If $x,y\in \mathbb{Z}$ then we caan clearly see that $(x,y)=(-1,k) \forall k \in \mathbb{Z}$. I think he meant $\displaystyle{x,y\in \mathbb{N}}$. We have the expression is equal to: $$(xy+1)^2+(xy+1)(x+1)$$Now WLOG suppose that, $$(xy+1)^2+(xy+1)(x+1)<(xy+2)^2$$we get, $$(xy+1)(x+1)<2xy+3$$Now Solve. Also, suppose, $$(xy+1)^2+(xy+1)(x+1)\geq(xy+2)^2$$$$(xy+1)(x+1) \geq 2xy+3$$Now again solve. Then we get the required answer.

IMO2019 wrote: If $x,y\in \mathbb{Z}$ then we caan clearly see that $(x,y)=(-1,k) \forall k \in \mathbb{Z}$. I believe that $(a^2-2,0)$ is a solution too.

Use Pell-type infinite descent.

IMO2019 wrote: If $x,y\in \mathbb{Z}$ then we caan clearly see that $(x,y)=(-1,k) \forall k \in \mathbb{Z}$. I think he meant $\displaystyle{x,y\in \mathbb{N}}$. We have the expression is equal to: $$(xy+1)^2+(xy+1)(x+1)$$Now WLOG suppose that, $$(xy+1)^2+(xy+1)(x+1)<(xy+2)^2$$we get, $$(xy+1)(x+1)<2xy+3$$Now Solve. Also, suppose, $$(xy+1)^2+(xy+1)(x+1)\geq(xy+2)^2$$$$(xy+1)(x+1) \geq 2xy+3$$Now again solve. Then we get the required answer. ?

The wrong solution posted here

^ $x,y$ are integers, not necessarily positive.

WizardMath wrote: ^ $x,y$ are integers, not necessarily positive. no, it is given that they are positive

The problem was apparently changed from when I saw it earlier.

Can you explain why $t^2-4(xy+1)t -(x+1)=0$ has integer solutions? I think it should be, $(xy+1)t^2-4(xy+1)t-4(x+1)$ so that the discriminant is perfect square.

AnArtist wrote: Can you explain why $t^2-4(xy+1)t -(x+1)=0$ has integer solutions. I think it should be, $(xy+1)t^2-4(xy+1)t-4(x+1)$ so that the discriminant is perfect square. Oh, yes, I found my mistake, Thanks. I will delete my solution

http://mp.weixin.qq.com/s?__biz=MzI4MzI4MTE2Nw==&mid=2247485136&idx=1&sn=a2247df57e937814cb385e9f030bfe35&chksm=eb8c54b0dcfbdda603eb073e4b311f6d0944d6c6a05608ad342320696ecc2c209ed37547fc13&mpshare=1&scene=23&srcid=0401Kv2YFfm2Jh5sRuvGooG7#rd

my answer

How is half of the AOPS supposed to understand it?

草书对你们来说是有点费劲 想办法翻译一下 意下如何

Hamel wrote: How is half of the AOPS supposed to understand it? 1. You can translate it using google 2. The equations can still be understood, though Diaoest has said the solution is quite messy, and he/she'll write it neater.

https://h5.qzone.qq.com/page/photo?init=photo.v7/common/viewer2/index&picKey=NDR03q1vlxKYwVodtXs7IgEAAAAAAAA!&ownerUin=2540678622&appid=4&topicId=V10ySpmo0HixtA_NDR03q1vlxKYwVodtXs7IgEAAAAAAAA!_0_0&pre=http%3A%2F%2Fa3.qpic.cn%2Fpsb%3F%2FV10ySpmo0HixtA%2F*17z5ms160BUaQyQyg0v6EUSZZsbmfySsZkbckhRy5g!%2Fm%2FdCIBAAAAAAAA%26ek%3D1%26kp%3D1%26pt%3D0%26bo%3DXQRGAgAAAAARFz0!%26t%3D5%26vuin%3D2540678622%26tm%3D1522634400%26sce%3D60-3-3%26rf%3D0-0&useqzfl=1&useinterface=1&noCloseBtn=0&inqq=1

