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Prove or Disprove: Divisibility and Integer Proofs

Question: Prove or disprove. Let $a,b,\text{ and }d\text{ be integers with }d\neq0.$ If $d\mid a$ and $d\mid b,$ then $d^2\mid ab.$

Proof

Assume $d\in\mathbb{Z}\setminus\{0\}.$

If $d\mid a$ and $d\mid b,$ then there exist $q,r\in\mathbb{Z}$ such that

$dq=a,\quad dr=b$

Then,

$ab=(dq)(dr)=d^2(qr)$

Since $qr\in\mathbb{Z},$

$d^2(qr)=ab\Longrightarrow d^2\mid ab$

Thus,

$d^2\mid ab$

■ (Q.E.D)

Section 2: Unique Factorization

Dudley, Underwood. Elementary Number Theory. 2nd ed. New York: W.H. Freeman and Company, 1989.

Prime Number
an integer that is greater than 1 and has no positive divisors other than 1 and itself

Composite Number
an integer that is greater than 1 but is not prime

Example 1. How many even primes are there? How many whose last digit is 5?

What is an even number greater than 1? An even number follows

$n=2k,\quad k\in\mathbb{N}$

What is an odd number greater than 1? An odd number follows

$n=2k+1,\quad k\in\mathbb{N}$

[Part 1] Solution — How many even primes are there?

Since a prime number is defined to be an integer greater than 1, the first prime number is 2, and then all even numbers greater than 2 are always divisible by 2, thus making 2 the only even prime number.

Answer: There is 1 even prime number.

[Part 2] Solution — How many whose last digit is 5?

The first prime number ending with 5 is 5. Then, we see that all numbers with the last digit of 5 are divisible by 5 so there is only one prime number with the last digit of 5.

Let

$n=10k+5$

be any number with the last digit of 5. Since

$5\mid(10k+5)$

iff there exists a $d\in\mathbb{N}$ such that

$5d=10k+5$

it is clear that every number ending in 5 is divisible by 5, and hence 5 is the only prime number with the last digit of 5.

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