Factorial and a Square

Rearrange the equation to get $(a!-1)(b!-1)=c^2+1 \tag{1}$ First note that $c^2+1$ doesn't have any prime divisor of the form $p=4k+3$ . Let's prove it. Assume that $c^2+1$ has a prime divisor with $p \stackrel{4}{\equiv} 3$ . Then $c^2 \stackrel{p}{\equiv} -1$ and by Fermat's little theorem $c^{p-1} \stackrel{p}{\equiv} 1$ . On the other hand $c^{p-1} \stackrel{p}{\equiv} (c^2)^{\frac{p-1}{2}} \stackrel{p}{\equiv} (-1)^{2k+1} \stackrel{p}{\equiv} -1$ This is a contradiction and completes the proof.

Now observe that if one of $a$ or $b$ is greater than $3$ , LHS of equation $(1)$ has at least one prime divisor in the form $4k+3$ . For example if $a \ge 4$ then $a!-1 \stackrel{4}{\equiv} 3$ and $a!-1$ does have a prime divisor of the form $4k+3$ , otherwise all of its prime divisors will be of the form $4k+1$ and $a!-1$ will be congruent to $1 \pmod 4$ . So we must have $a \le 3$ and $b \le 3$ . Trying few remaining cases for $a$ and $b$ we get $2$ solutions: $(a, b, c)=(2, 3, 2), (3, 2, 2)$

$a!b!=a!+b!+c^2$

$a!b!-a!-b!=c^2$

First consider $a,b \geq 5$

Both $a!,b!$ will end with zero.

So, $a!b!$ has $p+q$ number of zeroes where $a!$ and $b!$ has respectively $p$ and $q$ zeroes.

So, after subtraction the unequal number of zeroes i.e. $p+q$ and other side $p$ and $q$ zeroes..But in perfect square there should be even number of zeroes at least.

So, there is no solution for greater than 5

Now consider $1 \leq a,b \leq 4$

In this interval one can easily find the two ordered pair which satisfy the above equation.

Thy are $3,2,2$ and $2,3,2$

So, the answer is $\boxed {14}$

We shall consider the case a,b greater than equals 5.then the given equation may be written as (a!-1)(b!-1)=c^2 +1 so for our case lhs is 1 modulo 10 so c=10k.so c^2+1 is 100k^2+1.but this is a prime no for all positive integer k.but the lhs is a product of 2 nos.since our case is a,both greater than or equals 5 neither of the factor of lhs is 1.o the case does not yield solutions.now we can take a=5 and b less than a.checking out the possible case b={2,3,4} obviously not 1 do not yield any solution.a=b also yeilds a=b=2 and c=0.(not positive).observe that if a,b is a solution then is b,a. So we have only 3 cases to check out a=4,b=3 and a=4,b=2 and a=3 ,b=2.

(a!-1)(b!-1)=(c^2)+1 Both sides must be 1 mod c. This can happen when the product's terms are either both -1 or both +1 mod c. If a;b>=c, then the modular equations are satisfied, because both a! and b! are divisible by c, but the magnitude is clearly off when c>3 (c<(c-1)!). +1 mod c doesn't work; if a;b<c then they're either -k mod c (k>1), or c units below c, that is, 0. So, the only solutions have to have 1=<c=<3

Not done yet.