Let $a,b$ be positive integers.Prove that there are no positive integers on the interval $\bigg[\frac{b^2}{a^2+ab},\frac{b^2}{a^2+ab-1}\bigg)$.
Problem
Source: 2022 China Southeast Grade 10 P7
Tags: number theory
03.08.2022 09:51
The problem is equivalent to showing there are no solutions to the Diophantine equation \[\left\lceil \frac{b^2}{k} \right\rceil = a(a+b).\]If $b$ is even, then let $b = 2c$ so that \[c^2 + \left\lceil \frac{4c^2}{k} \right\rceil = (a+c)^2;\]this is impossible by IMO Shortlist 2019 N8.
03.08.2022 10:17
when the china grade 10 p7 is harder than an ISL N8
03.08.2022 10:24
samrocksnature wrote: when the china grade 10 p7 is harder than an ISL N8 china stronk world weak
27.05.2023 20:15
Actually, the problem phrased like this is an easier version of ISL 2019 N8, the picture instructs you much more what to do. For completeness, let us give the ISL self-contained proof (without switching firstly to the original problem) that $\left \lceil \frac{b^2}{k} \right\rceil = a(a+b)$ has no solution. I will write instead the equation $\left \lceil \frac{a^2}{b} \right\rceil = k(a+k)$, sorry. Suppose otherwise and let $(a,b,k)$ be a solution with minimal $k$. We have that $\frac{a^2}{b}$ exceeds $k(a+k) - 1 > ak$, so $a > kb$; on the other hand, $k(a+k)$ is at least $\frac{a^2}{b} > \frac{a^2-k^2}{b}$, thus $a<k(b+1)$. Hence we may write $a = kb + r$ for some $0 < r < k$. Substituting leads to $\left \lceil \frac{a^2}{b} \right \rceil = k^2b + 2kr + \left \lceil \frac{r^2}{b} \right \rceil$ and $k(a+k) = k^2b + 2kr + k(k-r)$, hence $\left \lceil \frac{r^2}{b} \right \rceil = k_0(k_0 + r)$, where $k_0 = k-r$. In the latter we have $r>0$ and $0 < k_0 < k$, contradicting the minimality of $k$.
28.05.2023 06:40
I think this problem is the same as ELMO SL 2012 N6 up to minor variable changes.