Polynomials

We factor the polynomial as follows: \(x^{4}+x^{3}+2 x^{2}+x+1=\left(x^{2}+1\right)\left(x^{2}+x+1\right)\).

So the roots are \(i,-i,\omega\) and \(\omega^2\).

(b) Both \( \;p\; \) and \( \;q\; \) must be irrational.
Since the discriminant is positive, both the roots must be real. Since \(c\neq 0\), the roots must be non-zero. We know that \(p+q=-b\) and \(pq=c\). If one root is rational and the other is irrational, then both \(b\) and \(c\) must be irrational. If both the roots are rational, then both \(b\) and \(c\) must be rational. Hence, both \(p\) and \(q\) must be irrational.

Let the third root be \(s\). By Vieta's formulas we have
\begin{align} -r+r+s &= -a \\ -rs + rs -r^2 &= b \\ -r^2s &= -8 \end{align} which implies that \(b=-r^2\) must be negative and \(ab=8\).

So the possible pairs of values of \((a, b)\) are \(\{ (-1,-8), (-2,-4), (-4,-2), (-8,-1) \} \).

(b) We have: \[f(x)=\frac{1-x^{n+1}}{1-x} \mbox{ for } x\neq 1\] If \(a\) is a root of \(f\), then \(a^{n+1}=1\). Since \(f(1)=n\), 1 cannot be a root of \(f\). So \(a\neq 1\). The only other possibility is \(a=-1\) with \(n\) being odd.

Suppose \(f(x)=c_{n} x^{n}+c_{n-1} x^{n-1}+\cdots+c_{0}\) Since \( f(0) \) and \( f(1) \) are odd numbers, \(c_0\) is odd and \[ \text{Parity of } \sum_{i=1}^n c_i \text{ is even}\] Suppose \(f(x)\) has an integer root \(r\). Let us see what happens to the parity of \( f(r) \). For \(n\geq 1\), the parity of \(r^n\) is the same as that of \(r\). So: \begin{align} \text{Parity of }f(r) &= \text{Parity of }r \times \text{Parity of} \sum_{i=1}^n c_i + \text{Parity of }c_0 \\ \text{Parity of }f(r) &= \text{odd} \end{align} Since \( f(r) \) is an odd number, \(r\) cannot be a root of the polynomial.

Complex roots come in conjugates, so \(-i-3\) must also be a root. So \( (x-i+3)(x+i+3) = (x^2+6x+10) \) must be a factor of \(p(x)\): \[ p(x) = (x^2+6x+10)(ax^2+bx+c) \] By comparing the coefficients, \(a=3,b=-8,\text{and }c=-3\). Hence the polynomial is: \[ (x^2+6x+10)(3x^2-8x-3) \] \[ = (x^2+6x+10)(x-3)(3x+1) \] Hence, the other roots are 3 and -1/3.

Solution due to Piyush Jha.
Suppose that \(f\) has three real roots, \(p, q,\) and \(r\). From Vieta's relation: \[p^{2}+q^{2}+r^{2}=1-2 c\] From Cauchy-Swarchz inequality on \( (p, q, r) \) and \( (1,1,1) \), we get \begin{align} (p+q+r)^{2} &\leq\left(p^{2}+q^{2}+r^{2}\right)\left(1^{2}+1^{2}+1^{2}\right)\\ 1 \leq (1-2 c)(3)\\ \frac{1}{3} \geq c \end{align} But this contradicts the fact that \(c\) was strictly greater than \(\frac{1}{3}\).

Since the function has an odd degree as the highest power, it has at least one real root. The function has exactly one real root since it is monotonic in \(\mathbb{R}\), as proved below.

Lemma. \(f(x)\) is a monotonic function in \( \mathbb{R} \).
Proof. Consider the derivative of \(f\). \[ f'(x) = 3x^2+2x+c \] The discriminant of the above quadratic, \(2^2-4\cdot3\cdot c\), is negative when \(c>1/3\). Hence, \(f'(x)\) is always positive in \(\mathbb{R}\). To see why \(f(x)\) is a strictly increasing function, consider any two points \(a\) and \(b\) in \(\mathbb{R}\) with \(b > a\). By mean value theorem, there exists an \(\alpha \in (a,b)\) such that: \begin{align} f^{\prime}(\alpha) &= \frac{f(b) - f(a)}{b-a} \\ f(b) - f(a) &= f^{\prime}(\alpha) (b-a) > 0 \\\\ & \Rightarrow\; f(b) > f(a) \quad\quad\square \end{align}

All of them are true.

