Method of infinite descent (fall)- assume $x\leqq\leq z$ are integers satisfying minimum sum such that any of $xy+1,yz+1,zx+1$ isn't perfect square. Consideration regarding number $t=x+y+z+2xyz\pm 2\sqrt{(xy+1)(yz+1)(zx+1)}$ (integer solutions of quadratic equation $t^2+x^2+y^2+z^2-2(xy+yz+zt+tx+zx+ty)-4xyzt-4=0$): We get the following identities $(x+y-z-t)^2=4(xy+1)(zt+1)$, $(x+z-y-t)^2=4(xz+1)(yt+1)$, $(x+t-y-z)^2=4(xt+1)(yz+1)$. From this system observe that not all $xt+1,yt+1,zt+1$ are the squares, while $(xt+1)(yt+1)(zt+1)$ is. Now any of products $(xy+1)(yt+1)(tx+1)$ is square. Let's prove that by this we got the least solution $(x,y,t)$: Since from the system is $zt+1\geq 0$,$t\geq -\frac1z$, for $z>1$ must be $t\geq 0$. It is easy to check that both cases $t=0$ and $t=1$ are impossible. Therefore all numbers $(x,y,t)$ are positive integers. But,according to assumption about minimality it must be $t\leq z$, and from Vieta's formulae we get $tt_1<z^2$ ($t_1$ is second solution of the quadratics).Contradiction. $\blacksquare$

Anonymous wrote: We get the following identities $ (x + y - z - t)^2 = 4(xy + 1)(zt + 1)$, $ (x + z - y - t)^2 = 4(xz + 1)(yt + 1)$, $ (x + t - y - z)^2 = 4(xt + 1)(yz + 1)$. could someone explain how these are found from what is said above? thank you for helping me to understand...

The identities follow from the assumption that $ t$ is a root of $ t^2+x^2+y^2+z^2-2(xy+yz+zt+tx+zx+ty)-4xyzt-4=0$

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Ok then, I'll rewrite it more detailed. The if part is trivial. For the only if part, assume $ x,y,z$ are such integers with $ xy + 1, yz + 1, zx + 1$ not all perfect squares, moreover pick them such that $ x + y + z$ is minimal. Without loss of generality $ xy + 1$ isn't a perfect square. Now let $ t$ be the smallest positive root of $ t^2 + x^2 + y^2 + z^2 - 2(xy + yz + zt + tx + zx + ty) - 4xyzt - 4 = 0$, an equation which rewrites equivalently as \[ \\ (x + y - z - t)^2 = 4(xy + 1)(zt + 1), \\ (x + z - y - t)^2 = 4(xz + 1)(yt + 1), \\ (x + t - y - z)^2 = 4(xt + 1)(yz + 1). \] But our quadratic equation has roots $ t = x + y + z + 2xyz\pm 2\sqrt {(xy + 1)(yz + 1)(zx + 1)}$, which is given to be integer, so the right hand side is a square each time. But then the left hand side needs to be a square, which gives that $ (xt + 1)(yt + 1)(xy + 1)$ is a perfect square. Now, since the centered equations above tell us that $ xt + 1\ge0,yt + 1\ge0,zt + 1\ge0$, we have $ t\ge - \frac1{\max(x,y,z)} > - 1$ (since $ x = y = z = 1$ isn't a solution). If $ t = 0$, then working out the formula for $ t$ gives $ (x + y + z)^2 = 4(xy + yz + zx + 1)$ or $ (x + y - z)^2 = 4(xy + 1)$, which means $ xy + 1$ is a square, contradiction, so we have $ t > 0$ and by the minimality of $ x + y + z$ we must have $ t\ge z$. But for the two roots $ t,t'$ of our quadratic equation we have by de Viete that $ tt' = x^2 + y^2 + z^2 - 2xy - 2yz - 2zx - 4 < z^2 - x(2z - x) - y(2z - y) < z^2$, contradiction. keywords: vieta

thanks!!

