Questions involving polynomial functions

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Questions that involve polynomial functions are slightly different from the usual type of functional equation problems. The biggest difference is that the functions must be in the form of polynomials i.e. {a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0}. The usual strategies in solving functional equations can still be useful in this type of problems. There are also other properties that one might one to take note of in solving this type of problems.

There are several useful angles which can help us deal with problems which involve polynomial functions. Most of the time, the problem will require us to write the function in the form of {a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0} and investigate its coefficients, possibly using Vieta’s theorem. One can also write the polynomial in the form of {A(x-x_1)(x-x_2)\cdots(x-x_n)} where {x_1, x_2, \cdots x_n} are the roots of the polynomial function so that we can analyse the properties of the roots. Often, one have to consider the bounds of the function on specific domains such as the domains between the roots of the function. Calculus may be useful too in determining the turning points of the function.

The concepts of Lagrange’s interpolation formula and difference sequences may be very useful in solving this type of problems too. Some problems are tailored just to test students on these properties. Otherwise, one will require lots of practice to get a hang on suitable assumptions and substitution which can be helpful in solving these problems. Personally I consider this one of the hardest field in algebra because of the lack of systematic approach in dealing with this sort of problems.

Consider the following problem which I have modified to standardise notations:

(Vietnam MO 1991) Let {f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0} be a polynomial with real coefficients such that {a_n\neq 0} and {f(x)f(2x^2)=f(2x^3+x)} for all {x}. Prove that {f(x)} has no real roots.

How can we prove that something has no real roots? One possible method is to treat this problem as a functional equation problem and attempt to see if we can derive {f(x)}. Readers can attempt to do so and may realise that {f(x)=x^2+1} is one of the solutions, but that is not sufficient in proving that all other possible polynomials which satisfy the above functional equation will have no real roots.

Perhaps we can arrive at some form of contradiction if we assume the contrary. Several forms of contradiction may arise in facing a problem with polynomial functions. One particular contradiction which is pretty common is when the number of roots does not tally with the degree of the polynomial. We all know that a polynomial with degree {n} will have at most {n} roots (prove this). If the number of roots exceed {n}, then a contradiction will arise. Now, we must either find more than {n} expressions which satisfy as the roots of the polynomial, or try to construct a recursive sequence which satisfy as the roots of the polynomial.

But before that, we still have to figure out how to make use of the functional equation given in the problem. Perhaps we can expand the left hand side of the functional equation and compare the coefficients with the right hand side of the equation. After all, it seems to be the most natural thing to do. Upon expansion, we have:

\displaystyle f(x)f(2x^2)=2^na^2_nx^{3n}+\cdots+a_0^2

\displaystyle f(2x^3+x)=2^na_nx^{3n}+\cdots+a_0

Upon inspection of the coefficient of the leading term as well as the constant, we do realise some interesting trends. It seems like {a_n=1} while {a_0} is either 1 or 0. We are more interested in whether the constant term {a_0} can be equal to zero. If that is the case, we consider the term with the lowest degree for the left hand side and the right hand side of the equation. We observe that the term with the lowest degree for the {f(x)f(2x^2)} will be three times of that of {f(2x^3+x)}. The terms in the polynomial function does not tally and this implies that {a_0=1}. The fact that the constant term is nonzero also implies that 0 is not a root of the polynomial. This fact can be useful when we construct the roots to the polynomial.

We suppose that a real root {\alpha} to the polynomial function exist. Substituting {\alpha} into the functional equation, we obtain:

\displaystyle f(2\alpha^3+\alpha)=0

which implies that {2\alpha^3+\alpha} is also another root. Based on the recursive sequence {\alpha_{n+1}=2\alpha_n^3+\alpha}, we can construct a sequence of roots to the function. The recursive sequence does have a fixed point at {\alpha=0}. However, since zero cannot be a root of the equation, the recursive sequence is either strictly increasing or strictly decreasing, which implies that we can construct infinitely many roots. This cannot be true for polynomials and a contradiction arises, which suggests that there cannot be any real roots to the equation!

In fact, it can be shown that the only solutions to {f(x)} is {f(x)=0, f(x)=1} and {f(x)=(x^2+1)^n}. Interested readers may refer to problem 3 of IMO Shortlist 1979.

Defining new functions can be very helpful in solving this genre of problems, either to change the degree of the polynomial, the nature of roots of the polynomial, etc. Common forms of definition include {h(x)=g(x)f(x), g(x)=f(x+1)-f(x)}, etc. We will see how this technique help us in solving one of the problems from SMO last year:

(SMO(O)2011 P4) Find all polynomials {P(x)} with real coefficients such that {P(a) \in \mathbb{Z}} implies that {a\in \mathbb{Z}}.

Let us guess some of the solutions to this problem. For a constant function, polynomials in the form of {P(x)=c} where {c \notin \mathbb{Z}} will satisfy the condition. Then, we look for linear functions which satisfy the condition. The linear function will be in the form of {P(x)=sx+t}. Obviously {s, t} cannot be irrational numbers. Suppose {t=\dfrac{p}{q}} for some positive numbers {p} and {q} where {(p,q)=1}. For an arbitrary integer {k}, we know that {\dfrac{k-\frac{p}{q}}{s}} must be an integer. This suggests that {s=\dfrac{1}{mq}} for another arbitrary constan {m}t. If we remove the condition that {(p,q)=}, this suggests that linear functions in the form of {P(x)=\dfrac{x}{q}+\dfrac{p}{q}} where {p, q \in \mathbb{Z}} and {q\neq 0} will satisfy the condition.

We can try and guess if any functions of higher degree can satisfy the condition, but these effort will lead to no avail. This leads us to guess that there aren’t any functions which can satisfy the condition if the degree of the function is larger than 1. The condition in the problem reminds us of the squeeze principle which we discussed in solving diophantine equations. It kinda makes sense to apply the squeeze principle in some way, since we are dealing with integers too. In applying squeeze principle, we will need to consider the values between {P(x) and P(x+1)}. Now we draw some mental pictures based on the condition given. If the difference between the values of {P(x)} and {P(x+1)} is too big, then there must be a lot of integers in between {P(x)} and {P(x+1)} that the function can satisfy. This cannot be the case because the maximum possible integers between {x} and {x+1} is two and there will be too many integers squeezed within this range if {|P(x)-P(x+1)|>2}. Hence, we must have {|P(x)-P(x+1)|\le 2}.

Now we can define {G(x)=P(x)-P(x+1)}. The degree of {g(x)} is one less than that of {P(x)}. Using the fact that {|G(x)|\le2}, we know that {G(x)} must be a constant function and {P(x)} must either be a linear function or a constant function. As such, the only possible solutions for {P(x)} are the two functions which we have found above.

Here are some practice problems that readers can try:

  1. (Vietnam MO 1992) Let {P(x)=1+x^2+x^9+x^{n_1}+\cdots +x^{n_s}+x^{1992}}, where {n_1, n_2, \cdots n_s} are integers and {9<n_1<n_2<\cdots<n_s<1992}. Prove that a real root (if there is any) of {P(x)} is at most {\dfrac{1-\sqrt{5}}{2}}.
  2. (IMO Shortlist 1997) Let {P(x)} be a real polynomial function, {P(x)=ax^3+bx^2+cx+d}. Prove that if {|P(x)|\le 1} for all {x} such that {|x|\le 1}, then {|a|+|b|+|c|+|d|\le 7}.

Cheers,

ksj

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