Question

cate numere prime p au propr. ca p+11 divide pe p(p+1)(p+2)?

Answer 1

Explicație pas cu pas:

p+11|p(p+1)(p+2)

p+11|(p²+p)(p+2)

p+11|p³+p²+2p²+2p =>

p+11|p³+3p²+2p

p+11|p+11=>p+11|p³+11p²

=>

p+11|p³+11p²-p³-3p²-2p =>

p+11|8p²-2p

p+11|p+11=>p+11|8p²+88p

=> p+11|8p²+88p-8p²+2p => p+11|90p

p+11|90p

p+11|p+11=>p+11|90p+990

=> p+11|90p+990-90p =>

p+11|990 => p+11€D990 inN (In N pt ca ne cere p-prim)

=> p+11€{11,15,18,22,30,33,45,55,66,90,99,110,165,198,330,495,990}

Nu am luat 1,2,.. pana la 11 pentru ca daca p este natural trebuie ca p+11>=11 pentru ca p>=0

=> p€{0,4,7,11,19,22,34,44,55,79,88,99,154,186,319,484,979}

cum p prim => p€{7;11,19,79}