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Find the total number of ordered pairs $(x,y)\in\mathbb{Z}^2$ which satisfy

$$ \tag1 x!y! = x!+y!+2 $$

My Attempt:

We can write $(1)$ as

$$ \begin{align} x!y!-x!-y!+1 &= 3\\ \left(x!-1\right)\left(y!-1\right) &= 3=1\times 3 = 3\times 1 \end{align} $$

So, without loss of generality, $(x!-1)=1$ and $(y!-1)=3$, which implies that $x!=2$ and $y!=4$. But there is no $y\in\mathbb{Z}$ such that $y!=4$. Thus no ordered pairs $(x,y)\in\mathbb{Z^2}$ exist which satisfy the given equation.

Is this solution correct? If not, why not?

Fabrosi
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juantheron
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    Your solution is fine, based on the fact that $3$ is prime (which you may want to mention). Alternatively, one might first solve $ab=a+b+2$ for integers $a,b$ and then notice that none of the solutions $(2,4)$, $(4,2)$, $(0,-2)$, $(-2,0)$ looks like a pair of factorials ... – Hagen von Eitzen Aug 13 '14 at 10:05
  • It seems valid to me. You could remark that $(x!-1)=1$ or $3$ because $(x!-1)$ is always an integer. – Jam Aug 13 '14 at 10:07

2 Answers2

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Seems correct.

Another way to prove this might be to say that your original equation implies $x!y!-x!-y!=2$ and that if wlog $x \leq y$, then the left side is divisible by $x!$ (because the three terms are) while the right side (2) is only divisible by $x!$ if $x=1$ or $x=2$ which both easily lead to a contradiction.

Frunobulax
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2

Here is another proof:

Assume $x\geq y\geq 3$. Then we have $$x!\,y!\geq x!\,3!>3\cdot x!=x!+x!+x!>x!+y!+2$$ The cases $0<x,y<3$ are checked by hand.

Ali Caglayan
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Wojowu
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