Numbers Do Factor

Let $n=1275$, and recall from above that $17\mid 1275$, so $n$ is definitely composite, $n=17 \cdot 75$. Next, $75$ is $5\cdot 15=5\cdot 5\cdot 3$. So, finally, $1275 = 3\cdot 5\cdot 5\cdot 17$.

Now suppose $n$ is any positive number. Then, just as above, $n$ can be written as a product of primes:

• If $n$ is prime, we are done.
• If $n$ is composite, then $n=ab$ with $a,b<n$. By induction, $a,b$ are products of primes, so $n$ is also a product of primes.

What if we had done something differently when breaking $1275$ apart as a product of primes? Could the primes that show up be different? Why not just try? We have $1275 = 5\cdot 255$. Now $255=5\cdot 51$ and $51=17\cdot 3$, so everything turned out the same. Will it always?

Incidently, there's an open problem nearby:

Unsolved Question: Is there an algorithm which can factor any given integer $n$ so quickly that its ``running time'' is bounded by a polynomial function of the number of decimal digits of $n$.

I think most people would guess ``no'', but nobody has yet proved that it can't be done (and told everyone...). If there were such an algorithm, then the cryptosystem that I use to send my girlfriend private emails would probably be easily broken.