Vietnam Mathematical Olympiad

The relation in the problem can be written in the form $$x!+y!=3^{n}n!(1)$$ Suppose that $(x,y,n)$ is a triple of natural numbers satisfying (1). It is easily seen that $n\ge 1$ and w.l.g. we can suppose that $x\le y$. We must now consider the following cases.

1) 1st case: $x\le n$ $$\text{It is clear that }(1)\iff 1+\frac{y!}{x!}=3^n\frac{n!}{x!}.(2)$$ (2) implies that $1+\frac{y!}{x!}\equiv 0(\mathrm{mod}3)$. Since the product of three consecutive integers is divisible by 3 and since $n\ge 1$, we have: $x<y\le x+2$.

a) If $y=x+2$, (2) implies $$1+(x+1)(x+2)=3^n\frac{n!}{x!}(3)$$ Since the product of two consecutive integers is divisible by 2, from (3), we deduce that $n\le x+1$. - If $n=x$, (3) implies $$1+(x+1)(x+2)=3^x,\text{i.e. }{x}^2+3x+3=3^x(4)$$ Since $x\ge 1$, (4) shows that $x\equiv 0(\mathrm{mod}3)$, therefore $x\ge 3$ and we get a contradiction $$3=x^2+3x-3^x\equiv 0(\mathrm{mod}9)$$ which proves that $n\ne x$. - If $n=x+1$, (3) implies that $$1+(x+1)(x+2)=3^n(x+1)$$ hence $x+1$ is a positive divisor of 1. Therefore $x=0$ and consequently $y=2,n=1$.

b) If $y=x+1$, (2) implies $$x+2=3^n\frac{n!}{x!}(5)$$ Since $x\ge 1$, (5) shows that $x\ge 1$ and then $x=n$ and $$x+2=3^x.(6)$$ $x=1$ is the unique natural number satisfying (6) therefore, in this case, if the triples $(x,y,n)$ satisfy (1) then $(x,y,n)=(0,2,1)$ or $(x,y,n)=(1,2,1)$.

2) 2nd case: $x>n$. It is clear that $$(1)\iff \frac{x!}{n!}+\frac{y!}{n!}=3^n.(7)$$ Since $n+1$ and $n+2$ can not be simultaneously the powers of 3, from (7), we deduce that $x=n+1$. Then (2) implies that $$n+1+\frac{y!}{n!}=3^n.(8)$$ Since $y\ge x$, we see that $y\ge n+1$. By putting $A=\frac{y!}{(n+1)!}$, we can write (8) in the form $$(n+1)(1+A)=3^n.(9)$$ It is clear that if $y\ge n+4$ then $A\equiv 0\mathrm{mod}3$, and $A+1$ can not be a power of 3. Hence (9) shows that $y\le n+3$. So $n+1\le y\le n+3$.

a) If $y=n+3$ then $A=(n+2)(n+3)$, and from (9) we get $$(n+1)(1+(n+2)(n+3))=3^n\text{ i.e. }(n+2{)}^3-1=3^n.(10)$$ It follows that $n>2$ and $n+2=1\mathrm{mod}3$. By putting $n+2=3k+1,k>2$, we can write (10) in the form $$9k(3k^2+3k+1)=3^{3k-1}.$$ It implies that $3k^2+3k+1$ is a power of 3. This contradiction proves that $y\ne n+3$.

b) If $y=n+2$ then $A=n+2$ and from (9) we get $$(n+1)(n+3)=3^n.(11)$$

c) If $y=n+1$ then $A=1$ and from (9) we get $$2(n+1)=3^n.$$ It is clear that there exist no $n$ satisfying this relation. So $y\ne n+1$.

Thus, if the triple $(x,y,n)$, with $x\le y$, satisfies (1) then $n\ge x$. Consequently, if $(x,y,n)$ is a triple of natural numbers satisfying (1) then $(x,y,n)=(0,2,1)$ or $(x,y,n)=(2,0,1)$, or $(x,y,n)=(2,1,0)$, or $(x,y,n)=(2,1,1)$. By direct verification, we see that the four mentioned triples satisfy (1). So these triples are all triples satisfying the conditions of the problem.