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### Factorials of Three Variables under Two Operations in One Equation

```Date: 07/24/2012 at 14:19:10
From: Jack
Subject: Finding all positive integer solutions for (a,b,c)

Find all positive integer solutions for (a, b, c) to the equation

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

I think that a doesn't equal b since then c! = 0 -- but there is no
integer value of c that satisfies that. I also know that either a! divides
c! or vice-versa; or that b! divides c! or vice-versa.

But I'm not sure if that helps; and generally, I am not sure where to
start with solving this equation. Should I try and make some simultaneous
equations that a, b, and c must satisfy?

```

```
Date: 07/26/2012 at 01:29:04
From: Doctor Vogler
Subject: Re: Finding all positive integer solutions for (a,b,c)

Hi Jack,

Thanks for writing to Dr. Math. That's an interesting problem.

I will prove that there is only one integer solution, namely,

a = 3, b = 3, and c = 4.

First of all, notice that if there is any solution where a > b, then you
can switch a and b to produce a solution with a < b. So it will be
sufficient to prove that there are no other integer solutions that have
a ≤ b.

Let's assume that a ≤ b and that

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

This is equivalent to

(a! - 1)(b! - 1) = c! + 1.

Now, if a = 0 or a = 1, then the equation becomes

0 = c! + 1.

This is clearly impossible, since c! > 0. And if a = 2, then the equation
becomes ...

b! = c! + 2,

... which implies that

b! > c!
b  > c
b  ≥ c + 1
b! ≥ (c + 1)*c!

Therefore,

c! + 2 ≥ (c + 1)*c!
2 ≥ c*c!
≥ c

So c ≤ 2. But neither c = 1 nor c = 2 has c! + 2 equal to a factorial. So
we conclude that a > 2. Therefore a ≥ 3, a! ≥ 6, and so

c! + 1  = (a! - 1)(b! - 1)
≥ 5(b! - 1)

However, since b! ≥ a! ≥ 6, that means that

c! + 1 ≥ b! + 4*b! - 5
≥ b! + 4*6 - 5
>  b! + 1

Therefore, c! > b!, so c > b.

So we have established that 3 ≤ a ≤ b ≤ c.

Now I will divide your original equation by b!:

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

Since c > b, that means that c!/b! is the product of integers from b + 1
to c. So, c!/b! is an integer. Therefore, this must also be an integer:

a!/b! = a! - 1 - c!/b!

But that means that b! divides a!, and that implies that b ≤ a. We
already know that a ≤ b, so we must conclude that a = b.

Now the equation becomes

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

Now, since a ≥ 3, that means that a! is divisible by 3. Therefore, this
cannot be divisible by 3:

c!/a! = a! - 2

But c!/a! is the product of all integers from a + 1 to c, so that means
that none of the integers between a + 1 and c, inclusive, can be divisible
by 3. And c > a, so there are either one or two integers in that range (if
there were three or more, at least one would be divisible by 3). So either
c = a + 1 or c = a + 2.

I will prove that there are no solutions with c = a + 2, and I will leave
it to you to prove that there is only one solution with c = a + 1.

If c = a + 2 and a = b, then we have

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

If a ≥ 5, then

a! = a*(a - 1)*...*3*2*1
≥ 3a(a - 1)
= a^2 + 2a^2 - 3a
≥ a^2 + 10a - 3a
= a^2 + 3a + 4a
≥ a^2 + 3a + 20
> a^2 + 3a + 4
= 1 + 1 + (a + 2)(a + 1).

Therefore, there are no solutions with c = a + 2 and a ≥ 5. So all
solutions with c = a + 2 must have a < 5. Checking the values with a < 5,
none of them solves

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

Now you do the c = a + 1 case, and that will finish the proof!

If you have any questions about this or need more help, please write back
and show me what you have been able to do, and I will try to offer further
suggestions.

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 07/26/2012 at 07:01:41
From: Jack
Subject: Thank you (Finding all positive integer solutions for (a,b,c))

Thanks for that. That's helped me a lot :)
```
Associated Topics:
High School Linear Equations

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