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Algebra Difficulty 7.1 National olympiad, round 2 Prove it Belarus

Let polynomials P(x)P(x) and Q(x)Q(x) have the same degree. Let PQ(x)P_Q(x) denote the polynomial such that its coefficients of even degree variables coincide with the corresponding coefficients of P(x)P(x) and its coefficients of odd degree variables coincide with the corresponding coefficients of Q(x)Q(x). (For example, if P(x)=x3+2x2+4x+1P(x) = x^3 + 2x^2 + 4x + 1 and Q(x)=3x3+x2+2Q(x) = 3x^3 + x^2 + 2, then PQ(x)=3x3+2x2+1P_Q(x) = 3x^3 + 2x^2 + 1, and QP(x)=x3+x2+4x+2Q_P(x) = x^3 + x^2 + 4x + 2.)

a) Prove that there exist P(x)P(x) and Q(x)Q(x) such that they have no real roots but both PQ(x)P_Q(x) and QP(x)Q_P(x) have at least one real root.

b) Find the smallest degree of P(x)P(x) and Q(x)Q(x) satisfying a).

Solution

a)
Show, for example, that the polynomials
P(x)=4x4+4x3+1andQ(x)=x4+4x+4 P(x) = 4x^4 + 4x^3 + 1 \quad \text{and} \quad Q(x) = x^4 + 4x + 4
satisfy the condition.
By Cauchy's inequality, for any real xx the following inequalities hold:
x4+x4+x4+14x4x4x414=4x3,(1) x^4 + x^4 + x^4 + 1 \ge 4\sqrt[4]{x^4 \cdot x^4 \cdot x^4 \cdot 1} = 4|x|^3, \quad (1)
x4+1+1+14x41114=4x.(2) x^4 + 1 + 1 + 1 \ge 4\sqrt[4]{x^4 \cdot 1 \cdot 1 \cdot 1} = 4|x|. \quad (2)
From (1) it follows that P(x)=4x4+4x3+1x4+4(x3+x3)0P(x) = 4x^4 + 4x^3 + 1 \ge x^4 + 4(|x|^3 + x^3) \ge 0, the last inequality turns to the equality only for x=0x = 0, but P(0)=1P(0) = 1, so the polynomial P(x)P(x) has no real roots.
Similarly, from (2) it follows that Q(x)=x4+4x+44(x+x)+1>0Q(x) = x^4 + 4x + 4 \ge 4(|x|+x) + 1 > 0, whence the polynomial Q(x)Q(x) has no real roots.
By definition,
PQ(x)=4x4+4x+1иQP(x)=x4+4x3+4 P_Q(x) = 4x^4 + 4x + 1 \quad \text{и} \quad Q_P(x) = x^4 + 4x^3 + 4
It is easy to verify that PQ(0.5)=0.75<0P_Q(-0.5) = -0.75 < 0 and PQ(0)=1>0P_Q(0) = 1 > 0. Also, QP(2)=12<0Q_P(-2) = -12 < 0 and QP(0)=4>0Q_P(0) = 4 > 0. Therefore each of the polynomials PQ(x)P_Q(x) and QP(x)Q_P(x) has at least one real root.

b)
It is evident that the polynomials P(x)P(x) and Q(x)Q(x) satisfying the condition of a) must have even degree (since any odd degree polynomial has at least one real root).
Show that the degree of these polynomials is greater than 2. Suppose, contrary to our claim, that there are the trinomials P(x)=a1x2+b1x+c1P(x) = a_1x^2 + b_1x + c_1 and Q(x)=a2x2+b2x+c2Q(x) = a_2x^2 + b_2x + c_2 having no real roots. Then at least one of the trinomials PQ(x)=a1x2+b1x+c1P_Q(x) = a_1x^2 + b_1x + c_1 and QP(x)=a2x2+b2x+c2Q_P(x) = a_2x^2 + b_2x + c_2 has no real roots. Indeed, without loss of generality we can assume that b2b1|b_2| \ge |b_1|. Since the discriminant of the trinomial Q(x)Q(x) is negative and b2b1|b_2| \ge |b_1|, we see that 0>b224a2c2b124a2c20 > b_2^2 - 4a_2c_2 \ge b_1^2 - 4a_2c_2. However, the last expression is a discriminant of the trinomial QP(x)Q_P(x), so this trinomial has no real roots.
The example of the polynomials from item a) shows the minimal degree of the polynomials is equal to 4.

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