Suppose a,binZ. If a∣b, then a^{2}∣b^{2}. If x is an odd integer, then x^{3} is odd.If a is an odd integer, then a^{2}+3a+5 is odd.

Hint: We just need to use definitions to solve each of them. If a∣b, then for some integer q, we have b=aq. Squaring both sides gives, ${\displaystyle b2=a^2{q}^2,}$ which be definition means ${\displaystyle a2\mid b2.}$
x is an odd integer means there is an integer nn such that ${\displaystyle x=2n+1.}$ Now, taking cube both sides gives ${\displaystyle x^3=2q+1,}$ where $q=4n^2+6n^2+3nq=4n+3n$ as shown below. This by definition means $x^3$ is odd.
$x^3=(2n+1{)}^3=(2n{)}^3+3(2n{)}^2(1)+3(2n)(1{)}^2+1^3$

$=8n^3+12n^2+6n+1=2(4n^3+6n^2+3n)+1$
Try using the definition as in the above question to prove this one.