Prove that the diophantine equation $(xz+1)(yz+1)=az^{3} +1$ has no solutions in positive integers $x, y, z$ with $z>a^{2} +2a$.

Question

Let $a$ be a positive integer that is not a perfect cube. From experimental data, it appears all solutions to $(xz+1)(yz+1)=az^{3} +1$ in positive integers $x, y, z$ occur when $z \le a^{2} +2a$ i.e it appears there are no solutions in $x, y,z$ with $z> a^{2} +2a$. Can this observation be proved?

To motivate the question, we shall prove that on the contrary if $a$ is a perfect cube, there are infinitely many positive integer solutions in $x, y, z$.

Proof. Let $a=m^{3} $ for some integer $m$. Using the identity $n^{3} +1 =(n+1)(n^{2}-n+1)$, we see that $az^{3} +1=(mz)^{3} +1= (mz+1)((mz)^{2}-mz+1) $.

A family of solutions is then given by $x=m$, $y=m^{2}z - m$ where $z$ takes on any positive integer.

How do I go about proving the striking observation: There are no positive integer solutions $x, y, z$ with $z>a^{2} +2a$ when the integer $a$ is not a perfect cube? Is there any counterexample?

Answer 1

A concise Proof based on an earlier proof:

The given equation $(xz+1)(yz+1)=az^3+1$ can be rewritten as $az^2-xyz-(x+y)=0$. We shall show that for any solution $(x,y,z)$, we have $z \le a^2+2a. \ $ Note that $z \ | \ x+y$, therefore $z \le x+y. \ $ Treating $x, y$ as constants, the only positive solution for $z \ $ is \begin{equation} z = \frac{xy+\sqrt{x^2y^2+4a(x+y)}} {2a} \end{equation} In order for $z$ to be rational, the discriminant must be a perfect square. Therefore $w^2 = x^2y^2+4a(x+y)$. We see that $w > xy$ and $w \equiv xy \ ( $mod$ \ 2)$. We can write $w = xy + 2t$, $t > 0$. Substituting $w$ above, $(xy+2t)^2 = x^2y^2+4a(x+y)$. Expanding and simplifying, $txy - ax - ay +t^2 = 0$. Multiplying through by $t$ and factoring, $(tx-a)(ty-a)=a^2 - t^3$. We must have $t \le a-1$ otherwise $RHS<0$ and $LHS \ge 0$. Because $a$ is not a perfect cube, $a^2 - t^3 \not = 0$. The remainder of the proof utilizes the result: If $ab = c \ $ where $a,b, c \not = 0$ are integers then $a+b \le c+1$ if $c>0$ and $a+b \le -(c+1)$ if $c < 0$. We now consider two cases:

Case $1: \ $ $ a^2 - t^3 >0 \ ;$ Using the result above on the factored equation, we have $(tx-a)+(ty-a) \le a^2 - t^3+1 \le a^2$. Hence, $z \le x+y \le (a^2+2a)/t \le a^2+2a \\$.

Case $2: \ $ $ a^2 - t^3 < 0 \ ;$ As in case $1$, we have $(tx-a)+(ty-a) \le t^3 - a^2-1 \ $, $x+y \le t^2 - (a^2-2a+1)/t < t^2$. Hence , $z \le x+y < t^2 \le (a-1)^2 < a^2 +2a$

Answer 2

Similar to finishing

Is it true that $f(x,y)=\frac{x^2+y^2}{xy-t}$ has only finitely many distinct positive integer values with $x$, $y$ positive integers?

where I had an acceptable bound but needed help from Gerry Myerson to improve to the sharp bound.

We have $$ (xz+1)(yz+1) = a z^3 + 1 $$

This becomes

$$ a z^3 - xyz^2 - (x+y)z=0$$ or $$ a z^2 - xyz - (x+y) = 0 $$

We get $$ z = \frac{ xy + \sqrt{ x^2 y^2 + 4a(x+y) } }{2a} $$

Let me also record

$$\color{fuchsia}{ z(az-xy) = x+y }$$ which follows directly from $ a z^2 - xyz - (x+y) = 0 $ Note also the simple $$\color{fuchsia}{ z \leq x+y }$$

It is necessary to have square discriminant to get a rational value for $z,$ take $$ w^2 = x^2 y^2 + 4a(x+y) $$ We have $$ w > xy $$ and $$ w \equiv xy \pmod 2. $$ Therefore we can define an integer $t,$ when it all works, with $$ w = xy+2t $$

