Vì a chia cho 5 dư 4
\(\Rightarrow a=-1\left(mod5\right)\)
\(\Rightarrow a^2=1\left(mod5\right)\)
Vậy \(a^2\)chia cho 5 dư 1( đpcm)
Ta có: \(a\equiv\left(-1\right)\left(mod5\right)\)
\(\Rightarrow a^2\equiv\left(-1\right)^2\left(mod5\right)\)
\(\Rightarrow a^2\equiv1\left(mod5\right)\)
\(\Rightarrow\)\(a^5\div5\)dư 1 \(\left(đpcm\right)\)
a chia 5 dư 4
=> a = 5k + 4 ( k thuộc N )
=> a2 = ( 5k +4)2 = 25k2 + 40k + 16
= 25k2 + 40k + 15 + 1
= 5( 5k2 + 8k + 3 ) + 1
=> a2 chia 5 dư 1
Ta có:
\(a\) :5 dư 4
\(\Rightarrow a=5k+4\left(k\in Z\right)\)
\(\Rightarrow a^2=\left(5k+4\right)^2\)
\(=25k^2+40k+16\)
Mà \(25k^2⋮5;40k⋮5;16:5\left(mod1\right)\)
=> \(a^2:5\) dư 1 (đpcm)
C2:
Vì a chia 5 dư 4 nên a có dạng \(5k+4\left(k\in N\right)\)
\(\Rightarrow a^2=\left(5k+4\right)^2=25k^2+40k+16=5\left(5k^2+8k+3\right)+1\)chia cho 5 dư1
Vậy \(a^2\)chia cho 5 dư 1