We shall prove that xy + 1, yz + 1, zy + 1 are pairwise relatively prime. Suppose there is a prime p1 such that p1 divides both xy+1 and yz+1. then p1 also divides z(xy+1) - y(zx+1) = z-x. Analogously if there are primes p2 and p3 such that p2 divides yz + 1, zy + 1 and p3 divides xy + 1,zy + 1 then p2 divides z-y and p3 divides x-y. Then there are numbers f1, f2, f3, such that p1f1= z-x; p2f2= z-y; p3f3=x-y. This gives us that p2f2 - p1f1= p3f3 and p3f3 - p2f2 = p1f1. Then p1f1 + p2f2 = p2f2 - p1f1 ==> p1f1=-p1f1 <==> p1f1=0. Contradiction. So these numbers are pairwise relatively prime hence result.

You have only shown that one pair is relatively prime. You assumed that every pair has a common prime factor, but what you really should assume is: one pair has a common prime factor.

I all of a sudden don't understand my own solution anymore... Can anyone tell me why $ z^2 - x(2z - x) - y(2z - y) < z^2$ in my last line? Maybe I'm just tired right now...

Zetax: in fact I proved it for a single pair of terms, but since the expressions xy + 1, yz + 1, xz + 1, are symmetric i could assume for a single pair without lost of generality. Also, the primes p1, p2, p3, i didnt say they were equal. maybe i cant see your point: can you explain?

You assumed that every pair out of $ xy+1,yz+1,zx+1$ has a common factor and got a contradiction. Thus you just proved that not _all_ pairs have a common factor, one or two pairs may still have. By the way, you also made a sign-mistake, p3f3 - p2f2 = p1f1 should in fact be p3f3 - p2f2 = -p1f1. This destroys your contradiction.

Peter wrote: Can anyone tell me why $ z^2 - x(2z - x) - y(2z - y) < z^2$ in my last line? Maybe I'm just tired right now... No, it's not because of your tiredness. The above solution is just not correct.. yet: You shouldn't suppose wlog that $ xy + 1$ is not a perfect square, but that $ z\ge x,y$ And you actually get $ (xt + 1)(yt + 1)(xy + 1)$ being a square by multiplying $ 4(xz + 1)(yt + 1)$, $ 4(xt + 1)(yz + 1)$ and $ (xy + 1)2$ and then dividing by $ 16(xy + 1)(xz + 1)(yz + 1)$, which is known to be a perfect square. You also know that $ yt + 1$ is a perfect square iff $ xz + 1$ is, and the same with $ xt + 1$ and $ yz + 1$ (and that's why you don't need xy+1 to be a perfect square). The $ t = 0$ case also implies $ xz + 1$ and $ yz + 1$ being a square, so you still have a contradiction there. And with all this, the answer to your question also becomes trivial.

everyone knows that A.1 is the hardest of PEN

A non-empty set of positive integers is defined "good" whenever the product of any two distinct elements is smaller of some squares by $1$: with this convention $\{x,y,z\}$ is good. Let's start proving that if $\{x,y,z\}$ is good, then $\{x,y,z,w\}$ is good too, whenever \[w=x+y+z+2xyz\pm 2\sqrt{(xy+1)(yz+1)(zx+1)}\] is a positive integer. The two values of $w$ are roots of the following equivalent quadratic equations: \[(x+y-z-w)^2=4(xy+1)(zw+1),\text{ }(1) \\ (x-y+z-w)^2=4(xz+1)(yw+1),\text{ }(2) \\ (x-y-z+w)^2=4(yz+1)(xw+1).\text{ }(3)\] And that's clear that $\alpha w+1$ is a integer that can be expressed as ratio of two squares, i.e. it's a square too, for all $\alpha \in \{x,y,z\}$. Since the problem is symmetric in $x,y,z$ it can be assumed without loss of generality that $x\le y\le z$. Let's suppose for the sake of contradiction that not all sets $\{x,y,z\}$ are good, i.e. there exists a set such that at least one between $xy+1,yz+1,zx+1$ is not a perfect square. Choose now such a set $\{x,y,z\}$ such that $x+y+z$ is minimal, and define $w$ using the negative square root part. It's claimed that $0<w<z$ and $xy+1,yw+1,wx+1$ are not all perfect squares, but their product it is. This will contradict the minimality of $x+y+z$, showing that $\{x,y,z\}$ has to be good by force. Then, multiplying (1), (2) and (3) we obtain that $(xy+1)(yw+1)(wx+1)$ is a perfect square. Moreover, e.g. $yz+1$ is a perfect square if and only if $xw+1$ is a square too, so not all $xy+1,yw+1,wx+1$ can be perfect squares. We are missed to show only thavt $0<w<z$. Let's rewrite (1) as \[zw+1=\frac{(x+y-z-w)^2}{4(xy+1)}\ge 0 \implies w \ge -z^{-1}.\] If $z=1$ then we are forced to $x=y=z=1$, but $(xy+1)(yz+1)(zx+1)$ would not be a square. That means that $w\ge 0$, as far as it is a integer. But if $w=0$ then looking at (1), (2) and (3) each term in $xy+1,yz+1,zx+1$ would be a perfect square, against our assumptions. That means that $w\ge 1$. Hence, proving that $w<z$ will be enough to reach our contradiction. Define $w^\prime$ the other (greater) root of (1); then \[w w^\prime=x^2+y^2+z^2-2(xy+yz+zx)-4<z^2-x(2y-x)-y(2z-y)<z^2,\] implying that $w<z$, as far as $0<w<w^\prime$, and it completes the proof.

