Question

Sa se arate ca numarul n=1 la puterea 1997+2 la puterea 1997+...+1997 la puterea 1997 nu este patrat perfect

Answer 1

Daca ultima cifra a numarului n ∈{2,3,7,8} => n≠patrat perfect

u(n)=u(1^1997+2^1997+3^1997+....+1997^1997)

1997 da restul 1 la impartirea la 4

=> u(n)=u(1^1+2^1+3^1+....+1997^1)=u(1+2+3+......+1997)

u(n)=u(1997·1998/2)=u(1997·999)=3

n are ultima cifra 3 => n ≠ patrat perfect