Diophantine Equation

Problem

Find all $ n $-tuples $ ( a_1 , a_2, \dots , a_n ) $ of positive integers such that

$$ (a_1! -1 )(a_2! -1 )\cdots(a_n! -1) - 16 $$ is a perfect square.

 ( Source : Problems From The Book, by Titu Andreescu and Gabriel Dospinescu )

Solution :

Let $ x \in \mathbb{N} $ such that $  (a_1! -1 )(a_2! -1 )\cdots(a_n! -1) - 16 = x^2 $.

Now observe that if $ a_i = 1 $ for some $ i \in \{ 1,2,\dots,n \} $ then we get $ - 16 = x^2 \geq 0 $( Absurd! )

So we deduce that $ a_i \geq 2 \ \forall \ i \in \{ 1,2,\dots,n\} $ 

Therefore in the LHS the term $  (a_1! -1 )(a_2! -1 )\cdots(a_n! -1) $ is always odd $ \Rightarrow x $ is odd.

Now the crucial observation is $ a_i < 4 \ \forall \ i \in \{1,2,\dots,n\} $. So let us proof it !

Suppose on the contrary $ \exists $ some $ i \in \{ 1,2,\dots,n \} $ such that $ a_i \geq 4 $.

Then the term $ (a_i! -1 ) \equiv 3 ( \mbox{ mod } 4 \ )$  hence $ \exists p \in \mathbb{P}_3 $ such that $ p \ | \ (a_i! -1 ) $

As $  (a_1! -1 )(a_2! -1 )\cdots(a_n! -1) = x^2 + 4^2 \Rightarrow p \ | \ x^2 + 4^2 $

Now using Properties of Primes of the form $ 4k+3 $ we get $ p \ | \ \gcd(x,4) $

$ \Rightarrow p \ | \ 4 \Rightarrow p = 2 $ ( Contradiction, as $p$ is odd ) 

Therefore we get $ a_i < 4 \ \forall \ i \in \{ 1,2,\dots,n \} $ 

$ \therefore a_i \in \{ 2,3 \} \forall \ i \in \{ 1,2,\dots, n \} $

Suppose $ k $ many of them are $ 3 $ and rest are $2$ then we get

$  (a_1! -1 )(a_2! -1 )\cdots(a_n! -1) = x^2 + 16 \Rightarrow 5^k = x^2 + 16 $

As $ x $ is odd, we have $ x^2 \equiv 1 ( \mbox{ mod } 8 \ )$

$ \Rightarrow 5^k \equiv 1 ( \mbox{ mod } 8 \ ) $

Now $ o_8(5) = 2 \Rightarrow 2 \ | \ k $ ( $ o_n(m) := $ order of $ m $ modulo $ n $ )

Let $ k = 2m $

So we have $ 5^{2m} - x^2 = 16 \Rightarrow ( 5^m - x )( 5^m + x )=16 $

Case 1 :  $ \boxed{ 5^m - x = 2  \mbox{ and }  5^m+x = 8 }$

We get $ 2 \cdot 5^m = 10 \Rightarrow m = 1 $

So $ k = 2 $

Therefore one possible solution of the given diophantine equation is $  ( a_1 , a_2, \dots , a_n ) = (3,3,\underbrace{2,\dots,2}_{ n-2 \mbox{ times}}) $ and its permutations.

Case 2 : $ \boxed{ 5^m - x = 1  \mbox{ and }  5^m + x = 16 } $

We get $ 2 \cdot 5^m = 17 $ ( Absurd! )

Therefore the only solution to the given diophantine equation is $ \boxed{ ( a_1 , a_2, \dots , a_n ) = (3,3,\underbrace{2,\dots,2}_{ n-2 \mbox{ times}}) }$ and its permutations.