# 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.

Suppose a,$b\in Z$. If a∣b, then ${a}^{2}\mid {b}^{2}.$
If x is an odd integer, then ${x}^{3}$ is odd.
If a is an odd integer, then ${a}^{2}+3a+5$ is odd.

You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

AGRFTr

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, $b2={a}^{2}{q}^{2},$ which be definition means $a2\mid b2.$
x is an odd integer means there is an integer nn such that $x=2n+1.$ Now, taking cube both sides gives ${x}^{3}=2q+1,$ where $q=4{n}^{2}+6{n}^{2}+3nq=4n+3n$ as shown below. This by definition means ${x}^{3}$ is odd.
${x}^{3}=\left(2n+1{\right)}^{3}=\left(2n{\right)}^{3}+3\left(2n{\right)}^{2}\left(1\right)+3\left(2n\right)\left(1{\right)}^{2}+{1}^{3}$

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