Formulating a Square

Algebra Level pending

Let

$N={ 2 }^{ k }\times{ 4 }^{ m }\times { 8 }^{ n }$

where $k$ , $m$ , $n$ are positive whole numbers.

When will N definitely be a square number?

k is odd but m + n is even k + n is odd k + n is even k is even

1 solution

Harri Bell-Thomas
Oct 11, 2014

Rewrite N as;

$N={ 2 }^{ k + 2m +3n}$

Note that ${2}^{x}$ is a square number when $x$ is a positive, even integer.

For example,

${2}^{4} = {4}^{2} = 16$

${2}^{6} = {8}^{2}=64$

${2}^{8} = {16}^{2}=256$

Therefore, if $k+2m+3n$ is even, N will be a square number.

Thus, the answer is k+n is even .