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:
[tex]n=qd+r[/tex]
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:
[tex]n^2=(2q)^2\\\\n^2=4q^2\\\\n^2=4k[/tex] (the square of an integer with divisor 2 is 4k)
with [tex]q^2=k[/tex]
[tex]n=2q+1\\\\n^2=(2q+1)^2=4q^2+2q+1[/tex]
[tex]n^2=2q(2q+1)+1[/tex]
but 2q + 1 = n
[tex]n^2=2qn+1\\\\n^2=4k+1[/tex]
where you have taken [tex]2qn=4k[/tex] (the product 2qn is another integer)