Jalil_Huseynov wrote:

Find all pairs $(a,b)$ of positive integers such that $a^3$ is multiple of $b^2$ and $b-1$ is multiple of $a-1$ .

Another bold action sol, during the test, i was just spamming stuff as i had like 1 hr remaining and i had no idea what to do, so i did prime factorization for bounding stuff, which supricingly ended up by working very well.
Case 1: $b=1$
Then here $a$ can be literaly any integer so the pair $(a,b)=(a,1)$ works.
Case 2: $b>1$
We divide in 2 subcases.
Case 2.1: $a \ne b$ (i.e. we clearly have $b>a$ )
Let $b=\prod_{i=1}^{n} p_i^{a_i}$ and $N=\prod_{i=1}^{n} p_i^{\left\lceil \frac{2a_i}{3} \right\rceil}$ , now using $b^2 \mid a^3$ we get that $v_{p_i}(a) \ge \left\lceil \frac{2a_i}{3} \right\rceil$ which means that for any $1 \le i \le n$ we get that $p_i^{\left\lceil \frac{2a_i}{3} \right\rceil} \mid a$ and using those along with the fact that all $p_i$ 's are different we get that $N \mid a$ , now from here we get that $N^3 \ge b^2$ and $a \ge N$ , now since $\gcd(a-1,N)=\gcd(-1,N)=1$ (using euclidean alg) we use the second condition in this way, $a-1 \mid b-1$ means $a-1 \mid b-a$ which means that $a-1 \mid \frac{b}{N}-\frac{a}{N}$ so $a \left(\frac{N+1}{N} \right) \le \frac{b}{N}+1$ which means that $aN+a \le b+N \le b+a$ which gives $aN \le b$ but now this means that $N^4 \le (aN)^2 \le b^2 \le N^3$ so $N=1$ which means that $b=1$ which is a contradiction!.
Case 2.2: $a=b$
All pairs with $a=b$ work.
Hence all the pairs that work are $\boxed{(a,b)=(a,1), (a,a)}$ thus we are done