Let $a_1, a_2, …,a_9$ be $9$ distinct positive integers. My question is, when(What properties should $a_i$ have) does there exist an integer $n$ so $a_1+n,a_2+n,…,a_9+n$ are all perfect squares?
EDIT 1
If there does exist an $n$ for which $a_1+n,a_2+n,…,a_9+n$ are all perfect squares, can we say there is an integer $n' < n$ so that $a_1 + n',a_2+n',…,a_9+n'$ are also all perfect squares?
EDIT 2
Here's some information about $a_1,a_2,…,a_9$:$1.$They all have the same parity. $2.$ If such $n$ exists, then $a_i+n \equiv 1$(mod 3).
Here's what I've tried:
Let $A = \{a_1,a_2,…,a_9\}$ and $x,y\in A$ where $x < y$. Now, obviously $y = x + r$ for some positive integer $r$ so we are looking for two squares whose difference is $r$. Let those two squares be $t^2$ and $(t + s)^2$ for some natural numbers $t$ and $s$. $(t+s)^2 – t^2 = 2ts + s^2$ so we need to be able to write $r$ as $2ts + s^2$ for some natural numbers $t$ and $s$ but I don't know when $r$ has this property or not.
UPDATE $1$(Before the edit)
$r = 2ts + s^2 = s(2t + 1)$ and if $s = 1$, then $r = 2t + 1$ so if $r$ is odd it can be written as $2ts + s^2$ and we're done?
UPDATE $2$ (Before the edit)
There's at least $5$ items of $A$ which have the same parity which results in $r$ being even for some $a_i$ and $a_j$ and therefore $s$ can't always be $1$ but will always be a divisor of every $r$.
Thanks in advance for any help!
Best Answer
I don't think the answer is easy to analyse.
Consider each mod. For example, squares are $\equiv0,1\pmod 4$ so there are at most $2$ residues $\pmod 4$ for your $9$ numbers. You can do the similar thing for all other mods.
On the other hand, choose any 9 squares and subtract by any number (as long as the smallest >0) will give such a set of $a_i$s.