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

Question

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?

Answer 1

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 $c$

Answer 2

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)

Answer 3

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, $a+1 = 2^2$

Answer 4

Without loss of generality, $b \ge a$.

This follows since the equation $a!b! = a! + b! + c!$ is symmetric in $a$ and $b$. If $a > b$ we could just relabel the variables $a$ and $b$ and switch them around. Just remember, if asked to give all solutions, that if we find $(a,b,c)=(p,q,r)$ with $p < q$ then we will need to list the symmetric solution $(a,b,c)=(q,p,r)$ as well, even though our method might not find it explicitly as the alternative solution violates our assumption $b \ge a$ .

There's a lower bound $b \ge a \ge 3$.

By dispensing of small cases: if $a=1$ then $b! = 1 + b! + c!$ so $c! = -1$, which is impossible. If $a=2$ then $2a! = a! + c! + 2$ so $a! = c! + 2$, but no factorials differ by two: the smallest non-zero gap is $2!-1!=1$ and for $n \ge 3$ consecutive factorials differ by $$n! - (n-1)! = n(n-1)! - (n-1)! = (n-1)(n-1)! \ge 2(2!) = 4$$

This imposes a lower bound on $c$, with $c \ge b+1 > b \ge a \ge 3$.

Since $b! \ge a! \ge 6$ we have $c! = a!b! - a! - b! \ge 6b! - b! - b! = 4b!$ so $c! > b!$ hence $c > b$.

(Adapted from Mike's answer, there's a different proof that $c \ge b+1$ that doesn't rely on establishing lower bounds for $a$ first. Suppose instead that $c \le b$ as well as $a \le b$. Then $b! < a!+b!+c! = a!b! \le 3(b!)$. Dividing by $b!$ gives $1<a!\le 3$ hence $a=2$, but this cannot give a solution for the reasons stated above. Therefore we must have $c > b$.)

$c!$ must be a multiple of $b!$, and so is $a!$.

From $c \ge b+1$ we know $c! = m(b!)$ where $m=c!/b!$ is the product of factors from $b+1$ to $c$. But $a! = a!b! - b! - c!$ so $a! = (a! - 1 - m)b!$.

Hence $a=b$ and we must solve $b! - 2 = c!/b!$.

From $b \ge a$ we already knew $a!$ divides $b!$. But we just showed $b!$ divides $a!$ also. Hence $a!=b!$ and (since $a\ge3$, so we don't need to worry about $0! = 1!$) we must have $a=b$. Our equation becomes $(b!)^2 = 2(b!) + c!$ so dividing by $b!$ gives $b! = 2 + c!/b!$.

$c \not\ge b+3$ so either $b! - 2 = b+1$ or $b! - 2 = (b+1)(b+2)$.

We know $b \ge 3$ so $b!$ is certainly divisible by $3$, and $b!-2 = c!/b!$ cannot be. If $c \ge b+3$ then $c!/b!$ would have three consecutive integer factors $(b+1)(b+2)(b+3)$ and so would be divisible by $3$, which is impossible. So we have bounded $c$ between $b+1 \le c \le b+2$ and $c!/b!$ is either $b+1$ or $(b+1)(b+2)$.

There are many ways to finish off from here. One approach is to note factorials grow much faster than polynomials, so $b! = b+3$ (for the case $c=b+1$) and $b! = b^2 + 3b + 4$ (for the case $c=b+2$) only need to be checked for a few low $b$ values.

We only need to check $b \ge 3$. In fact $b=3$ is a solution of $b! = b+3$, and for $b \ge 4$ we see $b! > b+3$. To prove this, note that for $b \ge 4$,

$$b! \ge b \times 3! > 2b = b + b > b + 3$$

Checking both $b=3$ and $b=4$ we find $b! < b^2 + 3b + 4$ but we see the inequality flips for $b \ge 5$. To prove this, note that for $b \ge 5$ we have $2b-2 > b+2$ so

$$b! \ge b(b-1) \times 3! > 4(b-1)^2 = (2b - 2)^2 > (b+2)^2 = b^2 + 4b + 4 > b^2 + 3b + 4 $$

Thus the only solution is $a=b=3$ with $c=b+1=4$, and we verify $3!3!=3!+3!+4!$ as required.

Another approach is to note that $b!-2$ cannot be divided by any integer from $3$ to $b$, and use this divisibility fact to rule out $c=b+2$ and solve $b!-2=b+1$.

If $c = b+2$ then $c!/b! = (b+1)(b+2)$ is the product of two consecutive integers, one of which is even. So either $(b+1)/2$ or $(b+2)/2$ is an integer factor of $c!/b!=b!-2$. Since $b\ge 3$ these potential factors are both at least $(3+1)/2=2$ but less than $(b+b)/2=b$. But every integer between $2$ and $b$ divides $b!$, so our factor $(b+i)/2$ (with $i\in\{1,2\}$) must divide both $b!$ and $b!-2$. Hence our factor must divide the difference, $b! - (b!-2) = 2$, and since our factor is at least $2$, it must be $2$. This can only happen when $b=3$, or else $(b+i)/2>2$ for $i=1,2$. Since $c=b+2$ we have $c!/b! = (b+1)(b+2) = 4 \times 5$ and $(b+1)/2=4/2=2$ is indeed a factor of $b!-2=3!-2=4$. But clearly $c!/b! \neq b!-2$ so this does not yield a solution.

Thus $c=b+1$ and we need to solve $b!-2=b+1$, or equivalently $b=b!-3$. So $b$ divides $b!-3$, but it also divides $b!$, hence it also divides the difference, $b! - (b!-3) = 3$. Since $b \ge 3$ and $b \mid 3$ we must have $b=3$, $a=b=3$, $c=b+1=4$ and we verify $3!3!=3!+3!+4!$ is a solution.

A third approach is to consider the implications of $b+1$, and potentially $b+2$, dividing $b!-2$, to think what kind of prime factor decomposition they must have.

Every positive integer $n$ is either one, prime, the square of a prime, or the product of two distinct nontrivial factors, $1 < f_1 < \sqrt{n} < f_2 < n$ with $n=f_1 f_2$.

We know $b!-2$ is either $b+1$ or $(b+1)(b+2)$ so $b+1$, and every factor dividing $b+1$ (as well as $b+2$ and every factor dividing $b+2$ if $c=b+2$), is also a factor of $b!-2$. We also know $b!-2$ is divisible by $2$ but not by any factor from $3$ to $b$.

Since $b \ge 3$, we can narrow those options for $b+1$ down to either being an odd prime, the square of a prime, or the product of two distinct nontrivial factors. If $b+1$ is an odd prime, we can't have $b!-2=b+1$ since the LHS is even. So we must have $c=b+2$ and $b!-2=(b+1)(b+2)$. Then $b+2$ is even, and so $b/2+1$ divides the RHS. But since $2 < b/2 + 1 < b$, it cannot divide the LHS. Hence $b+1$ is not an odd prime. But if $b+1$ is either the square of an odd prime, or the product of two distinct nontrivial factors, then $b+1$ must have a factor between $3$ and $b$, which is impossible for the same reason. The only possibility that remains is for $b+1$ to be the square of an even prime, i.e. $b=2^2-1=3$. Then $b!-2=3!-2=4$ equals $b+1=4$ but not $(b+1)(b+2)=4\times 5=20$. So our only solution has $c=b+1=4$, $a=b=3$, and $3!3!=3!+3!+4!$ as required.

This last approach is essentially @kotomord's answer but adapted so it doesn't rely on knowledge of Wilson's theorem to claim that $b+1$ divides $b!+1$ if $b+1$ is prime.