any solutions please

abbosjon2002 wrote: any solutions please Answer: there exist no such pairs. First of all, let us consider $d=(xy+1)(xy+x+2)=(x+1,y-1)$. There are coprime numbers $s,t$ such that $x=ds-1, y=dt+1$. After making this substitution in the original expression and dividing by $d^2$ we can see that $(dst+s-t)(dst+2s-t)$ should be a square of an integer. As factors in brackets are coprime and positive there are positive integers $k,l$ such that $$dst+s-t=k^2, \,\,\,dst+2s-t=l^2.$$This is equivalent to: $$s=l^2-k^2, \,\,\, yl^2-(y+1)k^2=t.$$So $(yl,k)$ is a solution of the next equation in variables $a,b$: $$a^2-Db^2=yt, \eqno{(1)}$$where $D=y(y+1)$. Let us consider solutions of above equation with $b\not =0$. Among them there is a solution $(a_1,b_1)$ with the least absolute value of the first coordinate. WLOG $b_1>0$ and, of course, $a_1>0$. Let us define $a_0=(2y+1)a_1-2Db_1, b_0=(2y+1)b_1-2a_1$. One can easily check that $(a_0,b_0)$ is also a solution of equation $(1)$. Looking at $(1)$ it is easy to see that $\frac{2D}{2y+2}<\frac{a_1}{b_1}<\frac{2D}{2y}$, so $|(2y+1)a_1-2Db_1|<a_1$. Therefore $b_0$ should be equal to $0$. This means $a_1=(y+1/2)b_1$, and after making this substitution in $(1)$ we get that $$b_1^2/4=yt.$$ We see that $yt$ is a full square, so $y=u^2, t=v^2$ because of $(y,t)=(dt+1,t)=1$. Hence "the least" solution of equation $(1)$ is $(a_1,b_1)=((2u^2+1)uv,2uv)$. Using Pell-style technic we could see that the only solutions of $(1)$ are $(a_n,b_n)$ with $$a_n+\sqrt{D}b_n = uv ((2u^2+1+2\sqrt{D})^n.$$ We remember that $(ly,k)$ is also a solution of equation $(1)$, so $uv|k$. We also remember that $k^2=dst+s-t=ys-t$ or, after substitutions: $$u^2s-v^2=m^2u^2v^2,$$where $k=muv$. As $(s,t)=(u,v)=1$, we can easily get that $u=v=1$. This means that $y=1$ but, of course, $(x+1)(2x+2)=2(x+1)^2$ is never a full square.

pavel kozlov wrote: abbosjon2002 wrote: any solutions please Answer: there exist no such pairs. First of all, let us consider $d=(xy+1)(xy+x+2)=(x+1,y-1)$. There are coprime numbers $s,t$ such that $x=ds-1, y=dt+1$. After making this substitution in the original expression and dividing by $d^2$ we can see that $(dst+s-t)(dst+2s-t)$ should be a square of an integer. As factors in brackets are coprime and positive there are positive integers $k,l$ such that $$dst+s-t=k^2, \,\,\,dst+2s-t=l^2.$$This is equivalent to: $$s=l^2-k^2, \,\,\, yl^2-(y+1)k^2=t.$$So $(yl,k)$ is a solution of the next equation in variables $a,b$: $$a^2-Db^2=yt, \eqno{(1)}$$where $D=y(y+1)$. Let us consider solutions of above equation with $b\not =0$. Among them there is a solution $(a_1,b_1)$ with the least absolute value of the first coordinate. WLOG $b_1>0$ and, of course, $a_1>0$. Let us define $a_0=(2y+1)a_1-2Db_1, b_0=(2y+1)b_1-2a_1$. One can easily check that $(a_0,b_0)$ is also a solution of equation $(1)$. Looking at $(1)$ it is easy to see that $\frac{2D}{2y+2}<\frac{a_1}{b_1}<\frac{2D}{2y}$, so $|(2y+1)a_1-2Db_1|<a_1$. Therefore $b_0$ should be equal to $0$. This means $a_1=(y+1/2)b_1$, and after making this substitution in $(1)$ we get that $$b_1^2/4=yt.$$ We see that $yt$ is a full square, so $y=u^2, t=v^2$ because of $(y,t)=(dt+1,t)=1$. Hence "the least" solution of equation $(1)$ is $(a_1,b_1)=((2u^2+1)uv,2uv)$. Using Pell-style technic we could see that the only solutions of $(1)$ are $(a_n,b_n)$ with $$a_n+\sqrt{D}b_n = uv ((2u^2+1+2\sqrt{D})^n.$$ We remember that $(ly,k)$ is also a solution of equation $(1)$, so $uv|k$. We also remember that $k^2=dst+s-t=ys-t$ or, after substitutions: $$u^2s-v^2=m^2u^2v^2,$$where $k=muv$. As $(s,t)=(u,v)=1$, we can easily get that $u=v=1$. This means that $y=1$ but, of course, $(x+1)(2x+2)=2(x+1)^2$ is never a full square. Why $uv$ is a divisor of $k$ ? To prove this , you must to prove that $l$ divisible by $uv$ . But i don't find anything about it in your solution. Can u explain more clearly