How does one think of a solution like this?? In particular, the value for $t$, the infinite descent, etc. Thanks!

Proving "if (xy+1)(yz + 1)(zx + 1) is a perfect square then xy + 1, yz + 1, and zx + 1 are all perfect squares". I am trying Proof by Contradiction, so... Assume ~(xy + 1, yz + 1, and zx + 1 are all perfect squares). Question: is the negative of this statement equivalent to "at least one is not a perfect square" in which case I would have to go through cases, looking for a contradiction in each case; or, is it enough to assume that exactly one is not a perfect square and then find the contradiction? Thanks for the help.

I have completed a possible proof of this theorem - it is attached. I do have one weak spot, possibly, it is clearly labeled in the proof. Please check to see if my logic holds. I look forward to your responses.

Attachments:
NT A1.doc (87kb)

I've updated my proof. The new draft is attached.

Attachments:
NT A1.doc (185kb)

thanks

Anonymous wrote: Consideration regarding number $t=x+y+z+2xyz\pm 2\sqrt{(xy+1)(yz+1)(zx+1)}$ (integer solutions of quadratic equation $t^2+x^2+y^2+z^2-2(xy+yz+zt+tx+zx+ty)-4xyzt-4=0$): Can someone explain to me how this person created a quadratic equation with the correct roots to solve the problem?

davisv wrote: I've updated my proof. The new draft is attached. Where is the case when all are not perfect squares?

equationcrunchor wrote: Anonymous wrote: Consideration regarding number $t=x+y+z+2xyz\pm 2\sqrt{(xy+1)(yz+1)(zx+1)}$ (integer solutions of quadratic equation $t^2+x^2+y^2+z^2-2(xy+yz+zt+tx+zx+ty)-4xyzt-4=0$): Can someone explain to me how this person created a quadratic equation with the correct roots to solve the problem? pell equality

