Question

Find all solutions in positive integers a, b, c to the equation
a! b! = a! + b! + c!

Answer 1

I can think of one solution a=3, b=3, c=4
but this looks like a very hard problem to solve mathematically.
maybe @mukushla can lend a hand here?

Answer 2

yes, I find a=3, b=3 and b=4 but I can't solve

Answer 3

@asnaseer ............... thank u for letting me about this question...

Equation is symmetric for \(a\) and \(b\) so without loss of generality we can assume \(a \le b\) ...

now suppose that \(c<b\) dividing both sides of equation by \(b!\) u have
\[a!=1+\frac{a!}{b!}+\frac{c!}{b!} \]\(LHS\) is an integer and \(RHS\) is not ....contradiction and we will have \(c \ge b\) so we must have

\(a \le b \le c\) ...

looking the equation again \(a! b!=a!+b!+c!\)...we can see \(b! | LHS\) consequencly \(b! | RHS\) so

\(b! | \ a!+b!+c! \) easily u can see \(b! | \ a!\) since \(b \le c\) then we must have \(a\ge b\) but we assumed

that \(a \le b\) so \(a=b\).

rearranging the equation with \(a=b\) we have \((a!)^2=2 a!+c! \) and \[a!=2+\frac{c!}{a!}\]so \(a\ge 3\) and \(c>a\) ... note that if \(c>a+3\) then.(\(LHS\equiv 0\mod 3 \)).but.(\(LHS\equiv 2\mod 3 \))

it means that \(c \le a+3 \) so \(a+1\le c\le a+3\) and we have 3 cases

case \(I\) ....... \(c=a+1\) \[a!=a+3\]\[a=3\]one solution from here \((a,b,c)=(3,3,4)\)

case \(II\) ....... c=a+2 \[a!=2+(a+1)(a+2)\]no solution

case \(III\) ....... c=a+3 \[a!=2+(a+1)(a+2)(a+3)\]no solution


the only solution for this equation \((a,b,c)=(3,3,4)\)

Answer 4

wow! very neat (as usual) @mukushla :)

Answer 5

:)

Answer 6

thank you very much.:)

Answer 7

@asnaseer.....*letting me know.............lol...I am not too bad at english...hahaha...

Answer 8

@mukushla hehe - np - I understood what you meant. :)