(i) Since the degree is even, \(p(x)\) goes to \(+\infty\) as \(x \rightarrow \pm \infty .\) So \(p(x)\) must have a global minimum somewhere.

(ii) The roots of \(p(x)\) are \(-a_{i}\)s. Since the coefficients of \(p(x)\) are positive, no nonnegative number is a root of \(p(x)\). Thus all the \(a_i\)s are positive.

(iii) and (iv) All 10 roots of \(p(x)\) are real and negative. There is a root of \(p^{\prime}(x)\) between consecutive roots of \(p(x)\) (this is valid even in case of multiple roots). So all 9 roots of \(p^{\prime}(x)\) are real and negative as well. For negativity, one can also note that all coefficients of \(p^{\prime}(x)\) are positive and apply the logic in (ii) to \(p^{\prime}(x)\).

(a) Non-real roots of a polynomial with real coefficients occur in conjugate pairs. \begin{align} \text{Let } p(x) &= (x-(\sqrt{2}+i))(x-(\sqrt{2}-i)) \\ &=x^{2}-2 \sqrt{2} x+3 \end{align} Since all the coefficients are real, the above quadratic is an example.

(b) The above polynomial \(p(x)\) has one irrational coefficient. We try squaring it as follows:

\begin{align} (x-(\sqrt{2}+i))(x-(\sqrt{2}-i)) & = 0 \\ x^{2}-2 \sqrt{2} x+3 & = 0 \\ x^{2}+3 & = 2 \sqrt{2} x \\ (x^{2}+3)^2 & = 8x^2 \end{align}

So \( q(x) = (x^2+3)^2 - 8x^2 \) is a polynomial with integer coefficients with \(\sqrt{2}+i\) as one of its roots. We found a polynomial with degree 4. We show that 4 is the least degree for which such a polynomial exists.

Lemma. There is no cubic polynomial \(q(x)\) with rational coefficients with \( q(\sqrt{2} + i) = 0 \).

Proof. Let \( \sqrt{2}+i \), \(\sqrt{2}-i\) and \(r\) be the roots of such a cubic, if it exists. We have: \begin{align} q(x) & = (x^{2}-2 \sqrt{2} x+3)(x-r) \\ & = x^{3}-(2 \sqrt{2}-r)x^2 + 2\sqrt{2}r x - 3r \end{align} If \(q\) has only rational coefficients, both \(2\sqrt{2}-r\) and \( 3r \) must be rational (Contradiction). \(\;\;\square\)

(c) Let \(f(x)\) be a polynomial with rational coefficients such that \(f(\sqrt{2}+i)=0\). Divide \(f(x)\) by \(q(x)\) using long division to get quotient \(a(x)\) and remainder \(b(x),\) both polynomials with rational coefficients. \[ f(x)=q(x)a(x)+b(x) \] Using \(f(\sqrt{2}+i)=0\) and \(q(\sqrt{2}+i)=0\) in the equation gives \(b(\sqrt{2}+i)=0\). Now if the remainder \(b(x)\) is a nonzero polynomial, then it would have rational coefficients, degree less than 4 and \(\sqrt{2}+i\) as a root. But we have proved in part (b) that such a polynomial does not exist. Hence, \(b(x)=0\) which implies that \(f(x)\) is a multiple of \(q(x)\).

Notation. Let \(f(x):=x^2-4\). Let \(\alpha_1\) and \(\alpha_2\) be the roots of the equation \(f(x)=x\). The values of the roots are: \[ \alpha_1 = \left( \frac{1+\sqrt{17}}{2} \right) \;\;\;\alpha_2 = \left( \frac{1-\sqrt{17}}{2} \right) \] Suppose \(p(x)\) is a quadratic polynomial that has the property \(\tau\). Let \(r_1\) and \(r_2\) be the roots of \(p(x)\). The constraint on the problem means that \(f(r_1) = r_1\mbox{ or } r_2\) and \(f(r_2) = r_1\mbox{ or }r_2\). We can find the candidate polynomials by enumerating all four cases. The polynomials obtained are numbered from (i) to (vi) in the table below.

