# Is $n$ a multiple of 5 if $n^2$ is?

Say $n$ is a natural number and $n^2$ is a multiple of 5. Does that mean $n$ is a multiple of 5?

This seems to be true. If we look at squares:

\begin{align*} 2^2 &= 4 & 3^2 &= 9 & 4^2 &= 16 & \mathbf{5^2} &\mathbf= \mathbf{25} \\ 6^2 &= 36 & 7^2 &= 49 & 8^2 &= 64 & 9^2 &= 81 \\ \mathbf{ {10}^2 }&\mathbf= \mathbf{100} & {11}^2 &= 121 & {12}^2 &= 144 & {13}^2 &= 169 \\ {14}^2 &= 196 & \mathbf{ {15}^2 }&\mathbf= \mathbf{225} & {16}^2 &= 256 & {17}^2 &= 289 \\ {18}^2 &= 324 & {19}^2 &= 361 & \mathbf{ {20}^2 }&\mathbf= \mathbf{400} & {21}^2 &= 441\\ \end{align*}

Seems to be that $n^2$ is a multiple of 5 only when $n$ is. Can we prove it?

If $n^2$ is a multiple of 5, then $n^2 = 5k$ for some $k \in \mathbb{N}$. Then $n = 5k/n = 5(k/n)$, which means $n$ is a multiple of 5 as long as $k/n$ is an integer… which is true if $k$ is a multiple of $n$. Is $k$ a multiple of $n$?

Well, $k = n^2/5 = n(n/5)$, which means $k$ is a multiple of $n$ if $n$ is a multiple of 5… ah, we’re in a loop here. Let’s try something else.

Maybe we need to use something specific about 5. Is this property true for all numbers? What about 4: is $n$ a multiple of 4 whenever $n^2$ is?

\begin{align*} \mathbf{2^2} &\mathbf= \mathbf{4} & 3^2 &= 9 & \mathbf{4^2} &\mathbf= \mathbf{16} & 5^2 &= 25 \\ \mathbf{6^2} &\mathbf= \mathbf{36} & 7^2 &= 49 & \mathbf{8^2} &\mathbf= \mathbf{64} & 9^2 &= 81 \\ \mathbf{ {10}^2} &\mathbf= \mathbf{100} & {11}^2 &= 121 & \mathbf{ {12}^2} &\mathbf= \mathbf{144} & {13}^2 &= 169 \\ \end{align*}

No! 4, 36, and 100 are multiples of 4, but 2, 6, and 10 are not multiples of 4.

There must be something different about 5 that makes this work for it. 5 is prime… can that help?

The fundamental theorem of arithmetic is often handy when dealing with primes. Let’s represent $n$ as its prime factorization:

$n = p_1p_2\cdots p_n$

Then:

$n^2 = p_1^2p_2^2\cdots p_n^2$

If $n^2$ is a multiple of 5, then 5 must be in $n^2$’s prime factorization. Which means 5 must be one of the primes $p_i$, which in turn means that $n$ itself must also have $5$ as a factor. Woohoo!

So we’ve proved this is true for 5. And, clearly, it’s true for any other prime number. We know it’s not true for 4. Is this true only for primes? Let’s look at 6:

\begin{align*} 2^2 &= 4 & 3^2 &= 9 & 4^2 &= 16 & {5^2} &= {25} \\ \mathbf{6^2} &\mathbf= \mathbf{36} & 7^2 &= 49 & 8^2 &= 64 & 9^2 &= 81 \\ {10}^2 &= {100} & {11}^2 &= 121 & \mathbf{12}^2 &\mathbf= \mathbf{144} & {13}^2 &= 169 \\ {14}^2 &= 196 & {15}^2 &= {225} & {16}^2 &= 256 & {17}^2 &= 289 \\ \mathbf{18}^2 &\mathbf= \mathbf{324} & {19}^2 &= 361 & { {20}^2 }&= {400} & {21}^2 &= 441\\ \end{align*}

Hm… seems like this actually might be true for 6, even though 6 isn’t prime. Can we prove it? It turns out we can make a pretty similar argument:

If $n^2 = p_1^2p_2^2\cdots p_n^2$ is a multiple of 6, 6 won’t be any of the $p_i$ (because 6 isn’t a prime). But 6 = 2*3, which means one of the $p_i$ must be 2 and another of the $p_i$ must be 3.

In case it isn’t clear why: if $m$ is a multiple of 6, then $m = 6k = 2\cdot3\cdot k$. We can prime-factorize $k = q_1q_2\cdots q_n$, which means $m = 2\cdot3\cdot q_1\cdot\cdots\cdot q_n$.

This means $n$’s prime factorization contains 2 and 3, which means $n$ is a multiple of 6. QED

So hang on… since every number has a prime factorization, can’t we make this argument for any number? What goes wrong with 4?

If $n^2 = p_1^2p_2^2\cdots p_n^2$ is a multiple of 4, since 4 = 2*2, one of the $p_i$ must be 2 and another of the $p_i$ must be… ah, 2 again. So we can’t actually prove that $n$ is a multiple of 4, but we can prove that it’s a multiple of 2.

4 happens to be a perfect square, but we’ll actually run into this problem for other numbers too. 12 = 2*2*3. If $n^2 = p_1^2p_2^2\cdots p_n^2$ is a multiple of 12, one of the $p_i$ is 2, one of the $p_i$ is 2 (again), and one of the $p_i$ is 3. So all we know for sure is that $n$ is a multiple of 2*3 = 6.

So it’s clear that this property holds for any $k$ that doesn’t have any repeats in its prime-factorization. These numbers are called square-free.

But even if our number isn’t square-free (like 4, or 12), we can still say something, which is that $n$ is a multiple of the number that you get when you get rid of all the duplicates in $k$’s prime factorization. The term for this is an integer’s radical:

\begin{align*} \operatorname{rad}({4}) &= \operatorname{rad}({2^2}) = 2 \\ \operatorname{rad}({5}) &= 5 \\ \operatorname{rad}({6}) &= \operatorname{rad}({2\cdot3}) = 2\cdot3 = 6 \\ \operatorname{rad}({12}) &= \operatorname{rad}({2^23}) = 2\cdot3 = 6 \\ \end{align*}

So the most general thing we can say is: if $n^2$ is a multiple of $k$, $n$ is a multiple of $\operatorname{rad}(k)$.

Inspired by Putnam 2017 A1.

- alec, May 23, 2020