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.

usagirl007A 2021-03-01 Answered

Suppose a,bZ. If a∣b, then a2b2.
If x is an odd integer, then x3 is odd.
If a is an odd integer, then a2+3a+5 is odd.

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AGRFTr
Answered 2021-03-02 Author has 95 answers

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=a2q2, which be definition means a2b2.
x is an odd integer means there is an integer nn such that x=2n+1. Now, taking cube both sides gives x3=2q+1, where q=4n2+6n2+3nq=4n+3n as shown below. This by definition means x3 is odd.
x3=(2n+1)3=(2n)3+3(2n)2(1)+3(2n)(1)2+13

=8n3+12n2+6n+1=2(4n3+6n2+3n)+1
Try using the definition as in the above question to prove this one.

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