The procedure is similar to the one followed in part (i). Let the function \(f\) and \(\alpha\)s be as defined in the previous problem. Let \(g(x)\) be a cubic polynomial with \(r_1,r_2\) and \(r_3\) as its roots.

The constraint on the problem means that:
\begin{align} f(r_1) &= r_1 \mbox{ or } r_2 \mbox{ or } r_3 \\ f(r_2) &= r_1 \mbox{ or } r_2 \mbox{ or } r_3 \\ f(r_3) &= r_1 \mbox{ or } r_2 \mbox{ or } r_3 \end{align}

We can find the candidate polynomials by enumerating all \(3^3=27\) cases.

Since zero cannot be a root of \(f(x)\), we focus only on the non-zero roots. Let us assume, for a contradiction, that \(f\) has a repeated root \(r\neq 0\).

Fact. If \(r\) is a repeated root both \(f(r)\) and \(f^{\prime}(r)\) must be zero.

On differentiating \(f(x)\), we get:
\[ f^{\prime}(x)=\frac{x^{n-1}}{(n-1) !}+\frac{x^{n-2}}{(n-2) !}+\cdots+x+1 \]

\[ f(x)-f^{\prime}(x)=\frac{x^{n}}{n!} \]

Therefore, \(r\) cannot be a repeated root since RHS \(\displaystyle = \frac{r^{n}}{n!} \neq 0 \).

(a) Write \(f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0} .\) Then \(a_{0}=f(0)=p^{k}\) for some prime \(p\) and integer \(k > 0\). Define \(g(x)=f(p x) .\) Then \(g(x)\) is a polynomial such that for each nonnegative integer \(n, g(n)=f(p n)=\) a perfect power of a prime number. This prime number has to be \(p,\) because by evaluating we see that \(g(n)=f(p n)\) is divisible by \(p\).

(b) Let \(g(x)=b_{n} x^{n}+b_{n-1} x^{n-1}+\cdots+b_{1} x+b_{0} .\) Then \(b_{0}=g(0)=p^{k} .\) Consider: \[ g\left(m p^{k+1}\right)=(b_{n}\left(m p^{k+1}\right)^{n}+b_{n-1}\left(m p^{k+1}\right)^{n-1}+\cdots+b_{1}\left(m p^{k+1}\right)+p^{k}\] Clearly for each non-negative integer \(m\), this expression is divisible by \(p^{k},\) but not by \(p^{k+1}\) (since it is \(p^{k}\) modulo \(p^{k+1}\) ). This forces \(g\left(m p^{k+1}\right)=p^{k}\) for all \(m,\) since it must be a perfect power of \(p .\) Thus the polynomial \(g\) takes the value \(p^{k}\) infinitely often, so it must be identically equal to \(p^{k}\). (Otherwise the polynomial \(g(x)-p^{k}\) would have infinitely many roots.) To finish the problem, note that since \(g(x)=f(p x)\) is constant, \(f(x)\) must be constant by the same logic.

False-True-False-True

(a) Counterexample: \( (x+1)^2 = x^2+2x+1 \)

(b) Complete the square to get \(f(x)=x^{2}+p x+q=\left(x+\frac{p}{2}\right)^{2}+\left(\frac{4 q-p^{2}}{4}\right)\). Since the roots are non-real the discriminant is negative, which implies \(4 q-p^{2} > 0\).

(c) Note that \(f(x) \rightarrow-\infty\) as \(x \rightarrow-\infty,\) so in particular \(f(x)\) takes negative values and hence can never be a sum of squares. This applies to any odd degree polynomial.

(d) Since all roots of \(f\) are non-real and occur in conjugate pairs, \(f(x)\) is a product of quadratic polynomials each of which is a sum of squares as shown in part (b).

Further reading. A generalization of part (d) is connected to Hilbert's 17th problem [pdf].

(a) Define \(f(0)=0, f(1)=1\) and \(f(-1)=-1\). For every other real number \(x,\) arbitrarily equate \(f(x)\) to 0,-1 or 1. Any such function satisfies the given condition.

(b) If \(f\) is a polynomial, then we make two cases.

