# Find all n-tuples $(a_1, a_2, ...,a_n)$ of positive integers such that $(a_1 ! – 1)(a_2 ! – 1)... (a_n! – 1) – 16$ is a perfect square.

Find all n-tuples $$(a_1, a_2,...,a_n)$$ of positive integers such that $$(a_1! – 1)(a_2! – 1)... (a_n! – 1)– 16$$ is a perfect square. https://photos.app.goo.gl/b98MVt1MwmTyecLz7

I've done this: $$N=k^2+4^2$$ this means that there is not any $$p$$ prime , $$p|N$$ , $$p=4m−1$$ => $$a_n!−1≠4m−1$$ => $$4$$ doesn't divide $$a_n!$$ => $$a_n<4$$ and also $$a_n≠1$$ (then $$N=0$$ which is wrong) $$a_n=2,3$$ we'll try some combinations of them $$(2!-1)=1$$ it doesn't change anything so all we need to do is find solutions for $$a_n=3$$ but I don't know can I use same $$a_n$$ two or more times?

• Welcome to Mathematics Stack Exchange !! . Please use Mathjax to edit and format math texts and symbols here :- mathjax.org. Jun 16 '20 at 5:37
• Hint: if $N-16$ is a perfect square, then $N$ is the sum of two squares, and therefore has particular factorization constraints. Jun 16 '20 at 5:59
• thanks I'll try now Jun 16 '20 at 6:14
• I've done something Jun 16 '20 at 7:20
• $N=(a_1!-1)(a_2!-1)...(a_n!-1)=k^2+16$ Jun 16 '20 at 7:21

Suppose there are $$n$$ and $$k$$ such that $$(a_1!-1)(a_2!-1)...(a_n!-1)=k^2+16=k^2+4^2\quad (*).$$ Notice that $$4$$ can not divide the LHS, and consequently $$\mathrm{gcd}(k,4)=1$$. Then by step $$4$$ here, every factor of $$k^2+4^2$$ is a sum of two squares. That is, for all $$i=1,\ldots,n$$, there are integers $$a$$ and $$b$$ such that $$a_i!-1=a^2+b^2.$$ If $$a_i\geq 4$$, then $$a_i!-1\equiv -1\pmod 4$$, but $$a^2+b^2\not\equiv -1 \pmod 4$$. Hence we must have $$a_i\leq 3$$ for all $$i=1,\ldots,n$$.
It follows that the LHS in $$(*)$$ is either zero, which is impossible, or equal to $$(3!-1)^m=5^m$$ for some $$0\leq m \leq n$$.
Now let's solve $$k^2+4^2=5^m.$$
• If $$m$$ is even, say $$m=2c$$, then $$k^2+4^2=5^{2c}$$ yields $$2^4=4^2=(5^c-k)(5^c+k)$$. Consequently $$5^c-k=2^{u}$$ and $$5^c+k=2^v$$ for some $$u,v\geq 0$$ with $$u+v=4$$. This is easy to solve, by just discussing cases.
• If $$m$$ is odd, say $$m=2c+1$$, then $$k^2+4^2=5\cdot 5^{2c}$$ yields $$k^2=((2\omega -1) 5^c-4)((2\omega -1) 5^c+4),$$ where $$w=\frac{1+\sqrt{5}}{2}$$.
The ring $$\mathbb{Z}[\omega]$$, which is the ring of integers of $$\mathbb{Q}(\sqrt{5})$$, is a UFD; see oeis.org/A003172. Moreover $$(2\omega -1) 5^c-4$$ and $$(2\omega -1) 5^c+4$$ are coprime (because if $$p$$ is irreducible and divides $$(2\omega -1) 5^c-4$$ and $$(2\omega -1) 5^c+4$$, then $$p\bar{p}\mid 8$$ and so $$p\bar{p}=2\mid k^2$$, which is impossible). Hence both factors are squares in $$\mathbb{Z}[\omega]$$, that is, $$(2\omega -1) 5^c-4=(e+\omega f)^2$$ for some $$e,f\in \mathbb{Z}$$. But using the fact that $$\omega^2= 1+\omega$$, we can see that the last equation has no solutions in $$e,f\in \mathbb{Z}$$.
Consequently all the solutions, if any (as I didn't do the computation), come from the case where $$m$$ is even.