Now $$ z = \frac{xy+w}{2a} = \frac{xy+xy+2t}{2a} = \frac{2xy+2t}{2a} = \frac{xy+t}{a} $$

$$ z = \frac{xy+t}{a} $$

There are always three flavors for any $a$ $$ t=a-1 \; , \; y = 1 \; , \; x = a^2 - 3a +1 \; , \; z = a-2 $$ $$ t=1 \; , \; y = 2a-1 \; , \; x = 2a +1 \; , \; z = 4a $$ $$ t=1 \; , \; y = a+1 \; , \; x = a^2 +a -1 \; , \; z = a^2+2a $$

From $$ x^2 y^2 +4a(x+y) = (xy+2t)^2 $$ we get $$ t xy - ax -ay + t^2 = 0, $$ $$ t^2 xy - tax -tay + t^3 = 0, $$

$$ \color{red}{(tx-a)(ty-a) = a^2 - t^3} $$

IF $a > 1$ and $t = a + \delta$ with $\delta \geq 0,$ we find

$$ ((a+\delta)x-a)((a+\delta)y-a) = a^2 - (a+\delta)^3 < 0 $$ since $a>1.$ However, the left hand side is non-negative, which is a contradiction.

$$ \color{red}{ t \leq a-1} $$

I will fill in the (lengthy) details in a bit.

I always have $x \geq y \geq 1$

IF $$ \color{blue}{ a^{2/3} < t \leq a-1} $$ we get

$$ (tx-a) (a-ty) = t^3 -a^2 > 0 $$ so $a-ty >0,$ $ty - a < 0,$ $$ ty < a $$ $$ y < \frac{a}{t} < a^{1/3} $$

$$ a - ty \geq 1 $$ $$ tx-a \leq t^3 - a^2 $$ $$ tx \leq t^3 - (a^2 - a)$$ $$ x \leq t^2 - \frac{a^2 - a}{t} $$

DETAIL: As $t$ increases, $t^2$ increases, while $\frac{1}{t}$ decreases. Then $\frac{-1}{t}$ increases. We have $a \geq 2$ so that $a^2 - a > 0,$ so that $\frac{-(a^2-a)}{t}$ increases. Together, $t^2 - \frac{a^2 - a}{t}$ increases and takes its maximum value at $t=a-1,$ that being $ a^2 - 3a + 1.$ Thus $$ \color{magenta}{x \leq a^2 - 3a + 1}$$ $$xy + t < a^{7/3} -3a^{4/3} + a + a^{1/3} -1 $$

$$ z < a^{4/3} -3a^{1/3} + 1 + a^{-2/3} -\frac{1}{a} $$

$$ \color{red}{ z < a^{4/3} } $$ when $a^{2/3} < t \leq a-1$

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

I always have $x \geq y \geq 1$

IF $$ \color{blue}{1 \leq t < a^{2/3} }$$

$$ (tx-a) (ty-a) = a^2 - t^3 > 0$$ $$ (tx-a) \leq a^2 -t^3$$ $$ tx \leq a^2 + a - t^3 < a^2 + a$$

$$ x \leq \frac{a^2 + a}{t} $$

Meanwhile $$ t^2 xy - ta(x+y)= -t^3 < 0 $$ $$txy < ta(x+y) \leq 2ax $$ $$ ty < 2a $$ y < \frac{2a}{t} $$

Together $$ xy < \frac{2 a^3 + 2a^2}{t^2} $$ $$ xy + t < \frac{2 a^3 + 2a^2}{t^2} + a^{2/3} $$ $$ z < \frac{2 a^2 + 2a}{t^2} + \frac{1}{a^{1/3}} $$

IF $t \geq 2$ then $z < \frac{a^2 + a}{2}$

IF $t=1$ we have $$ (x-a)(y-a) = a^2 - 1 > 0 $$

If $x>a$ then $y>a.$ Then $y-a \geq 1$ and $x-a \leq a^2 - 1$ When $t=1$ we have $x \leq a^2 + a - 1.$ In general, when we have real $p \geq 1, q \geq 1,$ and $pq=c,$

the maximum of $p+q$ occurs when $p=1$ and $q=c$ so that $p+q \leq 1+c$

With $ (x-a)(y-a) = a^2 - 1 $ we get $x-a+y-a \leq a^2.$ Thus $$ x+y \leq a^2 + 2a$$

