n chia 7 dư 4 thì \(n=7k+4\left(k\in Z\right)\)
Ta có:
\(n^2=\left(7k+4\right)^2=49k^2+56k+16=\left(49k^2+56k+14\right)+2=7\left(7k^2+8k+2\right)+2\)
Do đó \(n^2\)chia 7 dư 2
\(n^3=\left(7k+4\right)^3=343k^3+588k^2+336k+64=\left(343k^2+588k^2+336k+63\right)+1\)
\(=7\left(49k^3+84k^2+48k+9\right)+1\)
Do đó \(n^3\)chia 7 dư 1
Vậy \(n^2\)chia 7 dư 2 và \(n^3\)chia 7 dư 1