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.

  • $\begingroup$ 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

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:
$\{2,792\} \implies a=(2+792)/2 = 397, n=155497$

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

  • $\begingroup$ 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

Hint: With $w=z/2$, what can you say about $(y+w)(y-w)$?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.