With $t=1,$ we know $z = x+y.$ With $t=1$
$$ \color{red}{z \leq a^2 + 2a } $$

DETAIL $$\color{fuchsia}{ z(az-xy) = x+y }$$ and $$ z = \frac{xy+t}{a} $$ so that when $t=1,$ we get $az = xy+1$ or $az-xy = 1,$ so that $z(az-xy) = x+y$ tells us $z=x+y,$ when $t=1$

Answer 3

Let $a$ be a positive integer that is not a cube, and let $x$, $y$ and $z$ be positive integers such that $$(xz+1)(yz+1)=az^3+1.$$ Expanding the left hand side and rearranging a bit then shows that $$az^2-xyz-(x+y)=0,\tag{1}$$ so $z$ is an integral root of a quadratic with discriminant $x^2y^2+4a(x+y)$. In particular this discriminant is a perfect square, so there exists a positive integer $v$ such that $$x^2y^2+4a(x+y)=(xy+2v)^2,$$ and with a bit of rearranging we find the curious identity $$(a-xv)(a-yv)=a^2-v^3.$$ We see that $v<a$ as otherwise the right hand side is negative, whereas the left hand side is not. Applying the quadratic formula to $(1)$ shows that $$z=\frac{xy+\sqrt{x^2y^2+4a(x+y)}}{2a}=\frac{xy+(xy+2v)}{2a}=\frac{xy+v}{a},$$ where we have the $+$-sign because $z$ is positive. It follows that $$z<\frac{xy}{a}+1,$$ so now to prove that $z<a^2+2a$ it suffices to show that $xy<a(a+1)^2$.

Answer 4

Below is $a=47,$ I told it to write small $t$ first. Concentrating on $t=1$ we can see how $x+y=z$

This does not hold at all for $t\geq 2,$ where we see immediately that $x>z$

---

==================================  

 a: 47  t: 1 w 8837   x 95   y 93 z 188
 a: 47  t: 1 w 9166   x 116  y 79 z 195
 a: 47  t: 1 w 9871   x 139  y 71 z 210
 a: 47  t: 1 w 10012  x 143  y 70 z 213
 a: 47  t: 1 w 11657  x 185  y 63 z 248
 a: 47  t: 1 w 13631  x 231  y 59 z 290
 a: 47  t: 1 w 17767  x 323  y 55 z 378
 a: 47  t: 1 w 21997  x 415  y 53 z 468
 a: 47  t: 1 w 30551  x 599  y 51 z 650
 a: 47  t: 1 w 39152  x 783  y 50 z 833
 a: 47  t: 1 w 56401  x 1151 y 49 z 1200
 a: 47  t: 1 w 108242 x 2255 y 48 z 2303

 a: 47  t: 2 w 2305   x 59   y 39 z 49
 a: 47  t: 2 w 26980  x 1124 y 24 z 574

 a: 47  t: 3 w 11894  x 743  y 16 z 253

 a: 47  t: 4 w 568    x 28   y 20 z 12
 a: 47  t: 4 w 803    x 53   y 15 z 17
 a: 47  t: 4 w 1555   x 119  y 13 z 33
 a: 47  t: 4 w 6584   x 548  y 12 z 140

 a: 47  t: 6 w 2732   x 340  y 8 z 58

 a: 47  t: 7 w 994    x 140  y 7 z 21

 a: 47  t: 8 w 1324   x 218  y 6 z 28

 a: 47  t: 10 w 104   x 14   y 6 z 2
 a: 47  t: 10 w 245   x 45   y 5 z 5

 a: 47  t: 12 w 59    x 7    y 5 z 1
 a: 47  t: 12 w 200   x 44   y 4 z 4

 a: 47  t: 14 w 61    x 11   y 3 z 1

 a: 47  t: 15 w 156   x 42   y 3 z 3

 a: 47  t: 17 w 64    x 15   y 2 z 1

 a: 47  t: 22 w 304   x 130  y 2 z 6

 a: 47  t: 23 w 916   x 435  y 2 z 19

 a: 47  t: 24 w 71    x 23   y 1 z 1

 a: 47  t: 45 w 1079  x 989  y 1 z 22

 a: 47  t: 46 w 2161  x 2069 y 1 z 45

================================== 

Sat Mar 20 21:24:04 PDT 2021