Peter wrote: Ok then, I'll rewrite it more detailed. The if part is trivial. For the only if part, assume $ x,y,z$ are such integers with $ xy + 1, yz + 1, zx + 1$ not all perfect squares, moreover pick them such that $ x + y + z$ is minimal. Without loss of generality $ xy + 1$ isn't a perfect square. Now let $ t$ be the smallest positive root of $ t^2 + x^2 + y^2 + z^2 - 2(xy + yz + zt + tx + zx + ty) - 4xyzt - 4 = 0$, an equation which rewrites equivalently as \[ \\ (x + y - z - t)^2 = 4(xy + 1)(zt + 1), \\ (x + z - y - t)^2 = 4(xz + 1)(yt + 1), \\ (x + t - y - z)^2 = 4(xt + 1)(yz + 1). \]But our quadratic equation has roots $ t = x + y + z + 2xyz\pm 2\sqrt {(xy + 1)(yz + 1)(zx + 1)}$, which is given to be integer, so the right hand side is a square each time. But then the left hand side needs to be a square, which gives that $ (xt + 1)(yt + 1)(xy + 1)$ is a perfect square. Now, since the centered equations above tell us that $ xt + 1\ge0,yt + 1\ge0,zt + 1\ge0$, we have $ t\ge - \frac1{\max(x,y,z)} > - 1$ (since $ x = y = z = 1$ isn't a solution). If $ t = 0$, then working out the formula for $ t$ gives $ (x + y + z)^2 = 4(xy + yz + zx + 1)$ or $ (x + y - z)^2 = 4(xy + 1)$, which means $ xy + 1$ is a square, contradiction, so we have $ t > 0$ and by the minimality of $ x + y + z$ we must have $ t\ge z$. But for the two roots $ t,t'$ of our quadratic equation we have by de Viete that $ tt' = x^2 + y^2 + z^2 - 2xy - 2yz - 2zx - 4 < z^2 - x(2z - x) - y(2z - y) < z^2$, contradiction. keywords: vieta I did not understand why you checked $t=0$ case. You have already said $t$ is the smallest positive root of equation $ t^2 + x^2 + y^2 + z^2 - 2(xy + yz + zt + tx + zx + ty) - 4xyzt - 4 = 0$

Claim 1: If $\prod_{cyc} (xy+1) = (xyz+\frac{x+y+z}{2})^2$, I am done. If I expand, I get $(\sum_{cyc} xy) + 1 = \frac 12 \sum_{cyc} xy + \frac 14 \sum_{cyc} x^2$ $\sum_{cyc} x^2 - 2\sum_{cyc} xy + 4=0$ As a quadratic in $x$: $x^2-(2y+2z)x+((y-z)^2-4)=0$. Its discriminant is $4((y+z)^2 - (y-z)^2 +4) =16(yz+1)$ is a perfect square, as needed. Now suppose $\prod_{cyc} (xy+1) = (xyz+\frac{x+y+z-l}{2})^2$ where $l\in \mathbb{Z}$. It's clear $l\ge 0$. $\prod_{cyc} (xy+1) - (xyz+\frac{x+y+z}{2})^2 = -\frac14 \sum_{cyc} x^2 + \frac 12 \sum_{cyc} xy + 1$ $-\frac14 \sum_{cyc} x^2 + \frac 12 \sum_{cyc} xy + 1 = -\frac l2 (2xyz+x+y+z-\frac l2)$ $\sum_{cyc} x^2 -2 \sum_{cyc} xy -4 = l (4xyz+2x+2y+2z-l)$ $x^2 - (2y+2z)x +(y-z)^2 - 4 - 4lxyz - 2lx - 2ly - 2lz + l^2=0$ $x^2 - (2y+2z+4lyz+2l)x + (y^2+z^2+l^2 - 4 - 2yz -2ly - 2lz)=0$ Note this expression is fully symmetric: $x^2 - 2yx-2zx-4lyzx-2lx + (y^2+z^2+l^2 - 4 - 2yz -2ly - 2lz)=0$ so $x^2+y^2+z^2+l^2 - 4lxyz - 2(xy+xz+xl+yz+yl+zl) - 4=0$ We know $x,y,z,l\in \mathbb{Z}_{\ge 0}$. If $xyz=0$ the statement is clear. Let $(x,y,z,l)$ be the solution with $x+y+z+l$ minimal. I claim $\min\{x,y,z,l\}=0$. Furthermore WLOG $x$ is the largest. First we can assume $x<l+y+z+2xyz$ because otherwise $(2y+2z+4lyz+2l-x,y,z,l)$ gives a solution with smaller sum. Let $r=2y+2z+4lyz+2l-x$ We consider their product. Note $xr=y^2+z^2+l^2-2(yz+lz+yl) < 3x^2$, so it follows that $x\sqrt{3}>r$ so we can bound $(y^2+z^2+l^2 - 4 - 2yz -2ly - 2lz)=xr\ge \frac 29 (2y+2z+4lyz+2l)^2$ There is another case where $r<0$. In this case, $y^2+z^2+l^2 - 4 - 2yz -2ly - 2lz < 0$ and its absolute value is at least $2y+2z+4lyz+2l+1$, and we can get a contradiction.