sum 20736 =  .\r\n" );
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document.write( "And this progression provides the minimal positive sum which is the fourth degree of the integer number.\r\n" );
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document.write( "Check. The sum of the progression =  = 20736.\r\n" );
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document.write( " = 20736. ! Correct !\r\n" );
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document.write( " The number of terms in the progression =  = 384. ! Correct !\r\n" );
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document.write( " * * * Completed and solved ! * * *\r
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