(Romania Mathematical Olympiad). Let $a,b$ be positive integers such that exists a prime $p$ with the property $lcm(a,a+p)=lcm(b,b+p)$. Prove that $a=b$.
What I could do: WLOG $p|a, p \nmid b \implies a=pk, k \in \mathbb{Z}. lcm(pk,pk+p)=lcm(pk,p(k+1))=pk(k+1). lcm(b,b+p)=b(b+p)$
We want to prove that is impossible to $pk(k+1)=b(b+p)$, but I don't know how.
Answer after reading the comments:
WLOG $p \mid a, p \nmid b$.
This implies that $pk=a, k \in \mathbb{Z}$. Then, $lcm(pk, pk+p)=lcm(pk,p(k+1))=pk(k+1)$, and $lcm(b,b+p)=b(b+p)$
So it suffices to prove that it is impossible to $pk(k+1)=b(b+p)$, that is true because $p \mid pk(k+1)$ but not $b(b+p)$. Hence we are done.