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Integer Illation From Divisibility (Posted on 2016-02-13) Difficulty: 3 of 5
Determine all possible integers N such that:
X13 + X + 90 is divisible by X2 - X + N

No Solution Yet Submitted by K Sengupta    
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Solution Solution Comment 1 of 1
Let the quotient of the division be x^11 + Jx^10 + ... + Bx^2 + Ax + Z.  All the coefficients must be integers.

Multiplying the quotient by x^2 - x + N yields x^13 + (J-1)x^2 + ... + (B*N-A+Z)x^2 + (A*N-Z)x + Z*N.  This must equal x^13 + x + 90.

From the constant term: Z*N = 90 implies Z = 90/N, therefore N is a factor of 90.  There are 24: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90, -1, -2, -3, -5, -6, -9, -10, -15, -18, -30, -45, -90.

From the linear term A*N-Z = 1 implies A = (Z+1)/N. There are six values of N which have N, Z, and A integral: (N,Z,A) = (1,90,91), (2,45,23), (10,9,1), (-1,-90,89), (-2,-45,22), (-9,-10,1).

From the quadratic term: B*N-A+Z = 0 implies B = (A-Z)/N.  Only three of the six possible candidates make B an integer: (N,Z,A,B) = (1,90,91,1), (2,45,23,-11), (-1,-90,89,-179)

With only three possible N, trial division is feasible.  Of N=1, -1, and 2 the division works for only N=2.  Specifically x^13 + x + 90 = (x^2 - x + 2) * (x^11 + x^10 - x^9 - 3x^8 - x^7 + 5x^6 + 7x^5 - 3x^4 - 17x^3 - 11x^2 + 23x + 45).

  Posted by Brian Smith on 2016-02-13 12:56:33
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