(i) If \(f(x)=\) a constant \(c,\) then the given condition is equivalent to \(c=c^{2013},\) which happens precisely for three values of \(c,\) namely \(c=0,1,-1\) (since we have \(c\left(c^{2012}-1\right)=0,\) so \(c=0\) or \(c^{2012}=1\) ). Thus there are three constant functions with the given property.

(ii) If \(\mathrm{deg}(f)\geq 1\), then consider its range set \(A=\{f(x) \mid x \in \mathbb{R}\}\). Now for all \(a \in A,\) we have by the given property \(f(a)=a^{2013}\). So the polynomial \(f(x)-x^{2013}\) has all elements of \(A\) as its roots. Since there are infinitely many values in \(A\) (e.g. applying the intermediate value theorem because \(f\) is continuous), the polynomial \(f(x)-x^{2013}\) has infinitely many roots and thus must be the zero polynomial, i.e., \(f(x)=x^{2013}\) for all real number \(x\).

\(n=2018\). Note that \(\operatorname{deg} p_{0}(x)=2018\) and \(\operatorname{deg} p_{k}(x)=2018-k\). Define \(g_{n}(x)=p_{n}(x)-p_{n}(1)\), hence \(g_{n}(x)\) has degree \(2018-n\) and 5000 roots.

Let us define the binomial polynomial \( \binom{x}{k} \) as follows: \[ \binom{x}{k} := \frac{1}{k!} \times x\times (x-1)\times\cdots(x-(k-1)) \] for \( k>0 \) and 1 for \( k=0 \). Now consider the 7-degree polynomial \( g(x) \) defined as: \[ g(x) := \binom{x}{0} + \binom{x}{1} +\cdots + \binom{x}{7} \] The value of \( g(x) \) is same as \( p(x) \) for \(x=0\) to \(7\). Recall that if two \(n\)-degree polynomials agree on \(n+1\) points they must be identical. So, \( g(x) = p(x) \). \begin{align} p(10) & = g(10) = \binom{10}{0} + \binom{10}{1} +\cdots + \binom{10}{7} \\ & =\: 2^{10} - \left[ \binom{10}{8} +\binom{10}{9}+ \binom{10}{10} \right] \\ & =\: 2^{10} - \left( 45+10+1\right) \\ & = 968 \end{align}

Notes

(i) After expanding the powers of \((x-1),\) the degree \(n\) terms of \(f(x)\) and \(f(x-1)\) cancel out. The degree \(n-1\) term from \(f(x)\) cancels with the leading term of \(a_{n-1}(x-1)^{n-1}\). The only remaining term of degree \(n-1\) has to come from \(a_{n}(x-1)^{n}\). So \(\Delta f(x)=n a_{n} x^{n-1}+\mathrm{lower\; degree\; terms}\). This is similar to the derivative.

(ii) We use induction on the degree of \(f .\) If \(f(x)=a_{0}\) is constant, \(b_{0}=a_{0}\) works uniquely (base case).
Inductive hypothesis. Assume the result for polynomials of degree strictly less than \(n\) and let \(f\) be of degree \(n,\) so \(a_{n} \neq 0 .\) We are forced to take \(b_{n}=n ! a_{n}\) by comparing leading coefficients of \(f(x)\) and \(\sum_{i=0}^{n} b_{i} p_{i}(x) .\)
Now \(f(x)-b_{n} p_{n}(x)\) is a polynomial of degree \(d < n\) and hence by induction equals \(\sum_{i=0}^{d} b_{i} p_{i}(x)\) for unique \(b_{0}, \ldots, b_{d}\) Therefore \(f(x)=\sum_{i=0}^{n} b_{i} p_{i}(x),\) where \(b_{d+1}, \ldots, b_{n-1}\) are all \(0 .\) To see uniqueness of \(b_{i}^{\prime} s,\) let \(\sum_{i=0}^{n} b_{i} p_{i}(x)=\sum_{i=0}^{n} c_{i} p_{i}(x) .\) Subtract all terms with \(b_{i}=c_{i} .\) If any terms are remaining, compare the leading coefficients on each side to get a contradiction.

Alternate solution. If we are allowed to use linear algebra, it is a one-line proof. The \( p_i \)s form a basis for polynomial functions, so any polynomial has a unique representation.

