4
$\begingroup$

Find all positive integers $n$ and $m$ such that $$(2mn+1)^2-4mn(m+n)+n^2+m^2+(n-1)^2+(m-1)^2-3$$ is a perfect square.

Of course $n=m=1$ is a trivial solution and if you put $n=m+1$ (assume WLOG that $n \geq m$), the above expression is equal to $4n^4$ which is a perfect square - But I don't know if this is the only solution. I would really appreciate any help.

$\endgroup$
  • $\begingroup$ The polynomial can be expressed as $\frac{1}{4}\left((2m-1)^2+1\right)\left((2n-1)^2+1\right)-1$ $\endgroup$ – ccorn May 19 '13 at 10:59
  • $\begingroup$ To make this a square using positive integers $m,n$, one can employ a Pell equation. See my answer below. $\endgroup$ – Tito Piezas III May 20 '13 at 3:02
3
$\begingroup$

Given the OP's form,

$$(2mn+1)^2-4mn(m+n)+n^2+m^2+(n-1)^2+(m-1)^2-3 = w^2\tag{1}$$

ccorn and Alyosha showed it is equivalent to the more aesthetic forms,

$$\tfrac{1}{4}\big((2m-1)^2+1\big)\big((2n-1)^2+1\big)-1 =w^2$$

$$(m+n-2mn-1)^2+(n-m)^2-1 =w^2$$

respectively. There are infinitely many positive integer solutions to (1) other than $n=m+1$ as we can use a Pell equation $x^2-dy^2=1$ to solve it. In fact, we can use infinitely many $d$. For any positive integer $m>1$, the solution is then,

$$\begin{aligned} &n =(2m-3)x^2+4(m-1)^2xy-(m-2)\\ &w = 4(m-1)^2x^2+(2m-3)\big((2m-1)^2+1\big)xy-2(m-1)^2 \end{aligned}$$

where $x,y$ satisfy the Pell equation,

$$x^2-\big((2m-1)^2+1\big)y^2=1$$

For the simplest case when $m=1$, we have the alternative and simpler form,

$n = (x+1)/2,\;\; w=y,\;\; \text{where}\;\;x^2-2y^2=1.\tag{2}$

(Note: Since the $x$ of (2) is odd, then $n$ is an integer.)

$\endgroup$
  • $\begingroup$ Thank you. I was hiding under my only-a-comment, but really I had no idea how to progress. $\endgroup$ – Meow May 20 '13 at 17:16
  • $\begingroup$ Your welcome. Every time I come across a Diophantine equation with deg 2 (and sometimes deg 3 and 4) where one term $x_1 = 1$, I always try if a subset of solutions can be given by a Pell equation. (Doesn't always work, but sometimes it does.) $\endgroup$ – Tito Piezas III May 20 '13 at 17:24
4
$\begingroup$

(Too long for a comment) I think the question can be made clearer with a rearrangement. The $2(2mn)(m+n)$ smells like a quadratic: $$((m+n)-(2mn+1))^2=(2mn+1)^2-2(2mn+1)(m+n)+m^2+n^2+2mn$$ $$=(2mn+1)^2-4mn(m+n)-2(m+n)+m^2+n^2+2mn$$ Rearranging, $$=(2mn+1)^2-4mn(m+n)+m^2+n^2+2mn-2(m+n)$$ $$=[(2mn+1)^2-4mn(m+n)+m^2+n^2]+2mn-2(m+n)$$ And substituting, $$(2mn+1)^2-4mn(m+n)+m^2+n^2+(n-1)^2+(m-1)^2-3$$ $$=((m+n)-(2mn+1))^2-2mn+2(m+n)+(n-1)^2+(m-1)^2-3$$ $$=((m+n)-(2mn+1))^2-2mn+2m+2n+n^2-2n+m^2-2m-1$$ $$=((m+n)-(2mn+1))^2-2mn+n^2+m^2-1$$ $$=(m+n-2mn-1)^2+(n-m)^2-1$$

It's clear now why $n=m+1$ worked. It's equally clear that $m+n-2mn-1=1 \Rightarrow n=\frac{m-2}{2m-1}$ is a solution (although no positive integer solutions exist).

$\endgroup$

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.