Question

Aratati ca nr A= 3^2n+3•4^2n+3-2^2n+1•6^2n+3 este patrat perfect , unde n este nr natural.

Answer 1

A= 3^(2n+3)•4^(2n+3)-2^(2n+1)•6^(2n+3)

A=3^2n·3^3·2^4n·2^6 - 2^2n·2·2^2n·2^3·3^2n·3^3

A=3^(2n+3) ·2^(4n+6)- 3^(2n+3) ·2^(4n+4)

N=3^(2n+2) ·2^(4n+4)(3 ·4-3)

N=[3^(n+1) ·2^(2n+2)]^2 ·9

N=[3^(n+1) ·2^(2n+2) ·3]^2

sau N=[3^(n+2) ·2^(2n+2)]^2, deci e p.p.