Quote: Why $uv$ is a divisor of $k$ ? To prove this , you must to prove that $l$ divisible by $uv$ . But i don't find anything about it in your solution. Can u explain more clearly Here you could apply the same argument as above. Suppose, there are solutions of equation $(1)$ (now it is $a^2-Db^2=u^2v^2$ ) with $b\not | uv$. Let us choise a solution $(a_1,b_1)$ with the least absolute value of the first coordinate among all such solutions. The only difference is change of coordinates $$ a_0=(2y+1)a_1-2Db_1, b_0=(2y+1)b_1-2a_1$$leads us to contradiction strictly because of $a_0\equiv a_1 \mod uv$ and $b_0\equiv b_1-2a_1 \mod uv$, Here we just got a new solution $(a_0,b_0)$ with $|a_0|<a_1$ and $b_0\not |uv$

I realized that my solition is wrong because of second inequality in $\frac{2D}{2y+2}<\frac{a_1}{b_1}<\frac{2D}{2y}$ is not always true for solutions of $(1)$. One could check that $\frac{a_1}{b_1}\geq \frac{2D}{2y}$ is possible if $b_1^2< t$; here we couldn't apply an infinite descent argument. The least counterexample here is $y= 9 , t= 6 , b= 1 , a= 12$ while the least counterexample with $t|y-1$ is $y= 484 , t= 161, b= 4, a= 1958$. We don't get contradiction with original problem because $y\not | a$ in both cases. Here I understood that we need to deal with original equation: $$yl^2-(y+1)k^2=t. \eqno{(2)}$$ There is an easy finish here. Firstly, we notice that $l>k$. Then we use Pell's like trick, we set: $$l_1=(2y+1)l - (2y+2)k, \,\,\,k_1=(2y+1)k-2yl,$$then $(l_1,k_1)$ is also solution of $(2)$. There are three cases: 1) $l_1\geq l \Rightarrow yl\geq (y+1)k \Rightarrow t=yl^2-(y+1)k^2\geq (y+1)k(l-k)>y>t$ - we got contradiction. 2) $l_1\leq -l \Rightarrow (2y+2)l\leq (2y+2)k$ - that contradicts $l>k$. 3) The only possible case $|l_1|<l$: here we have a new solution of $(2)$ that "less" than $(l,k)$, and we use infinite descent that globally leads us to contradiction. So we got that there are no integer pairs $(k,l)$ satisfying $(2)$, hence $(xy+1)(xy+x+2)$ is never a perfect square.

sqing wrote: Find all pairs of positive integers $(x, y)$ such that $(xy+1)(xy+x+2)$ be a perfect square . Viete Jumping may work perfectly

Waterme123 wrote: sqing wrote: Find all pairs of positive integers $(x, y)$ such that $(xy+1)(xy+x+2)$ be a perfect square . Viete Jumping may work perfectly Vieta works for symmetric expressions I guess

