67th Romanian Mathematical Olympiad

If $d$ is a common divisor of $x$ and $x+p$, then $d\mid (x+p)-x=p$, hence $d=1$ or $d=p$. This leads to the cases:

i. $p\nmid a$. Then $p\nmid a+p$, $p\nmid [a,a+p]$, hence $p\nmid [b,b+p]$, $p\nmid b$, $p\nmid b+p$, the numbers $a,a+p$ and $b,b+p$ are co-prime and $[a,a+p]=a(a+p)$, $[b,b+p]=b(b+p)$. It follows that $a(a+p)=b(b+p)$, therefore $a=b$: if, for instance, $a>b$, then $a(a+p)>b(b+p)$.

ii. $p\mid a$. Then $p\mid a+p$, $p\mid [a,a+p]$, hence $p\mid [b,b+p]$, $p\mid b$, $p\mid b+p$, $a=p{a}^{\prime }$ and $b=p{b}^{\prime }$, so $[a,a+p]=p{a}^{\prime }(a^{\prime }+1)$, $[b,b+p]=p{b}^{\prime }(b^{\prime }+1)$. This yields $a^{\prime }(a^{\prime }+1)=b^{\prime }(b^{\prime }+1)$, hence $a^{\prime }=b^{\prime }$, therefore $a=b$.