# legal value for prime factorization

My understanding of mathematics is poor, although I'm trying to improve it, so I hope you will forgive me for asking a rather basic question.

I've written a computer code that prints the prime factors of a natural number. It works well. For instance, for the input $72$ it prints $2 \times 2 \times 2 \times 3 \times 3$. On youtube I saw somebody calculate the input $-72$ to $-1 \times 2 \times 2 \times 2 \times 3 \times 3$, so I adjusted my code to accept all whole numbers. But now I'm wondering what to do if the input is $0$.

Should I simply return $0$ (or $1$)? Should I refuse the input?

I tried some online prime factor calculators to see what would happen, but they either ignored/refused input of non-natural numbers, or, in one case, crashed my browser :-)

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This is the Fundamental Theorem of Arithmetic. Look at the prime factorization results from WA. Regards – Amzoti Jan 31 '13 at 17:30

Positive integers have a prime factorization (i.e., can be written as a product of zero or more factors, all of which are prime numbers.) You may already have a problem with the input $1$ as it does not have any prime factors (while writing $1 = 1$ just like $17=17$ might seem to imply that $1$ is prime). Prepending a factor of $-1$ for negative input is already a trick, as $-1$ is not a prime either. (In the ring of integers, these two exceptions $1$ and $-1$ are the so-called units). Technically, $0$ is a multiple of any prime, but it des not have a prime factorization. So before you start outputting $0 = 0\times 2\times 3\times 5$, I suggest you just output the non-factorization $0=0$ (as you would also output $-1=-1$ and $1=1$),
One could say that $1$ does not have any prime factors but does have a prime factorization, but $0$ doesn't have a prime factorization. – Michael Hardy Jan 31 '13 at 18:07
@MichaelHardy Yes, I wanted to imply that - but even $1$ poses the "problem" to write down the empty product – Hagen von Eitzen Feb 1 '13 at 7:29