Suppose $a_1, a_2, \cdots, a_n$ are positive integers, prove that if $a_1^2+a_2^2+\cdots+a_n^2$ is divisible by $a_1a_2\cdots a_n+1$, then there exists nonnegative integers $x_1, x_2, \cdots, x_{n-1}$ such that $$\frac{a_1^2+a_2^2+\cdots+a_n^2}{a_1a_2\cdots a_n+1}=x_1^2+x_2^2+\cdots+x_{n-1}^2.$$
Here is my approach. $n=1$ is trivial, and $n=2$ is already well-known $(1988/6)$. By Lagrange's four square theorem, $n\geqslant 5$ are also obvious. What's left is $n=3$ and $4$. For $n=3$, by the sum of two squares theorem, if I use the standard method of Vieta jumping, setting the quotient of $a_1^2+a_2^2+\cdots+a_n^2$ and $a_1a_2\cdots a_n+1$ as $k$, then we now have to prove there must not exist a prime $p$ s.t. $p^\alpha \,|\,k$ and $4\,|\,p-1$. I think achieving this result is much harder than the $n=2$ case because I may need to investigate the prime decomposition of $k$, which Vieta jumping might not be that helpful. For $n=4$, by the sum of three squares theorem, we have to prove that $k$ can be expressed as the form $k=4^\alpha(8s+7)$. Both of these cases are difficult to me and I haven't come up with a good way to approach this problem.
Partial Solution
(Very Little! Partial Solution for $n=3$ by Vieta Jumping)
Let $(a,b,c)$ be the minimal solution with $a\geqslant b\geqslant c\geqslant 1$. View it as a quadratic equation of $a$, then by Vieta's formula, we know that there exists $a'$ s.t. $a+a'=kbc$, $aa'=b^2+c^2-k$, so $$a'=\frac{b^2+c^2-k}{a}\leqslant 2a-\frac k a=2a-\frac{a+a'}{abc},$$ which is $$\left(1+\frac 1 {abc}\right)a'\leqslant 2a-\frac 1{bc}.$$ Therefore, if $a'>0$ $$a\leqslant a'\leqslant \frac{2abc-1}{bc}\,\frac{abc}{abc+1}=2a-\frac{3a}{abc+1}\leqslant 2a-\frac 3 {bc+1}.$$ We get $3a\leqslant a(abc+1)$, so $abc\geqslant 2$, which implies $a\geqslant 2$. Let $f(x)=x^2-kbcx+\left(b^2+c^2-k\right)$, then if $a=b$, $f(b)=0$. By $f(c)\geqslant 0$, we have $$b^2+2c^2-kbc^2-k-\left(2b^2+c^2-kb^2c-k\right)\geqslant 0 \implies kbc\geqslant b+c.$$
Hope
- I kinda want to prove a strict inequality $a>b>c$.
- If I proved all the solutions can be generated by a trivial solution $(x_1, x_2, \cdots, x_n)=(0, \square , \square, \cdots, \square)$, then we're done.
Generalizations
Will the generalizations help?
If we change the $1$ to $C$, i.e. $a_1a_2\cdots a_n+C\,|\,a_1^2+a_2^2+\cdots+a_n^2$, can we still make a similar conclusion? $$\frac{a_1^2+a_2^2+\cdots+a_n^2}{a_1a_2\cdots a_n+C}=\;???$$
This also reminds me of the Hurwitz equations, which states that if $C\in \mathbb N$, and $x_1, x_2, \cdots, x_n$ where $n>1$, then equation $$x_1^2+x_2^2+\cdots +x_m^2=Cx_1x_2\cdots x_n$$ has at least one solution if and only if $C\leqslant n$. Here, $C$ is not necessarily a square or a sum of a few squares (e.g. when $n=3$, $C=1,3$, $C=3$ is the Markov equation).
Edit (October 21, 2022)
I know how to solve the $n=3$ case now (strengthening to $k=\alpha^2+\beta^2$ where $\alpha, \beta \in \mathbb N$), but not the $n=4$ case. The proof doesn't incorporates the sum of two squares theorem.
