I've tested computationally if a natural number $n$ can be written as $n=c+m^2$, where $m$ is an integer greater than zero and $c$ is a composite with at least two different primefactors. It seems that just $11$ numbers can not be written in that form:
1 100000 ~ :| n | n test 0= ; intcond zet. {1,2,3,4,5,6,8,9,12,17,20} ok
Therefore, I would like to see a proof of the conjecture:
All integers $n$, large enough, can be written as $n=c+m^2$, where $m\in \mathbb Z_+$ and $\omega(c)>1$.
It's very plausible that there is a maximal number not satisfying 'test'. Otherwise there would be arbitrary large integers $n$ such that $n-m^2$ is a prime power for every $m$ such that $n<m^2$. But how to prove it?
Given an integer $n$ there are $\lceil\sqrt{n}\rceil-1$ perfect squares less than $n$, so there are $\lceil\sqrt{n}\rceil-1$ numbers $n-m^2$ that all must be prime powers in order to be an exception. Due to What's the asymptotic distribution of $p^n$ (powers of primes)? the number of prime powers less than a number $x$ is asymptotically equivalent with $\frac{x}{\ln x}$. Thus, the probability of $n-m^2$ to be a prime power is $\frac{1}{\ln (n-m^2)}$ and the probability $P_n$ for n to be an exception is: $$P_n=\prod_{m=1}^{\lceil\sqrt{n}\rceil-1}\Big(\frac{1}{\ln (n-m^2)}\Big)$$
Due to my calculations:
$P_{10}\approx 2.54007092542113\cdot 10^{-1}$
$P_{100}\approx 3.20233617274630\cdot 10^{-6}$
$P_{1000}\approx 7.55006999135314\cdot 10^{-25}$
$P_{10000}\approx 3.81090059510128\cdot 10^{-93}$
For me it isn't impossible that it exist a really large exception, but it is counter intuitive that there exist arbitrary large exceptions.