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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!

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  • $\begingroup$ Edit 2 in effect says $a_i=6n+1\ \text{or}\ a_i=6n+4$ $\endgroup$ – Keith Backman Jun 6 '19 at 1:20
  • $\begingroup$ @KeithBackman My bad. By $a_i \equiv 1$(mod 3), I meant $a_i+n \equiv 1$(mod 3). Fixed it. $\endgroup$ – Borna Ghahnoosh Jun 6 '19 at 1:56
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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.

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  • $\begingroup$ I updated the question. Could you please explain to me why I might be wrong(Since you said the answer might be difficult)? $\endgroup$ – Borna Ghahnoosh Jun 5 '19 at 12:56
  • $\begingroup$ @BornaGhahnoosh you said in your answer about two numbers. Consider add a third number in and see what happens. The point here is that maybe $t$ and $s$ can add 40 so they both become a square however $t$ and $k$ need to add 50 so they both become a square. $\endgroup$ – abc... Jun 5 '19 at 13:02
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You have found $s$ must be a factor of $r$. So, given, $a_1$ and $a_2$, there are only a few possible $n$ that will work. For each, test the other seven $n+a_i$.

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  • $\begingroup$ I actually can't do that because I have many, many, many sets of $9$ natural numbers and I need to see if there is any set for which that $n$ exists. $\endgroup$ – Borna Ghahnoosh Jun 5 '19 at 13:28
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Wlog. $a_1<a_2<\ldots <a_9$. If $a_i+n=s_i^2$, then $a_i-a_j=(s_i+s_j)(s_i-s_j)$. Hence all differences must be factorable into same-parity factors (i.e., cannot be $\equiv 2\pmod 4$). We also get a bound $a_9-a_1\ge s_9^2-(s_9-8)^2=16s_9-64$, i.e., $$s_9\le \frac{a_9-a_1}{16}+4.$$ This already leaves us with a limited choice for $0\le s_1<\ldots<s_9$.

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The question as posited assumes the numbers $a_i$ are arbitrarily chosen, and suitable $n$ is to be identified within that arbitrary setting. Within the confines of that assumption, the answer to the stated question is plainly "No". Without knowing more about the $a_i$, we cannot say that suitable $n$ do or do not exist. For example, if it happened that $a_i=i$, then the $9$ resulting $a_i+n$ would be consecutive numbers, and there are no nine consecutive squares. On the other hand, if the $a_i$ all happened to be $1$ less than a square, then $n=1$ would work. So the real question becomes, what are the limitations on $a_i$ that would permit an $n$ to be found? If such conditions were to be identified, then almost perforce that would identify suitable $n$ as well.

As to the question added in your edit, if the answer to that question were in general "Yes," then by descent, the proposition would be true for $n=1$. In that case, each of the $a_i$ would have to be (fortuitously) $1$ less than a perfect square. And if the $a_i$ were all $1$ less than a perfect square, then it is doubtful (I haven't looked in detail) that they would all be $n$ less than a different set of squares; that would require finding $9$ sets of two squares where the difference between the squares in each set was $n-1$.

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  • $\begingroup$ I just added some new information about $A$. Could you please update your answer according to that? $\endgroup$ – Borna Ghahnoosh Jun 5 '19 at 21:30

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