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November 3rd, 2012, 01:17 PM | #1 |
Math Team Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions | Roots of polynomial
There exists a quadratic polynomial P (x) with integer coefficients such that: (a) Both of its roots are positive integers, (b) The sum of its coefficients is prime, (c) For some integer k, P (k) = ?55. Show that one of its roots is 2, and find the other root. |
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November 3rd, 2012, 10:11 PM | #2 |
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond | Re: Roots of polynomial
P(x) = (x - 2)(x - 1 ![]() (x - m)(x - n) = x² - (m + n)x + mn, 1 - m - n + mn = p, n(m - 1) = p + (m - 1), [p + (m - 1)] (mod (m - 1)) ? 0 unless m - 1 = p, so n = 2, thus 2 must be a root if a = 1. |
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November 4th, 2012, 05:04 AM | #3 |
Global Moderator Joined: Dec 2006 Posts: 20,270 Thanks: 1958 |
Correct answer for P(x), but the deduction of it isn't quite right. One has p = a(m - 1)(n - 1), so one of the factors is p and each of the others is 1, which implies one of the roots is 2. Without loss of generality, one can assume that n = 2. It can easily be proved that the other root is not 2, so a = 1. Thus P(x) = (x - m)(x - 2), where p = m - 1 is prime. Since P(x) = -55 for some integer x, m cannot be odd, and so m > 3. One can now finish off by trying each way in which -55 can be factorized. |
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November 4th, 2012, 02:35 PM | #4 |
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond | Re: Roots of polynomial
Second attempt, P(x) = (x - 2)(x - 1 ![]() (ax - m)(x - n) = ax² - anx - mx + mn, a - an - m + mn = p, n(m - a) = p + (m - a), [p + (m - a)] (mod (m - a)) ? 0 unless m - a = p or m - a = 1 (m = 2 if a equals 1, and a cannot be greater than 1 if m - a = 1 since a must divide a + 1 (integer roots), which is only possible if a = 1), so, either n = 2 or m = 2, thus 2 must be a root. Assuming a = 1, (x - 2)(x - n) = x² - (n + 2)x + 2n, p = n - 1. p ? 5, so p ? {5, 7, 11, 13, 17, . . . } and we find n = 18 for the solution above. |
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November 4th, 2012, 06:30 PM | #5 |
Global Moderator Joined: Dec 2006 Posts: 20,270 Thanks: 1958 |
To avoid confusion, I suggest using a(x - m)(x - n) instead of (ax - m)(x - n). At some stage, you need to eliminate x² - 60x + 116 on the grounds that 1 - 60 + 116 = 57, which isn't prime. Alternatively, you could explain why primes beyond 17 won't work. |
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November 8th, 2012, 11:48 AM | #6 | |
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond | Re: Roots of polynomial Quote:
uv - nu -mv + mn = p n(m - u) = p + v(m - u) m - u divides p + v(m - u) only if m - u = 1 or m - u = p. If m - u = 1, m must be 2 since u must divide u + 1 (integer roots). If m - u = p, then np = p + vp p(n - v) = p so n - v = 1 so v = 1 (integer roots) and n must be two, thus at least one of the roots is 2. | |
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