fimo_2011_number_theory_p4.json

{
    "problem_name": "fimo_2011_number_theory_p4",
    "informal_statement": "For each positive integer $k$, let $t(k)$ be the largest odd divisor of $k$. Determine all positive integers $a$ for which there exists a positive integer $n$ such that all the differences\n\n$$\nt(n+a)-t(n), \\quad t(n+a+1)-t(n+1), \\quad \\ldots, \\quad t(n+2 a-1)-t(n+a-1)\n$$\n\nare divisible by 4.\n\nThe final answer is $a=1,3$, or 5 .",
    "informal_proof": "A pair $(a, n)$ satisfying the condition of the problem will be called a winning pair. It is straightforward to check that the pairs $(1,1),(3,1)$, and $(5,4)$ are winning pairs.\n\nNow suppose that $a$ is a positive integer not equal to 1,3 , and 5 . We will show that there are no winning pairs $(a, n)$ by distinguishing three cases.\n\nCase 1: $a$ is even. In this case we have $a=2^{\\alpha} d$ for some positive integer $\\alpha$ and some odd $d$. Since $a \\geq 2^{\\alpha}$, for each positive integer $n$ there exists an $i \\in\\{0,1, \\ldots, a-1\\}$ such that $n+i=2^{\\alpha-1} e$, where $e$ is some odd integer. Then we have $t(n+i)=t\\left(2^{\\alpha-1} e\\right)=e$ and\n\n$$\nt(n+a+i)=t\\left(2^{\\alpha} d+2^{\\alpha-1} e\\right)=2 d+e \\equiv e+2 \\quad(\\bmod 4)\n$$\n\nSo we get $t(n+i)-t(n+a+i) \\equiv 2(\\bmod 4)$, and $(a, n)$ is not a winning pair.\n\nCase 2: $a$ is odd and $a>8$. For each positive integer $n$, there exists an $i \\in\\{0,1, \\ldots, a-5\\}$ such that $n+i=2 d$ for some odd $d$. We get\n\n$$\nt(n+i)=d \\not \\equiv d+2=t(n+i+4) \\quad(\\bmod 4)\n$$\n\nand\n\n$$\nt(n+a+i)=n+a+i \\equiv n+a+i+4=t(n+a+i+4) \\quad(\\bmod 4) .\n$$\n\nTherefore, the integers $t(n+a+i)-t(n+i)$ and $t(n+a+i+4)-t(n+i+4)$ cannot be both divisible by 4 , and therefore there are no winning pairs in this case.\n\nCase 3: $a=7$. For each positive integer $n$, there exists an $i \\in\\{0,1, \\ldots, 6\\}$ such that $n+i$ is either of the form $8 k+3$ or of the form $8 k+6$, where $k$ is a nonnegative integer. But we have\n\n$$\nt(8 k+3) \\equiv 3 \\not \\equiv 1 \\equiv 4 k+5=t(8 k+3+7) \\quad(\\bmod 4)\n$$\n\nand\n\n$$\nt(8 k+6)=4 k+3 \\equiv 3 \\not \\equiv 1 \\equiv t(8 k+6+7) \\quad(\\bmod 4) .\n$$\n\nHence, there are no winning pairs of the form $(7, n)$."
}