3^420-1=2*(3^420-1)/2= 2*(3^420-1)/(3-1)=2 (3^419+3^418+....+3+1)=
2(1+3+9+27+81+243+.....+3^417+3^418+3^419)=
2 (1+3+9+27(1+3+9)+..+3^417 (1+3+9))=
=2(3^0 *13+3³*13+....+3^417 *13)=2*13*(3^0+3³+..+3^417) , divizibil cu 13
Se observa ca 3|417, pt ca 4+1+7=12, deci 417 este de forma 3k