Excellent problem. No such $(x,y)$: the assumption $k^2=(xy+1)(xy+x+2)$ rearranges to \[(y-1)^2=\big[2y(y+1)x+(3y+1)\big]^2-4y(y+1)k^2.\]Set $u=2y(y+1)x+(3y+1)$ and $v=k$, so it suffices to show that there is no solution to the Pell-like equation $(y-1)^2=u^2-4y(y+1)v^2$ with $u\equiv3k+1\pmod{2y(y+1)}$. Assume for contradiction such a solution exists. Define $N(u,v)=u^2-4y(y+1)v^2$ as the norm of $\textstyle u+v\sqrt{4y(y+1)}$. It is clear that the fundamental unit in $\textstyle\mathbb Z_{>0}\big[\sqrt{4y(y+1)}\big]$ is $\textstyle(2y+1)+\sqrt{4y(y+1)}$. In other words, the solutions $(u_i,v_i)$ to the Pell-like equation are described by the recurrence \[(u_{i+1},v_{i+1})=\Big( (2y+1)u_i+4y(y+1)v_i,\;u_i+(2y+1)v_i\Big).\]Furthermore one solution is $(2y^2-y-1,y-1)$. In this solution, $u\equiv-(3y+1)\pmod{2y(y+1)}$. Thus if we consider $u$ modulo $2y(y+1)$, there is a $t$ such that $-(3y+1)(2y+1)^t\equiv(3y+1)\pmod{2y(y+1)}$. However note that $(2y+1)^2\equiv1\pmod{2y(y+1)}$, hence $(2y+1)^t$ can only attain the values $2y+1$ and $1$. Neither of these obey $-(3y+1)(2y+1)^t\equiv(3y+1)\pmod{2y(y+1)}$, the desired contradiction.

: We claim there are no solutions at all! First, $\gcd(xy+1, xy+x+2) = \gcd(xy+1, x+1) = \gcd(y-1, x+1)$. So let $x+1=da$ and $y-1=db$, where $\gcd(a,b) = 1$. Then we have \[ xy+1 = d \cdot u^2 \quad \text{and} \quad xy+x+2 = d \cdot v^2 \]for some relatively prime $(u,v)$. Using $a = v^2-u^2$, we may rewrite this as $u^2 = (d \cdot b + 1) (v^2 - u^2) - b$ or \[ (d \cdot b + 1)v^2 - (d \cdot b + 2)u^2. \] So, the problem statement is asserting that this Pell equation has no solutions for any $(b,d)$. To prove this, note that $v > u$; then let $v = \frac{X+Y}{2}$ and $u = \frac{X-Y}{2}$ for positive integers $X$ and $Y$. Thus the equation becomes \[ X^2 - (4bd+6) XY + Y^2 + 4b = 0. \] We now employ Vieta jumping. Assume for contradiction there is a solution $(X,Y)$ in positive integers. Assume $X \ge Y$ by symmetry. Note that $\left( \frac{Y^2+4b}{X}, Y \right)$ is also a positive integer solution. If we repeat this process by decreasing the sum whenever possible, we must eventually reach a pair of pairs $(X_1, Y)$ and $(X_2, Y)$ with $X_1 > X_2 \ge Y$. This means that we should have \begin{align*} X_1 + X_2 &= (4bd + 6) \cdot Y \\ X_1 \cdot X_2 &= Y^2 + 4b. \end{align*}However, if $\min(X_1, X_2) > Y$ and $X_1 + X_2 = (4bd + 6) \cdot Y$, then $X_1 \cdot X_2 \ge Y \cdot (4bd+5)Y > Y^2+4b$, contradiction.

Can someone please help me check my solution? Its much simpler than all that vieta jumping happening above so I suspect there is something wrong, but i can't figure out what. Let $(xy+1)(xy+x+2)=k^2$ Note that since $x,y$ are positive, $(xy+1)^2<k^2<(xy+x+2)^2$ Let $k=xy+1+t$, where $0<t\le x$ Then $(xy+1)(xy+x+2)=(xy+1+t)^2 \iff x^2y+xy+x+1=t^2+2t+2txy$ This can be written as $(xy+1)(x+1-2t)=t^2$, and thus $x+1-2t$ is non-negative, in fact positive else $t^2=0$ and $t=0$, i.e. $t \le \frac{x}{2}$ Furthermore, the equation can also be written as $(x^2+x-2tx)y=t^2+2t-1-x$ Note if $RHS\le 0$ then $2t\ge x+1$, contradiction. Thus $RHS > 0$ Thus, $x^2+x-2tx|t^2+2t-1-x$. We next show that $\frac{t^2+2t-1-x}{x^2+x-2tx}$ must be exactly $1$. Proof: If $2(x^2+x-2tx)\le t^2+2t-1-x$ then $2x^2+3x+1\le t^2+2t+2tx \le \frac{x^2}{4}+x+x^2$, contradiction. Thus $t^2+2t-1-x=x^2+x-2tx$. But this means $x^2-(2t-2)x-(t^2+2t-1)=0$ whose discriminant happens to be $8t^2$, and thus $t=0$ contradiction. Hence no solution.