(iii) Substitute \(x=0,1,2, \ldots\) one by one in the equation \(f(x)=\sum_{i=0}^{n} b_{i} p_{i}(x)\) and solve for \(b_{0}, b_{1}, b_{2}, \ldots\) successively. \(x=0\) gives \(b_{0}=f(0) .\) Using \(x=1\) and 2 gives \(b_{1}=f(1)-b_{0}\) \(b_{2}=f(2)-b_{0}-2 b_{1} .\) In general, for all integers \(t, p_{i}(t)=\left(\begin{array}{c}t \\ i\end{array}\right)\) is an integer. Further, \(p_{i}(t)=0\) if \(0 \leq t < i\) and 1 if \(t=i .\) So \(b_{t}=f(t)-\sum_{i=0}^{t-1} b_{i}\left(\begin{array}{c}t \\ i\end{array}\right),\) which is an integer by induction.

(i) Since \((x^2-1)\) is a factor of \(P(x)\), we must have \( P(1)=P(-1)=0\). We get two equations: \begin{align} P(1) = 10 + a + b + 3 + 15 = 0 \\ P(-1) = 10 - a + b - 3 + 15 = 0 \end{align} By solving the above equations we get \(a=-3\) and \(b=-25\).

(ii) By Vieta's formulas, only the ultimate and the penultimate coefficients matter. The sum turns out to be \(-1/5\).

(i) The function has only even powers. Let us put \(z=x^{2}\). We want to know the remainder of \(7z^{16}+5z^{11}+3z^{6}+z\) when divided by \((z+1)\). By remainder theorem, \[ 7(-1)^{16}+5(-1)^{11}+3(-1)^{6}+(-1)=4 \]

So we have \(f(x)=q(x)\left(x^{2}+1\right)+4\), where \(q(x)\) is some polynomial.

(ii) \(x f(x)=x q(x)\left(x^{2}+1\right)+4x\), so the second remainder is \(4x\).

True-True-False-False

(i) If \(g(x)\) is linear, it is uniquely determined by any two values.

(ii) If \(g(x)\) is a polynomial then it is of the form \(q(x)(x-1)(x-2) \cdots(x-15) + 3x+5\), where \(q(x)\) is some polynomial. So \(g(x)\) cannot be a polynomial of degree 10.

The official solution states this fact without a proof. Why should the polynomial have a \(3x+5\) term? It is not clear. Where is the consecutive property of integers getting used?

(iii) \(g(x)\) can have any degree more than 15. To get 20, pick \(q(x)\) to be \(x^5\) in the expression in (ii).

(iv) Pick \(q(x)=\sin x\), for a counterexample.

Conditions (i) and (ii) imply that for some polynomials \(q(x)\) and \( r(x) \), we should be able to express \(p(x)\) as:

\begin{align} p(x) &= q(x)x^{100} + 1 \\ p(x) &= r(x)(x-2)^{3} + 2 \end{align}

Key observation. If we did not have the plus one terms at the end, we could have simply multiplied the two RHS terms. Taking the derivative makes the 1s go away. So let us look at \( p^\prime{(x)} \).

\begin{align} p^\prime(x) &= 100q(x)x^{99} + q^\prime(x) x^{100} \\ &= x^{99}\times (100q(x) + q^\prime(x) x^{100}) \end{align}

So \(p\prime(x)\) has \(x^{99}\) and \( (x-2)^{2} \) as factors. We may assume

\[ p^\prime(x) = Ax^{99}(x-2)^2 \]

for some constant \(A\). The assumption is justified since any polynomial whose derivative is divisible by \(x^{99}(x-2)^{2}\) will leave constant remainders when divided by either of \(x^{100}\) and \((x-2)^{3}\).

So let us find \( p(x) \) by integrating \( p^\prime (x) \):

\[ p(x)=A\left(\frac{x^{102}}{102}-\frac{4 x^{101}}{101}+\frac{4 x^{100}}{100}\right)+B \]

We use properties (i) and (ii) to calculate \(A\) and \(B\).

\begin{align} p(0)&=B=1 \\ p(2)&=A\left(\frac{2^{102}}{102}-\frac{4 \times 2^{101}}{101}+\frac{4 \times 2^{100}}{100}\right)+1=2 \end{align}

So \(A=2\) and \(B=1\).