Maths Olympiad Prep

Track / Stage 6 / 349 of 400 #1989 of 2604

Problem 1989

National Olympiad, first round
Algebra Difficulty 6.8 Find the answer Asia Pacific Math Olympiad

Find all polynomials P(x)P(x) with integer coefficients such that for all real numbers ss and tt, if P(s)P(s) and P(t)P(t) are both integers, then P(st)P(st) is also an integer.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Next problem →

Official solution

To find all polynomials P(x) P(x) with integer coefficients that satisfy the given condition, we analyze the condition: if P(s) P(s) and P(t) P(t) are integers for real numbers s s and t t , then P(st) P(st) must also be an integer.

### Step 1: Analyze the Degree of Polynomial

Assume P(x)=adxd+ad−1xd−1+⋯+a1x+a0 P(x) = a_d x^d + a_{d-1} x^{d-1} + \cdots + a_1 x + a_0 where ai a_i are integer coefficients.

The condition implies that for any real numbers s s and t t , if P(s) P(s) and P(t) P(t) are integers, then P(st) P(st) is also an integer. Consider the simplest cases:

- Constant Polynomial: If P(x)=c P(x) = c (a constant polynomial), then clearly P(s)=P(t)=P(st)=c P(s) = P(t) = P(st) = c , which is an integer. Thus, constant polynomials satisfy the condition.
- Linear Polynomial: Consider P(x)=ax+b P(x) = ax + b .
- If P(s)=as+b P(s) = as + b and P(t)=at+b P(t) = at + b are integers, P(st)=ast+b P(st) = ast + b must also be an integer. This imposes no new constraints as a,b a, b are integers.

### Step 2: Consider Higher Degree Polynomials

- If P(x)=axd+⋯+c P(x) = a x^d + \cdots + c with d≥1 d \geq 1 , analyze whether such a polynomial can satisfy the condition:
- Let P(s)=asd+…+c P(s) = a s^d + \ldots + c and P(t)=atd+…+c P(t) = a t^d + \ldots + c .
- The multiplication condition P(st) P(st) being an integer suggests that formulating such a polynomial while maintaining integer values involves specific form.

A key insight here is that the presence of cross-terms in the polynomial at higher degrees might violate integer preservation without specific structures.

### Step 3: Structure Imposition

If P(x)=xd+c P(x) = x^d + c or P(x)=−xd+c P(x) = -x^d + c , then:
- P(s)=sd+c P(s) = s^d + c and P(t)=td+c P(t) = t^d + c are integers assuming they yield integers separately.
- Consequently, if both P(s) P(s) and P(t) P(t) are integers, then:
P(st)=(st)d+c=sdtd+c P(st) = (st)^d + c = s^d t^d + c
remains an integer because sd s^d and td t^d are integers.

This structure ensures that P(x)=±xd+c P(x) = \pm x^d + c , thereby fulfilling the requirements.

### Conclusion

Thus, the form of the polynomial that satisfies the condition is:
P(x)=±xd+c P(x) = \pm x^d + c
where c c is an integer and d d is a positive integer.

Hence, the final answer is:
P(x)=±xd+c, where c is an integer and d is a positive integer. \boxed{P(x)=\pm x^d+c \text{, where } c \text { is an integer and } d \text{ is a positive integer.}}

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.