gghx wrote: Can someone please help me check my solution? Its much simpler than all that vieta jumping happening above so I suspect there is something wrong, but i can't figure out what. Let $(xy+1)(xy+x+2)=k^2$ Note that since $x,y$ are positive, $(xy+1)^2<k^2<(xy+x+2)^2$ Let $k=xy+1+t$, where $0<t\le x$ Then $(xy+1)(xy+x+2)=(xy+1+t)^2 \iff x^2y+xy+x+1=t^2+2t+2txy$ This can be written as $(xy+1)(x+1-2t)=t^2$, and thus $x+1-2t$ is non-negative, in fact positive else $t^2=0$ and $t=0$, i.e. $t \le \frac{x}{2}$ Furthermore, the equation can also be written as $(x^2+x-2tx)y=t^2+2t-1-x$ Note if $RHS\le 0$ then $2t\ge x+1$, contradiction. Thus $RHS > 0$ Thus, $x^2+x-2tx|t^2+2t-1-x$. We next show that $\frac{t^2+2t-1-x}{x^2+x-2tx}$ must be exactly $1$. Proof: If $2(x^2+x-2tx)\le t^2+2t-1-x$ then $2x^2+3x+1\le t^2+2t+2tx \le \frac{x^2}{4}+x+x^2$, contradiction. Thus $t^2+2t-1-x=x^2+x-2tx$. But this means $x^2-(2t-2)x-(t^2+2t-1)=0$ whose discriminant happens to be $8t^2$, and thus $t=0$ contradiction. Hence no solution. Something wrong happens here: If $2(x^2+x-2tx)\le t^2+2t-1-x$ then $2x^2+3x+1\le t^2+2t+2tx \le \frac{x^2}{4}+x+x^2$ It should be: If $2(x^2+x-2tx)\le t^2+2t-1-x$ then $2x^2+3x+1\le t^2+2t+4tx \le \frac{x^2}{4}+x+2x^2$ So there is no contradiction!(An irreparable mistake)

Very nice problem, I thought that this was very elegant; here goes my overly detailed solution. Before beginning the solution, one can rule out $1 \in \{x,y\}$ by noting that $(y+1)(y+3) = (y+2)^2-1$ is never a square for $y \in \mathbb{N}$ and that $(x+1)(2x+2) = 2(x+1)^2$ is never a square for $x \in \mathbb{N}$

