a.
\(20\equiv-1\left(mod7\right)\Rightarrow20^{2021}\equiv\left(-1\right)^{2021}\left(mod7\right)\Rightarrow20^{2021}\equiv-1\left(mod7\right)\)
\(22\equiv1\left(mod7\right)\Rightarrow22^{2022}\equiv1\left(mod7\right)\)
\(2021\equiv5\left(mod7\right)\Rightarrow2021^{2023}\equiv5^{2023}\left(mod7\right)\)
Lại có: \(5^{2023}=5.5^{2022}=5.\left(5^6\right)^{337}\)
Mà theo định lý Fermat nhỏ: \(5^6\equiv1\left(mod7\right)\Rightarrow\left(5^6\right)^{337}\equiv1\left(mod7\right)\)
\(\Rightarrow5^{2023}\equiv5\left(mod7\right)\)
Vậy A chia 7 dư: \(1+\left(-1\right)+5=5\)
b.
\(a^3+3a^2=5^b-5\)
\(a+3=5^c\Rightarrow3a+1=3.5^c+10\)
\(\Rightarrow a^3+3a^2+3a+1=5^b-5+3.5^c+10\)
\(\Rightarrow\left(a+1\right)^3=5^b+3.5^c+5\)
Mà \(5^b+3.5^c+5⋮5\Rightarrow\left(a+1\right)^3⋮5\Rightarrow a+1⋮5\Rightarrow a=5k-1\)
\(\Rightarrow a+3=5k+2⋮̸5\) (vô lý do giả thiết \(a+3=5^c⋮5\))
Vậy ko tồn tại các số nguyên dương a;b;c thỏa mãn yêu cầu
