# Solve $a!b!=a!+b!+c!$ where $a$, $b$ and $c$ are nonnegative integers.

My teacher in Math Team gave the following question to us.

Solve $$a!b!=a!+b!+c!$$ where $$a$$, $$b$$ and $$c$$ are nonnegative integers.

I found only one solution by trial and error and it is $$(a,b,c)=(3,3,4)$$.

Any help is appreciated!

• This was BMO 2002/3 Round 1 Problem 5. – J.G. Aug 6 at 7:28
• Did your teacher in Math Team tell you it was OK to farm the question out to the internet? – Gerry Myerson Aug 7 at 6:33
• He just told us to find the solution. – Culver Kwan Aug 7 at 9:11
• Fair enough. But I hope you'll acknowledge your sources when you turn your work in. – Gerry Myerson Aug 7 at 12:52

A sketch:

There is no solution with $$a\lt2$$ (this would imply $$c!=-1$$) or $$a=2$$ (this would imply $$b!-c!=2$$), so by symmetry $$a,\,b\ge3$$. Without loss of generality $$a\ge b$$. Since$$a!=\frac{a!}{b!}+1+\frac{c!}{b!}$$is an integer, $$c\ge b$$.

In the case $$a=b$$, $$b!=2+\frac{c!}{b!}$$ is a multiple of $$3$$, so $$c\in\{b,\,b+1,\,b+2\}$$. These subcases equate $$b!$$ to a polynomial in $$b$$, so only small $$b$$ can be solutions.

In the case $$a>b$$, since $$b!=1+\frac{b!}{a!}+\frac{c!}{a!}$$ is an integer, $$a>c\ge b$$. The above display-line equation also tells us $$\frac{a!}{b!},\,\frac{c!}{b!}$$ aren't both multiples of $$3$$, so $$c\in\{b,\,b+1,\,b+2\}$$. We can exhaust these using $$a!\ge 24,\,b!\ge6$$. The case $$c=b$$ gives $$b!=\frac{a!}{a!-2}$$, which gives no solutions; $$c=b+1$$ gives $$b!=\frac{a!}{a!-b-2}$$, while $$c=b+2$$ gives $$b!=\frac{a!}{a!-b^2-3b-3}$$.

• Can you add the case $a>b$? Or I can't accept your answer. – Culver Kwan Aug 7 at 4:46
• @CulverKwan That's just rude - he gave you the shortest (and arguably the best ) solution and you can't accept the answer??? There are also 2 similar answers above and you can't even be bothered to acknowledge the efforts of the authors? – asdf Aug 7 at 6:38
• I've edited my answer as requested, but it's still just a sketch. – J.G. Aug 7 at 7:33
• @asdf Thank you for defending me, but I do think it's fair to say answer acceptance required the case $a>b$. – J.G. Aug 7 at 7:33

Let $$x,y,z$$ be the highest powers of $$2$$ that divide $$a!, b!$$ and $$c!$$, respectively.Then $$x+y$$ is the highest power of $$2$$ that divides $$a!b!$$.

Also, since you clearly have that $$(a!-1)(b!-1)=c!+1$$ you can obtain that $$c>a$$ and $$c>b$$ (apart from some small values of $$a,b,c$$ which you've probably already ruled out using direct check)

We can now assume WLOG that $$a\leq b \leq c$$. Then we have:

$$a!+b!+c!=2^xa_1+2^yb_1+2^zc_1=2^x(a_1+2^{y-x}b_1+2^{z-x}c_1)$$ where $$a_1,b_1$$ and $$c_1$$ are odd numbers.

Since trivially $$x and $$2^{x+y}|\text{LHS}$$ then we must have that $$a_1+2^{y-x}b_1+2^{z-x}c_1$$ is even.

If now you assume that $$x\neq y$$ then $$2^{y-x}b_1$$ is even, hence we must have $$z=x$$, i.e. that $$c-a=0$$ or $$c-a=1$$ which implies that

$$(a!)^2\leq a!b!\leq a!+(a+1)!+(a+1)!$$ and that can hold for only small values of $$a$$ again.

Hence, we must have that $$x=y$$ hence $$b-a=0$$ or $$b-a=1$$.

If $$b-a=0$$ then we have $$(a!)^2=2a!+c!.$$ As @J.G. showed, this is equivalent to

$$a!=2+\frac{c!}{a!}$$ hence $$c-a=0, 1 \text{ or } 2$$ and it reduces to a check for small values.

Finally, if $$b-a=1$$ then you have $$(a!)^2(a+1)=a!(a+2)+c!$$ which is equivalent to

$$(a+1)!=a+2+\frac{c!}{a!}$$

Now, (unless $$c=a$$ which again reduces for to small values of a) we have that $$a+1| \text{ LHS }$$, and $$a+1| \frac{c!}{a!}$$, hence $$a+1|a+2$$ which is a contradiction.

Hopefully there aren't any mistakes, though you will have to fill in the gaps, namely do the small values checks.

I claim that $$a=b$$ or $$a=c$$. For the sake of this problem, assume $$b \geq a$$ (proving for $$b \leq a$$ is the exact same). Then $$b!=1+\frac{b!}{a!}+\frac{c!}{a!}.$$ Due to closure of the integers under addition/subtraction, we get that $$c\geq a$$.

Case 1: ($$b \leq c$$)

If $$b \leq c$$, then $$b!-1=\frac{b!}{a!}+\frac{c!}{a!}=(\frac{c!}{b!}+1)\frac{b!}{a!}.$$ But $$b> a$$ implies that $$b|(b!-1)$$, a contradiction, so $$b=a$$. Then this equates to solving $$(a!)^2=2a!+c!$$ or $$a!(a!-2)=c!$$. For $$a\geq 3$$, clearly $$3 \not|(a!-2)$$, so $$c=a+1$$ or $$c=a+2$$ (else $$c = (a+3)(a+2)(a+1)\dots$$ has $$3$$ as a factor). The equation $$a!(a!-2)=(a+1)!$$ has one solution at $$a=3$$ and the equation $$a!(a!-2)=(a+2)!$$ has none. Then the only solution of this form is $$(a,b,c)=(3,3,4)$$.

Case 2: ($$b>c$$)

If $$b>c$$, then $$b!-1=\frac{b!}{a!}+\frac{c!}{a!}=(\frac{b!}{c!}+1)\frac{c!}{a!}$$

Because $$c, $$[c(c-1)(c-2)\dots(c-a+1)] |b!$$, whence $$[c(c-1)(c-2)\dots(c-a+1)]\not| (b!-1)$$ for $$c > a$$. It follows that $$c=a$$, which then equates to solving $$a!b!=2a!+b!$$ Rearranging both sides gets $$a!(b!-2)=b!$$, which has no solutions because $$b!/(b!-2)$$ doesn't take an integer value for $$b \geq 3$$. Then there are no solutions of the form $$(a,b,c) = (a,b,a)$$.

Conclusion:

From all of this, we get that $$(3,3,4)$$ is the only solution to $$a!b! = a!+b!+c!$$. (Note that I used guessing and checking to prove that $$a,b,c > 2$$)

• In your "Case 2:", you state $b \lt c$ in the title & the start of the first paragraph. However, this is almost the same as "Case 1:" (apart from the equal). I believe you meant it to say $c \lt b$ instead, e.g., as your second paragraph starts with "Because $c \lt b$, ...". – John Omielan Aug 7 at 2:46
• @JohnOmielan ah you are right – Hyperion Aug 7 at 6:19