Part 1: Reduction to Nicer Equation Assume for the sake of contradiction that there exists some $(x,y) \in \mathbb{N}$ such that $(xy+1)(xy+x+2)$ is a perfect square. Firstly, let $s = core_2(xy+1) = core_2(xy+x+2)$, then we can see that $s \mid gcd(xy+1, xy+x+2)$ and therefore $s \mid x+1, y-1$ and hence there exist $m,n \in \mathbb{N}$ such that $x =sm-1$ and $y = sn+1$ which means that $$S^2 = \frac{xy+1}{s} = smn+m-n$$and $$(S+t)^2 = \frac{xy+x+2}{s} = smn+2m-n$$for some $S,t \in \mathbb{N}$. Then we have that $t(2S+t) = m$ meaning that $t \mid m$, define $d = \frac{m}{t}$, then we can conclude that $2St = (m-t)^2 = t(d-t)$ and consequently $$4(sdtn+2dt-n) = 4S^2 = (d-t)^2$$then redefining $U = d, V = t$, we have that there is a solution $s,n,U,V \in \mathbb{N}$ to $$U^2-(4sn+6)UV+V^2+4n = 0 \qquad (\clubsuit)$$ Part 2: Vieta Jump We will prove that $(\clubsuit)$ has no solutions which clearly suffices, for this, fix $s,n$ and notice that $(\clubsuit)$ is symmetric in $U,V$. We now Vieta Jump as follows; firstly assume that $V_1 \in \mathbb{N}$ is the minimal positive integer such that there is $(U_1,V_1) \in \mathbb{N}$ satisfying $(\clubsuit)$ which must exist as $\mathbb{N}$ is well-ordered. Looking at $(\clubsuit)$ as a quadratic in $U$, we can notice by Vieta's Relations that the second root of the quadratic satisfies $U_1+U_2 = V_1(4sn+6)$ meaning that $U_2 \in \mathbb{Z}$ and $U_1U_2 = V_1^2+4n > 0$ and hence $U_2 > 0$, then we know that $U_1, U_2 \geq V_1$ because $V_1$ is the smallest positive integer satisfying the condition. We now do some straightforward bounding which I have tried to present in the most linear and cleanest way possible. $\textbf{Lemma 1:}$ $V_1 \not\in \{U_1,U_2\}$ $\textbf{Proof)}$ If $U =V$ satisfies $(\clubsuit)$, then $n = U^2(sn+1) > n$ which is a contradiction, hence $V_1 \neq U_1, V_1 \neq U_2$. $\blacksquare$ $\textbf{Lemma 2:}$ $U_1 + U_2 \geq 2V_1+4n$ $\textbf{Proof)}$ We know that $$U_1+U_2 = V_1(4sn+6) \geq 4nV_1+6V_1 > 4n(V_1)+2V_1 \geq 4n+2V_1$$which proves the inequality. $\blacksquare$ $\textbf{Lemma 3:}$ $U_1U_2 > V_1(U_1+U_2-V_1)$ $\textbf{Proof)}$ Let $f(x) = x(U_1+U_2-x)$ and notice that $f$ reaches its local minimum at $x = V_1$ and $x = U_1+U_2-V_1$ over $x \in [V_1,U_1+U_2-V_1]$ which proves the inequality because we know that $$U_1U_2 = U_2(U_1+U_2-U_2) = f(U_2) > V_1(U_1+U_2-V_1)$$as $U_1 \in (V_1,U_1+U_2-V_1)$ by $\textbf{Lemma 1}$ and the fact that $U_1 \geq V_1$, as desired. Now, notice by $\textbf{Lemma 2}$ and $\textbf{Lemma 3}$ that $$V_1^2+4n = U_1U_2 > V_1(U_1+U_2-V_1) \geq V_1(V_1+4n) = V_1^2+V_1(4n) \geq V_1^2+4n$$which is a contradiction, as desired. $\blacksquare$

BlizzardWizard wrote:

Nice solution!How it was motivated？

Solved with GoodMorning. There are no solutions . Suppose there existed a solution. Let $d = \gcd(xy + 1, xy + x + 2) = \gcd(x+1, xy + 1 )= \gcd(x+1, y-1)$. Notice that $\frac{xy+1}{d}$ and $\frac{xy + x + 2}{d}$ are perfect squares, so let $xy + x + 2= dm^2, xy + 1 = dn^2$, where $\gcd(m,n) = 1$. By subtracting the equations, we get that $x + 1 = d(m^2 - n^2)$. Additionally let $\frac{y-1}{d} = k$. We have \[dn^2 = xy + 1 = y (d(m^2 - n^2) - 1) + 1 ,\]so $d y (m^2 - n^2) - (y-1) = dn^2$, now dividing both sides by $d$ and simplifying gives \[(dk + 1)(m^2 - n^2) - k = n^2\implies (dk + 1)m^2 - (dk + 2)n^2 = k\]We want to show this has no solutions. Let $a = m+n, b = m-n$ ($a$ and $b$ are positive integers because $m>n$). We have $m = (a+b)/2$ and $n = (a-b)/2$. Plugging into the equation and multiplying both sides by $4$ gives $(a+b)^2 (dk + 1) - (a-b)^2 (dk+2) = 4k$, so \[a^2 - ab(4dk + 6) + b^2 + 4k = 0\]Now we use Vieta Jumping. Consider the positive integer solution $(a,b)$ where $\min(a,b)$ is minimal, WLOG $a\ge b$. Note there exists a positive integer solution \[(b(4dk + 6) - a, b) = \left( \frac{b^2 + 4k}{a}, b\right )\]by Vietas. If $a = b(2dk + 3)$, then $a^2 > b^2 + 4k$, so $b(4dk + 6) - a \ne (b^2 + 4k)/a$, absurd. Thus, $a\ne b(2dk + 3)$, so $b(4dk + 6) - a \ne a$. Hence we have two different solutions, say $(a_1, b), (a_2, b)$, where $a_2 > a_1 \ge b$. Using Vietas we have $a_1 + a_2 = b(4dk + 6)$ and $a_1 \cdot a_2 = b^2 + 4k$. Since $a_1$ and $a_2$ are both greater than or equal to $b$, the minimal possible value of $a_1\cdot a_2$ is \[b\cdot b(4dk + 5) > b^2 + 4k,\]contradiction. Therefore there are no solutions for $x,y$.

