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Find all the integers $n$ such that $n + 2112$ and $4n + 2112$ are perfect squares

First,for $$ n + 2112 = y^2 $$ The possible values for $n$ here are $\{...,-87,4,97,192,...\}$ and it's a quadratic series of form $x_1^2+90x_1-87$

Then, for $$4n + 2112=z^2$$ I factorized $2112$ as $11*3*2^6$ so $$4n + 11*3*2^6= 4(n + 528)$$ So $(n + 528)$ has to be a square number, then $n=\{...,-44,1,48,97,... \}$ and it's a quadratic series of form $x_2^2+44x_2-44$.

Then all the values of $n$ are given when $$x_1^2+90x_1-87=x_2^2+44x_2-44$$ An example is $n=97$, with it, $x_1=2$ and $x_2=3$, But i didn't figure how to get all the possible values with that equality. So is this way correct? is there other ways to do it? thanks.

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  • $\begingroup$ For $n + 2112 = m^2$ $$\sum_{n = 1}^{p} n^3 = \bigg(\sum_{n = 1}^{p} n\bigg)^2$$ $$\therefore \text{if 2112 can be expressed as the sum of cubed integers then we can find the value(s) of}\ n$$ However it can't, so we know that $n \neq k^3$ $\endgroup$ – George N. Missailidis Aug 18 '17 at 4:08
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You have $a^2:=n+2112$ and $b^2:=n+528$. Then $(a^2-b^2) = 1584 = 2^4\cdot3^2\cdot 11$.

$1584$ thus has $5\cdot 3\cdot 2=30$ factors, but for $(a+b)(a-b)$type factors we need both even, so only $3\cdot 3\cdot 2=18$ factors $\implies 9 $ factor pairs:
$\{2,792\} \implies a=(2+792)/2 = 397, n=155497$
$\{4,396\}$
$\{8,198\}$
$\{6,264\}$
$\{12,132\}$
$\{24,66\}$
$\{18,88\}$
$\{36,44\}$
$\{22,72\}$

and so on to find the $9$ possible values of $n$. Your $n=97$ comes from the last of these.

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  • $\begingroup$ How do you know that there are 18 even factors of 1584? i mean, i know how to get a number's factors but not only the even as you did without searching them manually $\endgroup$ – SonodaUmi Aug 18 '17 at 4:31
  • $\begingroup$ you need to have the powers of $2$ split 1:3, 2:2 or 3:1 - that 's the first $3$ in the multiplication of options there. $\endgroup$ – Joffan Aug 18 '17 at 4:36
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Hint: With $w=z/2$, what can you say about $(y+w)(y-w)$?

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