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I think I have proved that that a cannot be greater than b (ie a & b must be equal) then using this fact I got a quadratic in terms of a and c, trying out some values I got that (3,3,4) is a solution, but how do I prove this is the only solution or if this is not the only solution then what are the other solutions and how do I find them?

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    For completeness, I assume that your proof for $a=b$ is like this: If $a$a!$, hence so is $c!$, i.e., $c\ge a$. The left hand side is evenly divisible by $(a+1)!$, but on the right we get either $\frac1{a+1}$ or $\frac 2{a+1}$ plus an integer, hence $a+1\mid 2$ – 2017-01-31
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    If b is is less than a, you have two cases to analyse **c>b>a** or **b>c>a** you can express both **a** & **c** in terms of **b** then then you have a quadratic in terms of **b**,in the end you find the solution to this quadratic is not an integer (for both cases) and you are done. – 2017-02-01

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If you prove that a=b, c>a

$a!(a!-2)=c!$ => $a+1|a!-2$

if $a+1$ is prime, $a+1|a!+1$ =>$a+1|3$ => $a= 2$, but it is not an example

if $a+1 = a_1a_2, 1< a_1 < a_2 < a $, $a+1| a!$ => $a+1|2$ => a=1, but it is not an example

So, $a+1 = p^2$, p is prime;

if p>2, $a! = C*p*(2p)$, so $a+1| a!$ => as in previous case

So, $a+1 = 2^2$

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    $a!-2=k_1(a+1), a!+1 =k_2(a+1)$ => $3=(k_2-k_1)(a+1)$ – 2017-01-31
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    Yeah,I just noted that you *used* $a+1\mid a!+1$, not concluded it – 2017-01-31
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If you have already proved that $a=b$, then

$a!b!=a!+b!+c!$

$(a!)^2=2.a!+c!$

$(a!)^2-2.a!-c!=0$

$(a!-1)^2-c!-1=0$

$c!=(a!-1)^2-1$

So, there is only value for $c$

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HINT

Let $b \le a, 2 \le a$

$a! = \frac{a!}{b!} + 1 + \frac{c!}{b!}$

So, $a=b$ or $a=b+1$ or $c=b$ or $c = b+1$ (in another case left part is even, but right is odd)

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    Yeah well I have proven that a can only b, but why must c=b+1? – 2017-01-31
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    case if c = 2*k + 1 and a>b – 2017-01-31