Talk:2012 USAMO Problems/Problem 3

The answer is the set of all integers that are at least $3$ .

For composite $n$ where there are two primes $p_1$ and $p_2$ such that $\frac{n}{2}<p_1<p_2<n$ , here's your construction:

Pick maximal integers $j_1$ and $j_2$ such that $p_1^{j_1}p_2^{j_2}$ divides $i$ .

Pick a minimal positive integer s such that $\frac{n(n+1)}{2}+(s-1)p_1 \equiv 0$ (mod $p_2$ ). (You know it exists since $p_1$ and $p_2$ are relatively prime.)

Pick an integer t such that $\frac{n(n+1)}{2}+(s-1)p_1+(t-1)p_2=0$ . (It exists because of how we defined s. It also must be negative.)

Then $a_i=(s^{j_1})(t^{j_2})$ .

For n=4:

$a_i=(-1)^{j_1+j_2}$ , where $2^{j_1}3^{j_2}$ divides i.

For n=6:

$a_i=2^{j_1}(-5)^{j_2}$ , where $3^{j_1}5^{j_2}$ divides i.

For n=10:

$a_i=2^{j_1}(-9)^{j_2}$ , where $5^{j_1}7^{j_2}$ divides i.

[I don't know LaTeX, so someone else can input it.]

--Mage24365 09:00, 25 April 2012 (EDT)

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Talk:2012 USAMO Problems/Problem 3