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 Post subject: Exactly one squarePosted: Fri Jan 15, 2016 10:54 pm

Joined: Mon Nov 09, 2015 11:52 am
Posts: 76
Location: Limassol/Pyla Cyprus
Let $x$ and $y$ be positive integers such that $x+xy$ and $y+xy$ are both perfect squares.
(a) Show that exactly one of $x$ and $y$ is a perfect square.
(b) Can you characterise all such pairs?

Edit after the solution: This was problem 10213 from the American Mathematical Monthly proposed by P. G. Walsh from the University of Waterloo. The solution appeared in the issue of February 1995.

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 Post subject: Re: Exactly one squarePosted: Fri Jan 15, 2016 10:55 pm

Joined: Wed Nov 11, 2015 6:18 am
Posts: 4
First of all, we note that $x$ and $y$ cannot be both squares. Indeed, suppose that $x$ is a square. Since $xy+x=x(y+1)$ is a square, $y+1$ must also be a square. But then (since $y >0$ ) $y$ cannot be a square.

Suppose now that $x$ and $y$ are positive integers such that both $x(y+1)$ and $y(x+1)$ are squares. We can write $x = au^2$ and $y= bv^2$ for some positive integers $a, b, u, v$, where $a$ and $b$ are square-free. We need to show that $a =1$ or $b =1$.

Since $au^2(bv^2 +1)$ and $bv^2(au^2 +1)$ are perfect squares, there exist positive integers $s$ and $t$ such that $a\left( {b{v^2} + 1} \right) = {\left( {as} \right)^2}$ and $b\left( {a{u^2} + 1} \right) = {\left( {bt} \right)^2}$. It follows that $\boxed{a{s^2} - b{v^2} = 1}$ and $\boxed{a{u^2} - b{t^2} = - 1}$. Note that $\gcd \left( {a,b} \right) = 1$.

Let $z = st - uv$. Then, we have:

$\left( {b{v^2}} \right)\left( {a{u^2}} \right) = ab{\left( {st - z} \right)^2} \Leftrightarrow \left( {a{s^2} - 1} \right)\left( {b{t^2} - 1} \right) = ab{\left( {st - z} \right)^2} \Leftrightarrow$

$\Leftrightarrow ab{s^2}{t^2} - a{s^2} - b{t^2} + 1 = ab{s^2}{t^2} - 2abstz + ab{z^2} \Leftrightarrow$

$\Leftrightarrow a{s^2} + b{t^2} + ab{z^2} - 1 = 2abstz$ $\bf \color{red} {(1)}$.

If we set $m = as^2$, $n = bt^2$ and $p = abz^2$, then $\bf \color{red} {(1)}$ is equivalent to
$\boxed{m + n + p - 1 = 2\sqrt {mnp} } \bf \color{red} {(\bigstar)}.$
Claim : At least one of the numbers $m, n$ and $p$ is a perfect square.

It follows from the Claim that at least one of the numbers $a, b$ and $ab$ is a square. But $a$ and $b$ are square-free and $\gcd \left( {a,b} \right) = 1$, so $a =1$ or $b = 1$, answering the first part of the problem.

It remains to prove the Claim. Rewrite equation $\bf \color{red} {(\bigstar)}$ as

${\left( {m + n + p - 1} \right)^2} = 4mnp \bf \color {red} {(2)}$.

Let $\alpha = 2m - 1$, $\beta = 2n -1$ and $\gamma = 2p-1$. Then, equation $\bf \color {red} {(2)}$ is equivalent to

${\left( {\alpha + \beta + \gamma + 1} \right)^2} = 2\left( {\alpha + 1} \right)\left( {\beta + 1} \right)\left( {\gamma + 1} \right) \Leftrightarrow$

$\Leftrightarrow {\alpha ^2} + {\beta ^2} + {\gamma ^2} - 2\alpha \beta \gamma = 1 \bf \color {red} {(3)}$.

Now equation $\bf \color {red} {(3)}$ is equivalent to each of the equations

$\left( {{\alpha ^2} - 1} \right)\left( {{\beta ^2} - 1} \right) = {\left( {\alpha \beta - \gamma } \right)^2}$

and

$\left( {{\beta ^2} - 1} \right)\left( {{\gamma ^2} - 1} \right) = {\left( {\beta \gamma - \alpha } \right)^2}$,

from which we infer that there exist non-negative integers $d, e, f, g$, with $d$ not a perfect square, such that

${\alpha ^2} - 1 = d{e^2}$, ${\beta ^2} - 1 = d{f^2}$, ${\gamma ^2} - 1 = d{g^2}$,

$\left| {\alpha \beta - \gamma } \right| = def$ and $\left| {\beta \gamma - \alpha } \right| = dfg$.

