Show that every $n$ can be written uniquely in the form $n = ab$, with $a$ square-free and $b$ a perfect square

Question

I need to show that every positive integer $n$ can be written uniquely in the form $n = ab$, where $a$ is square-free and $b$ is a square. Then I need to show that $b$ is then the largest square dividing $n$.

The problem I have here is that I can't even see how this is true. How can $1$ be represented this way? Is $1$ a square number? If not, then I cannot see how this is possible. If it is, then I cannot see how $1$, $2$, or $3$ can be represented this way.

Please, any help you can offer giving me a push in the right direction here is greatly appreciated.

Answer 1

$1=1^2$, so $1$ is square. There are two natural ways to prove this result. One is using the prime factorization, i.e., the fundamental theorem of arithmetic. The other is by induction: if $n$ does not have a square factor, then $n$ is squarefree and $n$=$n\cdot 1$. Otherwise, divide $n$ by a square factor and use the induction hypothesis on the quotient.

Answer 2

As others have mentioned, the number 1 is both squarefree and a square. When $n=1$, we have $a=b=1$. When $n$ is a prime, we have $a=n$ and $b=1$. This is perfectly valid.

Let me expand on the first method mentioned by lhf. In general, let $$N=\prod_{\text{primes }p_i}p_i^{e_i}$$ be the prime decomposition of a number $N$. Note that $N$ is squarefree if and only if each $e_i$ is either 0 or 1, and $N$ is a square if and only if every $e_i$ is even. So if we want to write $n=ab$ where $a$ is squarefree and $b$ is a square, then if their prime factorizations are as follows, $$n=\prod_{\text{primes }p_i}p_i^{x_i}\hskip0.3in a=\prod_{\text{primes }p_i}p_i^{y_i}\hskip0.3in b=\prod_{\text{primes }p_i}p_i^{z_i}$$ then we want to solve $$n=\prod_{\text{primes }p_i}p_i^{x_i}=\prod_{\text{primes }p_i}p_i^{y_i+z_i}=ab$$ such that each $y_i$ is either 0 or 1, and each $z_i$ is even. Can you see why this uniquely specifies all the $y_i$'s and $z_i$'s, and hence uniquely specifies $a$ and $b$?

Answer 3

HINT $\ $ The problem is multiplicative so it suffices to show that it's true for a prime power $\rm\ P^N\:.\ $ But that's trivial: $\rm\ P^{2N} =\ (P^N)^2,\ \ P^{\:2N+1} =\ P\ (P^N)^2\:,\ $ uniquely.