It's not actually that easy

We know that when we square a number ending with $3$ , the result is a number ending with $9$ , which will be subtracted by the previous prime ending with $3$ , and again subtracted by $1$ , so the number will end with $5$ . The smaller primes ending with $3$ are $3$ and $13$ . But obviously, $3$ doesn't works because it yields the prime $5$ . So all we need is to check the primes $5$ , $7$ and $11$ , and if they don't work, then the answer will be the substitution of $13$ . They dont work, and hence the least composite result is $169 - 13 - 1 = \boxed {155}$ .

I'm not sure whether there's a better way of doing it, but when I first thought of this, I was trying to solve something else. If you factor this to $n(n-1)-1$ , you're looking for 1 less than twice a triangular number. There must be other ways of solving this, because that's a stupid way. Yeah, but anyways, when I started trying out primes, none of them seemed to work! I thought I had discovered a property of primes, but then I tried 13. $169-13-1=155$ which is composite. And that is the least possible value.

13*13-13-1=155