Now we invoke the theory of the Pell equation. Let $\left( {{x_1},{y_1}} \right)$ be the fundamental solution of the Pell equation ${X^2} - d{Y^2} = 1$. If we set ${z_1} = {x_1} + {y_1}\sqrt d$, then it is known that all its solutions $\left( {{x_k},{y_k}} \right)$ (where $k \ge 0$ ) are given by
${x_k} = \frac{1}{2}\left( {z_1^k + z_1^{ - k}} \right), \mbox{ } {y_k} = \frac{1}{{2\sqrt d }}\left( {z_1^k - z_1^{ - k}} \right).$
It follows that there exist non-negative integers $k_1, k_2$ and $k_3$, such that
$\alpha = \frac{1}{2}\left( {z_1^{{k_1}} + z_1^{ - {k_1}}} \right),\beta = \frac{1}{2}\left( {z_1^{{k_2}} + z_1^{ - {k_2}}} \right),\gamma = \frac{1}{2}\left( {z_1^{{k_3}} + z_1^{ - {k_3}}} \right).$
Assume, without loss of generality, that $m \ge n \ge p$. Then, we have $k_1 \ge k_2 \ge k_3$ and $\alpha \beta - \gamma = def$. Hence,
$\gamma = \alpha \beta - def =$
$=\frac{1}{4}\left( {z_1^{{k_1}} + z_1^{ - {k_1}}} \right)\left( {z_1^{{k_2}} + z_1^{ - {k_2}}} \right) - \frac{1}{4}\left( {z_1^{{k_1}} - z_1^{ - {k_1}}} \right)\left( {z_1^{{k_2}} - z_1^{ - {k_2}}} \right) =$
$= \frac{1}{2}\left( {z_1^{{k_1} - {k_2}} + z_1^{ - \left( {{k_1} - {k_2}} \right)}} \right).$
We conclude that $k_3 = k_1 - k_2$, so not all of the numbers $k_1, k_2$ and $k_3$ are odd. Suppose that $k_1$ is even. Then, we can write
$m = \frac{{\alpha + 1}}{2} = {\left[ {\frac{1}{2}\left( {z_1^{\frac{{{k_1}}}{2}} + z_1^{ - \frac{{{k_1}}}{2}}} \right)} \right]^2},$
so $m$ is a perfect square and the Claim follows.

To answer the second question. i.e. to characterize all pairs $(x,y)$ of positive integers with the property that both $x(y+1)$ and $y(x+1)$ are squares, we may proceed as follows: Without loss of generality, we may assume that $x$ is a square. Then, $y+1$ will also be a square, so there exist positive integers $r$ and $s$ such that $x = r^2$ and $y+1 = s^2$. We also have that $y(x+1) = (r^2 +1)(s^2 -1)$ is a square. Let $d = \gcd \left( {{r^2} + 1,{s^2} - 1} \right)$. Then, there exist positive integers $t$ and $u$ such that

${r^2} + 1 = d{t^2} \Longleftrightarrow r^2 - dt^2 =-1$

and

${s^2} - 1 = d{u^2} \Longleftrightarrow s^2-du^2 =1$.

Thus, the pair $(r,t)$ of positive integers is a solution of the negative Pell equation $X^2 - dY^2 = -1$. Depending on $d$, this has either no solutions or infinitely many solutions.

According to the theory of the Pell equation, if $d$ is such that the above negative Pell equation is solvable in the positive integers, then we can find positive integers $r,s,t$ and $u$ such that ${r^2} + 1 = d{t^2}$ and ${s^2} - 1 = d{u^2}$. Then the pair $\left( {x,y} \right) = \left( {{r^2},d{u^2}} \right)$ has the desired property.

One could possibly characterize (via continued fractions) all $d$ such that the negative Pell equation $X^2 - dY^2 = -1$ has (infinitely many) solutions and therefore obtain a more precise answer to the second part of the problem.

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 Post subject: Re: Exactly one squarePosted: Fri Jan 15, 2016 11:02 pm

Joined: Mon Nov 09, 2015 11:52 am
Posts: 76
Location: Limassol/Pyla Cyprus
This was problem 10213 from the Monthly. I will update my original post with the details. In the proposed solution by Robin Chapman it shows that the negative Pell equation $X^2 - abY^2 = -1$. has a solution. For example, with Vangelis' notation $(aus+bvt,uv+st)$ is such a solution. Since now $(2x+1)^2 - ab(2cd)^2 = 1$ it follows by the theory of Pell's equation that $2x+1 + 2cd\sqrt{ab} = (p+q\sqrt{ab})^2$ for some $p, q$ with $p^2 - abq^2 = \pm 1.$. With the choice of $-1$ this gives $x=p^2$ and with the choice of $+1$ this gives $x+1=p^2$ in which case $y$ is a perfect square.

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