Peter wrote: Find all triples $(a, b, c)$ of positive integers to the equation \[a! b! = a!+b!+c!.\] Assume WLOG that $a \geq b$ $a!b!-a!-b!=c!$ $(a!-1)(b!-1)=c!+1$ (3,3,4) Not sure how to prove that it's the only solution.

It isn't the only solution either. There's (2,2,1) and I think a few others.

http://www.mathlinks.ro/Forum/viewtopic.php?t=169922 I have post solution with other method.

Peter wrote: Find all triples $(a, b, c)$ of positive integers to the equation \[a! b! = a!+b!+c!.\] a!b!=a!+b!+c! CASE 1 IF, a<b<c Then a!q=b! and a!k=b!i=a!qi=c! ………(1) substituting in the equation,we get (a!)^2 q=a!(1+q+k) (a!)q =1+q+k q divides 1+q+k q divides k+1 ………..(2) But by (1), a!k=a!qi k=qi q divides k contradicting (2) Similarly, for all other permutations of a, b and c we get the same contradiction. CASE 2 a =b=c Substituting in the equation we get, (a!)2 =3(a!) (a!) =3 ,and we know there is no such ‘a’. CASE 3 b=c Substituting, in the equation we get a!b! = a! +b! + b! a!b! = a! +2b! b!(a!-2) =a! a!-2 divides a! which we may see as impossible. Similar for a=c. Now a=b, Substituting this in the equation we get, (a!)^2 =2(a!) +c! a!(a!-2) =c! c>a a! -2 =(a+1)(a+2)(a+3)…….(a+n) As 2 divides a, (a+1)……(a+n) must contain only 2 factor. We know 4 consecutive numbers will have 4 as a factor . If a+1=a!-2 a=a!-3 a divides 3 a =(1,3) a =3 ……(3) If a!-2 =(a+1)(a+2) a!-2 =a^2+3a+2 a! -4=a(a+3) a divides 4 a= (1,2,4) here a has no solution. Similarly a! -2 =(a+1)(a+2)(a+3) will also have no solution. Therefore solution set is (a,b,c) = (3,3,4) by (3)