Question

Sa se arate ca numarul n=2 la puterea 1999 - 2 la puterea 1998 - 2 la puterea 1997 - 2 la puterea 1996 este patrat perfect.

Answer 1

n=2^1999-2^1998-2^1997-2^1996

n=2^1996(2^3-2^2-2-1)

n=2^1996(8-4-2-1)

n=2^1996

n=(2^998)^2, deci n= patrat perfect

Answer 2

$2^{1999}-2^{1998}-2^{1997}-2^{1996}$ patrat perfect

$2^{1999}-2^{1998}-2^{1997}-2^{1996} =2^{1996}(2^{3}-2^{2} -2^{1}-1)=\\ \\2^{1996}(8-4-2-1)=2^{1996} =\\ \\(2^{2})^{998}= (2^{998})^{2}$