Problem 252

Factoring gives $112=2\times56=2\times2\times28=2\times2\times2\times14=2\times2\times2\times2\times7$.

So, the prime factorization of 112 is $2\times2\times2\times2\times7$ or $2^4\times7$.
For every perfect square, each of its prime factors occurs an even number of times (see the Note at the end of part (d) for a brief explanation of this).

From part (a), the prime factorization of 112 is $2^4\times7$.

We are asked for the smallest value of the positive integer $u$ so that $112\times u$ or $2^4\times7\times u$ is a perfect square.

The prime factor 2 already occurs an even number of times, (four times), in the factorization of 112.

Thus, no additional factors of 2 are needed to make $2^4\times7\times u$ a perfect square.

However, the prime factor 7 occurs only once.

Since all prime factors must occur an even number of times, then at least one additional factor of 7 is needed for $2^4\times7\times u$ to be a perfect square.

Therefore, the smallest positive integer $u$ that makes the product $112\times u$ a perfect square, is 7: $$112\times u=2^4\times7\times u=2^4\times7\times 7=2^4\times7^2=(2^2\times 7)\times (2^2\times 7).$$
Since $5632=512\times11$ and $512=2^9$, then the prime factorization of 5632 is $2^9\times11$.

Again, every perfect square has each of its prime factors occurring an even number of times.

We are asked for the smallest value of the positive integer $v$ so that $5632\times v$ or $2^9\times11\times v$ is a perfect square.

The prime factor 2 occurs an odd number of times, (nine times), in the factorization of 5632.

Thus, at least one additional factor of 2 is needed to make $2^9\times11\times v$ a perfect square.

The prime factor 11 occurs only once.

Thus, at least one additional factor of 11 is needed for $2^9\times11\times v$ to be a perfect square.

Therefore, the smallest positive integer $v$ that makes the product $5632\times v$ a perfect square, is $2\times11$ or 22: $$5632\times v=2^9\times11\times v=2^9\times11\times 2\times 11=2^{10}\times11^2=(2^5\times 11)\times(2^5\times 11).$$
For every perfect cube, the number of times that each of its prime factors occurs is a multiple of 3 (see the Note below for a brief explanation of this).

From part (a), the prime factorization of 112 is $2^4\times7$.

We are asked for the smallest value of the positive integer $w$ so that $112\times w$ or $2^4\times7\times w$ is a perfect cube.

The prime factor 2 occurs four times in the factorization of 112.

Thus, the smallest number of additional factors of 2 needed to make $2^4\times7\times w$ a perfect cube is two (since 6 is the smallest multiple of 3 that is greater than 4).

The prime factor 7 occurs only once.

Thus, the smallest number of additional factors of 7 needed to make $2^4\times7\times w$ a perfect cube is two (since 3 is the smallest multiple of 3 that is greater than 1).

Therefore, the smallest positive integer $w$ that makes the product $112\times w$ a perfect cube, is $2^2\times 7^2$ or 196: $$112\times w=2^4\times7\times w=2^4\times7\times 2^2\times 7^2=2^6\times7^3=(2^2\times 7)\times (2^2\times 7)\times (2^2\times 7).$$

Note: Every positive integer greater than 1 can be written as a unique product of prime numbers (this is known as the Fundamental Theorem of Arithmetic!).

Every perfect square, $P$, is the product of a positive integer, $n$, with itself.

That is, $P=n\times n$.

By the Fundamental Theorem of Arithmetic, $n$ can be written as a product of prime numbers.

Since $P=n\times n$, the prime factors of $P$ are matching pairs of prime factors of $n$.

Thus, the prime factors of every perfect square occur an even number of times.

This argument similarly extends to every perfect cube, $C$.

Since $C=n\times n\times n$ for some positive integer $n$, then the prime factors of $C$ occur in sets of three matching prime factors of $n$.

Thus for every perfect cube, the number of times that each of the prime factors occurs is a multiple of 3.