Given a positive integer $n$, we define $\lambda (n)$ as the number of positive integer solutions of $x^2-y^2=n$. We say that $n$ is olympic if $\lambda (n) = 2021$. Which is the smallest olympic positive integer? Which is the smallest olympic positive odd integer?
Problem
Source: Spain Mathematical Olympiad 2021 P2
Tags: Spain, number theory, number of divisors
11.05.2021 11:11
04.07.2021 14:29
FibonacciMoose wrote:
If $n$ is a square numbers then do your formala correct? If $n=2^2\cdot 3^2$ then $\lambda(n)=1$. I found formula of $\lambda(n)$: $n=2^\alpha\cdot n'$ with $\alpha\geqslant 1$ and $n'$ be a odd number $$\lambda(n)=\left\{\begin{array}{ll} \frac{(\alpha-1)\cdot\tau(n')}{2}&n\mbox{ is not square number}\\ \frac{(\alpha-1)\cdot\tau(n')-1}{2}&n\mbox{ is square number} \end{array}\right.$$else $n$ is a odd number then $$\lambda(n)=\left\{\begin{array}{ll} \frac{\tau(n)}{2}&n\mbox{ is not square number}\\ \frac{\tau(n)-1}{2}&n\mbox{ is square number} \end{array}\right.$$
05.07.2021 10:56
Yeah you are right, I did not exhaust the formula at all. Since the problem was only concerned with 2021, I did only count solutions for not square, odd numbers(Nevertherless I pointed out in the solution that the formula only works for these numbers). But thanks for pointing it out!
14.12.2021 20:01
Easy problem