Does there exist a positive integral solution $(x, y, n)$ to $(xy+1)(xy+x+2)=n^2$? If there doesn't, how does one prove that?
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2$\begingroup$ Do you mean, does there exist an $n \in \mathbb N$ such that...? $\endgroup$– Michael StollCommented Oct 21, 2018 at 9:22
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$\begingroup$ @MichaelStoll Yes. Or just simply, does there exist (x, y) such that the expression can become a square number $\endgroup$– Benkerd22Commented Oct 21, 2018 at 9:54
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3$\begingroup$ Alternatively, you could state it as: does there exist a positive integral solution (x, y, n) to...? $\endgroup$– Robin ZhangCommented Oct 21, 2018 at 21:06
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$\begingroup$ No solutions below $10^4$. $\endgroup$– WojowuCommented Oct 21, 2018 at 21:27
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9$\begingroup$ Where did this problem originate? $\endgroup$– Yaakov BaruchCommented Oct 22, 2018 at 4:18
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3 Answers
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It looks that Vieta jumping helps.
For fixed positive integer $y$ choose a minimal positive integer $x$ for which $(xy+1)(xy+x+2)$ is a perfect square.
Denote $4(xy+1)(xy+x+2)=4n^2=(2xy+x+3-z)^2$ for some integer $z=2n-2xy-x-3$, this yields $0<z<x+3$ and rewrites as $z^2-2z(2xy+x+3)+x^2+2x+1=0$. Note that $x$ must divide $z^2-6z+1$, for each $z\leqslant 5$ this gives several variants for $x$ for which it is straightforward to check that $y$ does not appear to be a positive integer. If $z\geqslant 6$, we may replace $x$ to $x'=(z^2-6z+1)/x>0$ (which is another root of the same quadratic equation in $x$.) This contradicts to the minimality since $z^2-6z+1<(z-3)^2<x^2$.
Remark: for the new pair $(x',y)$ we have different value of $z$, as $2x'y+x'+3-z$ becomes negative, but it is still true that $(x'y+1)(x'y+x'+2)$ is a perfect square.
answered Oct 21, 2018 at 22:02
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$\begingroup$ Do you mean $z \geq 6$ instead of $z \geq 3$? (The cases $z \in \{3,4,5\}$ are also easily ruled out.) $\endgroup$ Commented Oct 21, 2018 at 22:36
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This is not really an answer, but you can at least observe that $xy+1 \neq xy + x + 2$ if $x$ is positive.
Then consider what kind of common prime factors they could have, since $(xy + 1)(xy + x + 2)$ is a square if and only if the square-free parts of $xy + 1$ and $xy + x + 2$ are the same. Perhaps it is relevant that since their difference is $x + 1$, we know that $\gcd(xy + 1, xy + x + 2)$ divides $x + 1$?
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2$\begingroup$ Since $xy+1 = (x+1)y -(y-1),$ you can continue to say that ${\rm gcd}(xy+1,xy+x+2)$ divides ${\rm gcd}(x+1,y-1)$ ( and then note that those two gcd's are in fact equal). $\endgroup$ Commented Oct 21, 2018 at 21:28
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$4n^2=4(xy+1)(xy+x+2)=(3+x+2xy)^2-(1+x)^2$.
For positive even $k$ and $ka^2+b^2=c^2$, when $a,b,c$ is positive integers, and $a=2m,b=km^2-1,c=km^2+1$, when $m$ is any positive integer. Then $n=2m,1+x=4m^2-1,3+x+2xy=4m^2+1$, and then $2m^2=1$, i.e. $m$ is not integer - contradiction. Means equation $n^2=(xy+1)(xy+x+2)$ has no positive integer solution.
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3$\begingroup$ So if $ka^2=c^2 - b^2 $, we must have $c-b=2$? But why? $\endgroup$ Commented Mar 13, 2019 at 16:07
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$\begingroup$ $4\times15^2=900=1156-256=34^2-16^2$ and $34-16\ne2$. $\endgroup$ Commented Mar 14, 2019 at 0:49
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$\begingroup$ Yes, it is not full parametrization. Answer not true, sorry. $\endgroup$ Commented Mar 14, 2019 at 3:33