amparvardi wrote: Does there exist a number $n=\overline{a_1a_2a_3a_4a_5a_6}$ such that $\overline{a_1a_2a_3}+4 = \overline{a_4a_5a_6}$ (all bases are $10$) and $n=a^k$ for some positive integers $a,k$ with $k \geq 3 \ ?$ $n\ge 4$ and so $a>1$ and so $k\le 19$ (since $2^{20}>10^6$) Let $\overline{a_4a_5a_6}=x$ : $n=1000(x-4)+x\equiv 4\pmod{1001}$ $a^k\equiv 4\pmod 7$. Which implies : either $k=7,13,19$ and $a\equiv 4\pmod 7$ either $k=8,14$ and $a\equiv 2\pmod 7$ or $a\equiv 5\pmod 7$ either $k=4,10,16$ and $a\equiv 3\pmod 7$ or $a\equiv 4\pmod 7$ either $k=5,11,17$ and $a\equiv 2\pmod 7$ $a^k\equiv 4\pmod {11}$. Which implies : either $k=11$ and $a\equiv 4\pmod{11}$ either $k=12$ and $a\equiv 2\pmod{11}$ or $a\equiv 9\pmod{11}$ either $k=3,13$ and $a\equiv 5\pmod{11}$ either $k=4,14$ and $a\equiv 3\pmod{11}$ or $a\equiv 8\pmod{11}$ either $k=6,16$ and $a\equiv 4\pmod{11}$ or $a\equiv 7\pmod{11}$ either $k=7,17$ and $a\equiv 9\pmod{11}$ either $k=8,18$ and $a\equiv 5\pmod{11}$ or $a\equiv 6\pmod{11}$ either $k=9,19$ and $a\equiv 3\pmod{11}$ $a^k\equiv 4\pmod {13}$. Which implies : either $k=13$ and $a\equiv 4\pmod{13}$ either $k=14$ and $a\equiv 2\pmod{13}$ or $a\equiv 11\pmod{13}$ either $k=5,17$ and $a\equiv 10\pmod{13}$ either $k=7,19$ and $a\equiv 4\pmod{13}$ either $k=10$ and $a\equiv 6\pmod{13}$ or $a\equiv 7\pmod{13}$ either $k=11$ and $a\equiv 10\pmod{13}$ From these three sets, we get that $k\in\{7,11,13,14,17,19\}$ $k=7$ implies $a\le 7$ and $a\equiv 9\pmod{11}$ and so no solution $k=11$ implies $a\le 3$ and $a\equiv 4\pmod{11}$ and so no solution $k=13$ implies $a=2$ and $a\equiv 4\pmod 7$ and so no solution $k=14$ implies $a=2$ and $a\equiv 3\pmod{11}$ or $a\equiv 8\pmod{11}$ and so no solution $k=17$ implies $a=2$ and $a\equiv 9\pmod{11}$ and so no solution $k=19$ implies $a=2$ and $a\equiv 4\pmod 7$ and so no solution Hence no such $n$

my solution is same with pco's solution! But, if $k\ge 2$? For example $n=856^2=732736$, find all for $k=2$.