fimo_2011_number_theory_p6.json

{
    "problem_name": "fimo_2011_number_theory_p6",
    "informal_statement": "Let $P(x)$ and $Q(x)$ be two polynomials with integer coefficients such that no nonconstant polynomial with rational coefficients divides both $P(x)$ and $Q(x)$. Suppose that for every positive integer $n$ the integers $P(n)$ and $Q(n)$ are positive, and $2^{Q(n)}-1$ divides $3^{P(n)}-1$. Prove that $Q(x)$ is a constant polynomial.",
    "informal_proof": "First we show that there exists an integer $d$ such that for all positive integers $n$ we have $\\operatorname{gcd}(P(n), Q(n)) \\leq d$\n\nSince $P(x)$ and $Q(x)$ are coprime (over the polynomials with rational coefficients), EucLid's algorithm provides some polynomials $R_{0}(x), S_{0}(x)$ with rational coefficients such that $P(x) R_{0}(x)-$ $Q(x) S_{0}(x)=1$. Multiplying by a suitable positive integer $d$, we obtain polynomials $R(x)=$ $d \\cdot R_{0}(x)$ and $S(x)=d \\cdot S_{0}(x)$ with integer coefficients for which $P(x) R(x)-Q(x) S(x)=d$. Then we have $\\operatorname{gcd}(P(n), Q(n)) \\leq d$ for any integer $n$.\n\nTo prove the problem statement, suppose that $Q(x)$ is not constant. Then the sequence $Q(n)$ is not bounded and we can choose a positive integer $m$ for which\n\n$$\nM=2^{Q(m)}-1 \\geq 3^{\\max \\{P(1), P(2), \\ldots, P(d)\\}} .\n$$\n\nSince $M=2^{Q(n)}-1 \\mid 3^{P(n)}-1$, we have $2,3 X M$. Let $a$ and $b$ be the multiplicative orders of 2 and 3 modulo $M$, respectively. Obviously, $a=Q(m)$ since the lower powers of 2 do not reach $M$. Since $M$ divides $3^{P(m)}-1$, we have $b \\mid P(m)$. Then $\\operatorname{gcd}(a, b) \\leq \\operatorname{gcd}(P(m), Q(m)) \\leq d$. Since the expression $a x-b y$ attains all integer values divisible by $\\operatorname{gcd}(a, b)$ when $x$ and $y$ run over all nonnegative integer values, there exist some nonnegative integers $x, y$ such that $1 \\leq m+a x-b y \\leq d$\n\nBy $Q(m+a x) \\equiv Q(m)(\\bmod a)$ we have\n\n$$\n2^{Q(m+a x)} \\equiv 2^{Q(m)} \\equiv 1 \\quad(\\bmod M)\n$$\n\nand therefore\n\n$$\nM\\left|2^{Q(m+a x)}-1\\right| 3^{P(m+a x)}-1 .\n$$\n\nThen, by $P(m+a x-b y) \\equiv P(m+a x)(\\bmod b)$ we have\n\n$$\n3^{P(m+a x-b y)} \\equiv 3^{P(m+a x)} \\equiv 1 \\quad(\\bmod M)\n$$\n\nSince $P(m+a x-b y)>0$ this implies $M \\leq 3^{P(m+a x-b y)}-1$. But $P(m+a x-b y)$ is listed among $P(1), P(2), \\ldots, P(d)$, so\n\n$$\nM<3^{P(m+a x-b y)} \\leq 3^{\\max \\{P(1), P(2), \\ldots, P(d)\\}}\n$$\n\nwhich contradicts (1)."
}