Question

Aratati ca:
$a) \: {2007}^{5} - 4 \times 2007 \: nu \: e \: prim \\ b) \: {2}^{31} \times {5}^{30} + 1 \: e \: compus \\ c) \: {2}^{30} \times {5}^{31} + 1 \: e \: compus$
Va rog!!! URGENT!!!!

Answer 1

a)2007⁵-4*2007=2007*(2007⁴-1) deci e compus fiind produsul a doua numere diferite de 1.
b)2³¹ * 5³⁰+1=2 * 2³⁰ * 5³⁰+1=2 * 10³⁰+1= 2 * (9+1)³⁰+1= M9+2+1=M9+3 ,care e divizibil cu 3,deci e compus
c)2³⁰ * 5³¹ +1 = 5 *2³⁰ * 5³⁰+1=5 * 10³⁰+1= 5 * (9+1)³⁰+1= M9+5+1=M9+6 ,care e divizibil cu 3 ,deci e compus.
Observatie: Am folosit identitatea: (a+b)ⁿ=Mₐ+bⁿ, n∈N