[tex]\overline{abc}+\overline{bca}+\overline{cab}=p^2\\ \\ daca \ si \ numai \ daca \ 100a+10b+c+100b+10c+a+100c+10a+b=p^2\\ \\ daca \ si \ numai \ daca \ 111a+111b+111c=p^2\\ \\ 111(a+b+c)=p^2 \ daca \ si \ numai \ daca \ a+b+c=111\cdot k^2, \ k \in \mathbb{N^*}\\ \\ Cum \ a, \ b \ si \ sunt \ cifre \Rightarrow maxim(a+b+c)=27\\ \\ \Rightarrow nu \ exista \ niciun \ numar \ de forma \ \overline{abc} \ astfel \ incat \ \overline{abc}+\overline{bca}+\overline{cab}=p^2[/tex]
