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| | == Solution == | | == Solution == |
| − | Let polynomial <math>P(x)</math> be a [[monic polynomial]] such that it's roots are only all of <math>a_1</math> through <math>a_k</math>. Therefore, the sum and product of the roots is n, and the constant term of <math>P(x)</math> is <math>\pm n</math>. From the [[Rational Root Theorem]], all <math>a_i</math> are divisors of n, and integral. We split this into cases:
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| − | ''Case 1: n is prime''
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| − | If n is prime, the only divisors of n are 1 and n. We must have an n in <math>a_i</math> so that <math>a_1 \cdot a_2 \cdot \cdots a_k = n</math>, but then <math>a_1+a_2+...+a_k>n</math>, since <math>k\geq 1</math>. We have a contradiction, therefore n may not be prime.
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| − | ''Case 2: n is composite''
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| − | Let two divisors of n(not necessarily distinct) be <math>d_1</math> and <math>d_2</math>, such that <math>d_1\cdot d_2 = n</math> and <math>d_1,d_2<n</math>. We will prove that <math>d_1+d_2 \leq n</math>:
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| − | We subtract <math>d_1+d_2</math> from <math>d_1d_2</math>: <math>d_1d_2-d_1-d_2</math>. Now we add 1: <math>d_1d_2-d_1-d_2+1=(d_1-1)(d_2-1)</math>. Since <math>d_1</math> and <math>d_2</math> are greater than 1, <math>d_1 -1</math> and <math>d_2 -1</math> are positive. Therefore, <math>d_1d_2\geq d_1+d_2</math>, and <math>d_1+d_2\leq n</math>.
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| − | [[WLOG]], we let <math>a_1=d_1</math> and <math>a_2=d_2</math>. If <math>n-a_1-a_2>0</math>, we can let the rest of the numbers be ones. Therefore, there are such k when n is composite.
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| − | ''Case 3: n=1''
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| − | Therefore, k=1, but <math>k\geq 2</math>, so that is impossible.
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| − | Therefore, there are such <math>k\geq 2</math> such that <math>a_1+a_2+...+a_k = a_1 \cdot a_2 \cdot \cdots a_k = n</math> only when n is composite.
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| | == See Also == | | == See Also == |
Revision as of 12:12, 28 January 2008
such that there are 
positive rational numbers 
satisfying 
.
