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$(1)\;:$ How many ordered triplets of positive integers $(x,y,z)$ are there are such that $z! = x!+y!$

$(2)\;:$ How many ordered triplets of positive integers $(x,y,z)$ are there are such that

$w! = x!+y!+z!$

$\bf{Solution::}$ Given as ::Using $\bf{WLOG\;,}$ Let $z=A\times B \times C\;,\; y=B\times C\;,\; x=C$

So $AB=B+1\Rightarrow \displaystyle A=1+\frac{1}{B}.$ So $1<A\leq 2$

So the only possibility is $A=2,$ which is $(1,1,2)$

Similarly for $(b):$ Using $\bf{WLOG\;,}$

Let $w=A\times B \times C \times D\;,z=B\times C\times D\;,y=C\times D\;,x=D$

So $ABC=BC+C+1\Rightarrow \displaystyle A = 1+\frac{1}{B}+\frac{1}{BC}$

So $1<A \leq 3$. So $A=2$ is not possible . So only possibility is $A=3\;,$ which is $(3,2,2,2)$

Would anyone like to explain me the above solutions.

Thanks.

juantheron
  • 53,015

2 Answers2

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Whatever the solution(s), you will always be able to name the variables so that $y\ge x$, and naturally, $z> y\ge x$, so that $z!$ is a proper multiple of $y!$ (implying $z!/y!\ge z$) and $y!$ is a multiple of $x!$ (implying $y!/x!\ge1$).

Divide $z!=y!+x!$ by $y!$, and you get

$$\frac{z!}{y!}=1+\frac{x!}{y!}.$$ The RHS cannot exceed $2$ and neither can $z$. The only candidate is $\color{green}{(2,1,1)}$.

Similarly, $$\frac{w!}{z!}=1+\frac{y!}{z!}+\frac{x!}{z!}.$$ The RHS cannot exceed $3$ and neither can $w$. The only candidates are $(2,1,1,1)$, $(3,2,1,1)$,$(3,2,2,1)$ and $\color{green}{(3,2,2,2)}$.

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A simpler approach:

[Lemma: When $a<b$, $b!\ge ba!$]

Wlog, let $x\le y<z$. Thus $z!=y!+x!\le 2y!$. Since $z!\ge zy!$, $z$ cannot exceed $2$.

Wlog, let $x\le y\le z<w$. Thus $w!=z!+y!+x!\le 3z!$. Since $w!\ge wz!$, $w$ cannot exceed $3$.