<s>[INST] Minimize $a+b+c$ where $a!=b! \cdot c^2$ and $c \geq 10$, $a,b,c \in \Bbb N^+$ For $c=10$, we get $a=100$ and $b=99$.
However, I am wondering if it is possible to write a square of a number as a multiplication of consecutive integers. So, if 
$c^2 = a (a-1) (a-2) \cdots (a-n)$
then we would possibly get a smaller $a+b+c$.
$b=1$, $\ \  $ ($n=a-1$) is not possible because the only square factorials are $0!$ and $1!$.
These are my thoughts on the problem. Thanks for your help.
 [/INST] This is a variant of Xpw's answer, intended to reduce the number of separate cases that need to be examined in order to show that $(a,b,c)=(100,99,10)$ minimizes the sum $a+b+c$ if $c\ge10$.
Suppose $a!=b!\cdot c^2$ with $c\ge10$ and $a+b+c\lt209$. Let's write $a=b+n$.  Then we have 
$$c^2=(b+1)(b+2)\cdots(b+n)\le(208-2b-n)^2$$
We cannot have $n=2$, since any two consecutive integers are relatively prime, hence each would have to be a square in order for their product to be a square, but positive squares differ by at least $3$.
For $n\ge3$ we have the following crude inequality:
$$b^n\le205^2\lt50000$$ 
(You can be less crude if you like, but there's little to be gained from it.) Since $40^3=64000$ and $20^4=160000$, we see that $b\lt40$ if $n=3$ and $b\lt20$ if $n\ge4$.  Let's do the $n\ge4$ case first.
Any string of $4$ or more consecutive numbers starting at a number less than or equal to $20$ contains at least one prime within the final $4$ numbers in the string (the first prime gap greater than $4$ occurs at $29-23$). The product of such a string contains that prime to the first power only, hence cannot be a square.
Any string of three consecutive numbers contains exactly one number divisible by $3$.  In order for their product to be a square, the number divisible by $3$ must be divisible by an even power of $3$.  For strings beginning at $40$ or less, the only possibilities are $9$, $18$, and $36$.  Since $7$, $11$, $17$, $19$, and $37$ are primes, the only strings to check are $8\cdot9\cdot10$ and $34\cdot35\cdot36$, neither of which is a square.
</s>