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November 3rd, 2012, 12:17 PM   #1
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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, 09:11 PM   #2
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Re: Roots of polynomial

P(x) = (x - 2)(x - 1, P(13) = -55. ax + bx + c = x - 20x + 36 = 0, a + b + c = 17.

(x - m)(x - n) = x - (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, 04:04 AM   #3
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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, 01:35 PM   #4
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Re: Roots of polynomial

Second attempt,

P(x) = (x - 2)(x - 1, P(13) = -55. ax + bx + c = x - 20x + 36 = 0, a + b + c = 17.

(ax - m)(x - n) = ax - 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) = x - (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, 05:30 PM   #5
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To avoid confusion, I suggest using a(x - m)(x - n) instead of (ax - m)(x - n).

At some stage, you need to eliminate x - 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, 10:48 AM   #6
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Re: Roots of polynomial

Quote:
Originally Posted by skipjack
To avoid confusion, I suggest using a(x - m)(x - n) instead of (ax - m)(x - n).
(ux - m)(vx - n) = uvx - nux - mvx + mn.
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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