## It’s as easy as a, b, c

Posted by Chris on October 7, 2014 – 12:30 pm

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

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

Posted by Chris on October 7, 2014 – 12:30 pm

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

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

October 8th, 2014 at 4:44 pm

Divide throughout by a! to get:

b! = 1 + b!/a! + c!/a!

Since there are no odd factorials then b! is even so that either b = a + 1 (with b odd) or c = a + 1 (with c odd).

By inspection, even for small a we find that c >> a+1. So b = a+1 and our equation now becomes:

(a+1)! = 1 + (a+1) + c!/a!

c! = a!(a+1)! – a!(a+2)

This is where it gets beyond me. Even if I generate solutions to this equation for successive (even) values of a, how do I test for the result being a factorial number?

October 9th, 2014 at 2:43 pm

Divide throughout by a! to get:

b! = 1 + b!/a! + c!/a!

Divide throughout by b! to get:

a! = 1 + a!/b! + c!/b!

Since factorials are integers, so a! = b!.

Now we can convert the equation into

a!a! = 2a! + c!

let a! = x

x^2 – 2X – c! = 0

Determinant must be a square number,

square root of 4 + 4c! is 2(1+c!)^1/2;

so c! + 1 must be a square number.

1! = 1, 1 + 1 = 2, invalid

2! = 2, 2 + 1 = 3, invalid

3! = 6, 6 + 1 = 7, invalid

4! = 24, 24 + 1 + 25, valid

5! = 120, 121 valid

….

If c = 4, then a = b = 3

If c = 5 , a and b cannot be found.

(This is the only solution I can find, and I don’t know if c could be a large number)

October 10th, 2014 at 8:17 am

continuing from above

we have another choice that is b = a and c= a + 1 and c is even

we can divide by a! to get c = (a+1) or b = a+ 1

combining the 2 we get b = a and c = a+ 1

so (b!)^2 = 2b! + (b+1) ! = b!(b+3)

so b! = b + 3

we get b = 3 or solution set = a = b = 3 and c = 4

check 3! * 3! = 36 = 6+6+ 24 = 3! + 3! + 4!

October 10th, 2014 at 8:34 am

above is not 100% correct as if b = a the c need not be a +

so b! = 2 + c!/b!

b = 3 , c = 4 is a solution

b = 4 LHS = 24 c = 5=> RHS = 7

c = 6 => RHS = 30

no solution

I cannot work out

October 16th, 2014 at 12:44 pm

Here is a complete solution.

Without any loss of generality, assume that b ≥ a

First, observe that if a = 0 or a = 1 then the equation becomes b! = 1 + b! + c!, which is impossible.

Also if a = 2 we have 2b! = 2 + b! + c! which implies that b! = 2 + c! which is also impossible.

Conclusion: b ≥ a ≥ 3

Second, if c ≤ b then a!b! ≤ a! + 2b! ≤ 3b! which implies that a! ≤ 3 which is impossible since a ≥ 3.

In summary : c > b ≥ a ≥ 3

As Catherine said:

Divide throughout by a! to get: b! = 1 + b!/a! + c!/a!

Divide throughout by b! to get: a! = 1 + a!/b! + c!/b!

implies that a = b and the equation simplifies to

a!a! = 2a! + c!

The trick is to replace c = a + d with d ≥ 1 and to divide throughout by a!. We get

a! = 2 + (a+1)(a+2)…(a+d).

Next, divide both sides by 3. a!/3 is an integer. 2/3 is not an integer. Therefore (a+1)(a+2)…(a+d)/3 is not an integer, which implies that d ≤ 2 . There are only 2 cases to consider

Case d=1: c = a+1

—————

a! = 2 + (a+1) = a+3

(a-1)! = 1 + 3/a and the only integer solution is a = 3 (and b=3, c=4)

Case d=2: c = a+2

—————

a! = 2 + (a+1)(a+2) = a^2 + 3a + 4

(a-1)! = a + 3 + 4/a and the only integer solution would be with a=4, but

3! differs from 4 + 3 + 1

CONCLUSION: There is a unique solution (a,b,c) = (3,3,4)

October 25th, 2014 at 6:12 am

Hi Zorglub. Thanks for that. I didn’t have an official solution. Yours look spot on.