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PrintSilk Road Mathematics Competition
algebra
Problem
Determine all polynomials with real coefficients such that for any rational (number) the equation has a rational solution.
Solution
Answer: , where , are rational numbers.
Obviously, is not constant, so .
First of all we prove that the coefficients of are rational numbers. Let , consider distinct rational numbers and rationals such that . Then by the Lagrange's interpolation theorem we have and it easily follows all coefficients of the polynomial are rational numbers.
Multiplying by appropriate integer we can assume that has integer coefficients (and, of course, it satisfies the condition of the problem).
Let be the leading coefficient of , w.l.o.g. and let be the last coefficient of .
Consider numbers of the kind , where is a prime number. There exist an integer and a positive integer with such that . Then it is easy to see that and , hence, .
There are infinitely many different , so taking them into consideration we obtain infinitely many different corresponding pairs .
We state that among these pairs there are infinitely many , for which , since otherwise , but can take very large values; then at some moment the equation is impossible.
W.l.o.g. assume that there are infinitely many pairs with (if sign is negative solution is similar). Since the number of divisors of is finite, there are infinitely many prime numbers and some constant (the divisor of ) with .
Then consider another polynomial . So, there are infinitely many numbers for which and, therefore, . Finally, we have that is a linear polynomial.
Obviously, is not constant, so .
First of all we prove that the coefficients of are rational numbers. Let , consider distinct rational numbers and rationals such that . Then by the Lagrange's interpolation theorem we have and it easily follows all coefficients of the polynomial are rational numbers.
Multiplying by appropriate integer we can assume that has integer coefficients (and, of course, it satisfies the condition of the problem).
Let be the leading coefficient of , w.l.o.g. and let be the last coefficient of .
Consider numbers of the kind , where is a prime number. There exist an integer and a positive integer with such that . Then it is easy to see that and , hence, .
There are infinitely many different , so taking them into consideration we obtain infinitely many different corresponding pairs .
We state that among these pairs there are infinitely many , for which , since otherwise , but can take very large values; then at some moment the equation is impossible.
W.l.o.g. assume that there are infinitely many pairs with (if sign is negative solution is similar). Since the number of divisors of is finite, there are infinitely many prime numbers and some constant (the divisor of ) with .
Then consider another polynomial . So, there are infinitely many numbers for which and, therefore, . Finally, we have that is a linear polynomial.
Final answer
P(x) = a x + b, where a and b are rational and a ≠ 0
Techniques
Polynomial interpolation: Newton, LagrangeIrreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein