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Showing posts from November, 2024

RMO 2024: Discussing Solutions

Hello everyone!  Congratulations to everyone who attempted the RMO 2024. As you might know, we had an amazing livesolve of the paper with Archit, Adhitya, Abel and Kanav which you can check out  here . We also have question wise video solutions to all the problems, thanks to Nanda, Om and Shreya!  We had a lot of people interested in solutions for the KV/JNV paper, which is what this blog post will be about. Without further ado, let's get started! Problem 1:  Find all positive integers $x,y$ such that $202x+4x^2=y^2$. Solution:  Notice that $y>2x$. Let $y=2x+k$ for some integer $k>0$. Thus, the given equation reduces to $$202x=4xk+k^2\implies x=\frac{k^2}{202-4k}\cdots (1)$$ This tells us that $202-4k|k^2,$ or that $101-2k|2k^2\implies 101-2k|101k$. However, since 101 is a prime, $\gcd(101-2k,\,101)=1\implies 101-2k|k$ or that $101-2k|2k\implies 101-2k|101\implies k=50$. Substituting in $(1)$, we get that $x$ must be $$\frac{50^2}{202-4(50)}=50\cdot 25=12...