Question

Chứng minh các số sau không phải là số chính phương:

a) 2001²⁰⁰² + 23

b) 19²ⁿ + 5ⁿ + 2001

c) (16²)¹⁹⁹⁶ + (17²)¹⁹⁹⁶ - (13²)¹⁹⁹⁶ + 1

Answer 1

\(\left(3k+1\right)^2=9k^2+6k+1chia3du1\)

\(\left(3k+2\right)^2=9k^2+12k+4chia3du1\)

\(\Rightarrow\left\{{}\begin{matrix}\left(16^2\right)^{1996}\equiv1\left(mod3\right)\\\left(17^2\right)^{1996}\equiv1\left(mod3\right)\\\left(13^2\right)^{1996}\equiv1\left(mod3\right)\end{matrix}\right.\Rightarrow\left(16^2\right)^{1996}+\left(17^2\right)^{1996}-\left(13^2\right)^{1996}+1\equiv1+1-1+1\equiv2\left(mod3\right)\Rightarrow dpcm\)

Answer 2

Ta co:

\(2001⋮3\Rightarrow2001^{2002}⋮3\) mà \(23\) chia 3 dư 2

\(\Rightarrow2001^{2002}+23\) chia 3 dư 2 \(\Rightarrow dpcm\)

b,

\(+,n=0\Rightarrow19^{2n}+5^n+2001=1+1+2001=2003\left(notscp\right)\)

\(+,n>0\Rightarrow19^{2n}+5^n+2001=361^n+5^n+2001=\left(...1\right)+\left(....5\right)+2001=\left(...7\right)\Rightarrow klscp\)