Question

Use the quotient-remainder theorem with divisor equal to 2 to prove the following statement.

Answer 1

Answer:

Please read the enswer below

Step-by-step explanation:

The quotient-remainder theorem establishes that given an integer n, there are unique integers d and r, with 0≤r<d, and such that:

$n=qd+r$

q: the quotient

d: the remainder

d: divisor = 2

By the quotient-remainder theorem with divisor 2, you have:

n = 2q

n = 2q + 1

Then, for both cases you have:

$n^2=(2q)^2\\\\n^2=4q^2\\\\n^2=4k$   (the square of an integer with divisor 2 is 4k)

with $q^2=k$

$n=2q+1\\\\n^2=(2q+1)^2=4q^2+2q+1$

$n^2=2q(2q+1)+1$

but 2q + 1 = n

$n^2=2qn+1\\\\n^2=4k+1$

where you have taken $2qn=4k$   (the product 2qn is another integer)