(a) Let x, y, z be integers such that x+ y +z = 0. Show that if d is an integer

Question

(a) Let x, y, z be integers such that x+ y +z = 0. Show that if d is an integer which divides two of r, y, z, then it also divides the third of these numbers. Note: your proof should (b) State a generalization of part (a) to n integers ai, a2,..., an whose sum is zero. No proof (c) Suppose that the quadratic equation r2 + br + c = 0 has an integer solution, r. Prove involve three cases is required. to assume in the proof that 0. polynomial A monic polynomial is one in which the coefficient of the highest power of (d) State a generalization of part (c) to the equation p(z)-0, where p(x) is the monic p(x)az -1az2+... + an-1z a equals 1

Explanation / Answer

(a) If d divides (say ) x and y

Then x = dt, y = dr for t, r in Z

Then dt + dr + z =0

=> z = d(-t-r)

=> z=-d(t+r)

=> d divides z

(b) Let a1, a2, a3 , ..., an be n integers such that a1 + a2 + a3 + ...+an = 0

If d is an integer that divides any (n-1) integers from set {a1, a2,...,an} then d must divide the nth integer

(c)Since x2 + bx + c =0 has integer solution r; we have :

Then r+r1 = -b, rr1 = c

Here as r is integer so we get rr1 = c => r|c

(d) suppose the monic polynomial p(x) = xn + a1xn-1 + a2xn-2+...+an =0 has an integer solution r then r| an

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