Proof. Suppose $(a_0, b_0, c_0)$ is the minimal solution of the equation $$\frac{a^2+b^2+c^2}{abc+1}=k$$ where $k$ is an positive integer and $a_0\geqslant b_0\geqslant c_0\geqslant 1$. By Vieta's formulas we know that there exists $a'\in\mathbb Z$ s.t. $a_0+a'=kb_0c_0$ and $a_0a'=b_0^2+c_0^2-k$.
Case 1: $a'<0$. Then $a'b_0c_0+1\leqslant 0 \implies k(a'b_0c_0+1)\leqslant0 <a'^2+b_0^2+c_0^2$, which is a contradiction.
Case 2: $a'=0$. $k=b_0^2+c_0^2$. Correct.
Case 3: $a'>0$. By the minimality of $a_0$, we must have $a'\geqslant a_0\geqslant b_0\geqslant c_0\geqslant1$, so $$a'b_0\leqslant a_0a'=b_0^2+c_0^2-k\leqslant 2b_0^2-k<2b_0^2 \implies a'<2b_0\,,$$ $\implies 4b_0=2b_0+2b_0>a_0+a'=kb_0c_0$ $\implies 4>kc_0$. If $k$ cannot be represented as a sum of two squares $k$ must be $3$, and $c_0=1$. Then, $b^2\leqslant a_0a'=b_0^2+c_0^2-k=b_0^2-2$, which is also a contradiction.
I don't know if a geometric perspective, or some advanced number theory techniques would help.
the $n=4$ case.
Proof. Suppose $(a_0, b_0, c_0,d_0)$ is the minimal solution of the equation $$\frac{a^2+b^2+c^2 + d^2}{abcd+1}=k$$ where $k$ is an positive integer and $a_0\geqslant b_0\geqslant c_0\geqslant d_0 \geq 1$. By Vieta's formulas we know that there exists $a'\in\mathbb Z$ s.t. $a_0+a'=kb_0c_0d_0$ and $a_0a'=b_0^2+c_0^2 + d_0^2-k$.
Case 1: $a'<0$. Then $a'b_0c_0d_0+1\leqslant 0 \implies 0 \geq k(a'b_0c_0d_0+1) =a'^2+b_0^2+c_0^2 + d_0^2$, which is a contradiction.
Case 2: $a'=0$. $k=b_0^2+c_0^2 + d_0^2$. We like this one.
Case 3: $a'>0$. By the minimality of $a_0$, we must have $a'\geqslant a_0\geqslant b_0\geqslant c_0\geqslant d_0 \geq 1$, so $$a'b_0\leqslant a_0a'=b_0^2+c_0^2 + d_0^2-k\leqslant 3b_0^2-k<3b_0^2 \implies a'<3b_0\,,$$ $\implies 6b_0=3b_0+3b_0>a_0+a'=kb_0c_0d_0$ $\implies 6>kc_0 d_0$. At this point we know $k \leq 5$ and $k$ is the sum of three squares.
In case of interest, one may check all the cases of $kc_0 d_0 \leq 5$ for more detail. Each case leads to a binary, for example $k=3, c_0 = d_0 = 1$ leads to $a_0^2 - 3a_0 b_0 + b_0^2 = 1.$ This does have solutions such as $a_0 = 3, b_0 = 1.$ Another one with solutions is $k=1, c_0 = 3, d_0 = 1$ leading to $a_0^2 - 3a_0 b_0 + b_0^2 = -9$ including $a_0 = b_0 = 3$
For larger $n$ the final case $a' \geq a_0$ becomes $2n-3 \geq k c_0 d_0 e_0..$ This is quite restrictive and is similar to formula (15) in Hurwitz 1907. The conclusion is that all but a finite number of these, for a specific $n,$ have $k = b_0^2 + c_0^2 + d_0^2 + e_0^2 \ldots$