a)Có 122n+1=12.144n≡12.11n(mod133)⇒11n+2+122n+1≡121.11n+12.11n=133.11n(mod133) chia hết cho 133 (đpcm)b) n lẻ ⇒n có dạng 2k+1
46n+296.13n=462k+1+296.132k+1=46.2116k+296.13.169k≡46.169k+3848.169k=3894.169k=1947.2.169k(mod1947)
chia hết cho 1947(đpcm)