Maths Olympiad Prep

Track / Stage 6 / 349 of 400 #1829 of 2444

Problem 1829

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+ad1xd1++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 d1 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.