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.

when the china grade 10 p7 is harder than an ISL N8

samrocksnature wrote: when the china grade 10 p7 is harder than an ISL N8 china stronk world weak

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 $k$.

I think this problem is the same as ELMO SL 2012 N6 up to minor variable changes.