megarnie wrote: Solved with GoodMorning. There are no solutions . Suppose there existed a solution. Let $d = \gcd(xy + 1, xy + x + 2) = \gcd(x+1, xy + 1 )= \gcd(x+1, y-1)$. Notice that $\frac{xy+1}{d}$ and $\frac{xy + x + 2}{d}$ are perfect squares, so let $xy + x + 2= dm^2, xy + 1 = dn^2$. By subtracting the equations, we get that $x + 1 = d(m^2 - n^2)$. Additionally let $\frac{y-1}{d} = k$. We have \[dn^2 = xy + 1 = y (d(m^2 - n^2) - 1) + 1 ,\]so $d y (m^2 - n^2) - (y-1) = dn^2$, now dividing both sides by $d$ and simplifying gives \[(dk + 1)(m^2 - n^2) - k = n^2\implies (dk + 1)m^2 - (dk + 2)n^2 = k\]We want to show this has no solutions. Let $a = m+n, b = m-n$ ($a$ and $b$ are positive integers because $m>n$). We have $m = (a+b)/2$ and $n = (a-b)/2$. Plugging into the equation and multiplying both sides by $4$ gives $(a+b)^2 (dk + 1) - (a-b)^2 (dk+2) = 4k$, so \[a^2 - ab(4dk + 6) + b^2 + 4k = 0\]Now we use Vieta Jumping. Consider the positive integer solution $(a,b)$ where $\min(a,b)$ is minimal, WLOG $a\ge b$. Note there exists a positive integer solution \[(b(4dk + 6) - a, b) = \left( \frac{b^2 + 4k}{a}, b\right )\]by Vietas. If $a = b(2dk + 3)$, then $a^2 > b^2 + 4k$, so $b(4dk + 6) - a \ne (b^2 + 4k)/a$, absurd. Thus, $a\ne b(2dk + 3)$, so $b(4dk + 6) - a \ne a$. Hence we have two different solutions, say $(a_1, b), (a_2, b)$, where $a_2 > a_1 \ge b$. Using Vietas we have $a_1 + a_2 = b(4dk + 6)$ and $a_1 \cdot a_2 = b^2 + 4k$. Since $a_1$ and $a_2$ are both greater than or equal to $b$, the minimal possible value of $a_1\cdot a_2$ is \[b\cdot b(4dk + 5) > b^2 + 4k,\]contradiction. Therefore there are no solutions for $x,y$. I believe u probably read MONT.

I am not sure if Pell Equation is popular in 2018, but it is not hard now. If there exists $(x,y).$ take the one with $x$ take its minimum value. Let $xy+1=du^2,xy+x+2=dv^2$ where $d$ is square-free. then $x=d(v^2-u^2)-1,$ $y=\frac{du^2-1}{d(v^2-u^2)-1}.$ Now $$yv^2-(y+1)u^2=\frac{y-1}d:=t<y.$$Technique to face these equations are popular. First consider the basic solution $(a_0,b_0)$ of $a^2-y(y+1)b^2=1.$ Obviously $(2y+1,2)$ is the solution. Now because $$[(\sqrt yv+\sqrt{y+1}u)(x_0-\sqrt{y(y+1)}y_0)]\cdot [(\sqrt yv-\sqrt{y+1}u)(x_0+\sqrt{y(y+1)}y_0)]=t.$$$$\Rightarrow [(vx_0-(y+1)uy_0)\sqrt y+(ux_0-yvy_0)\sqrt{y+1}]\cdot [(vx_0-(y+1)uy_0)\sqrt y+(ux_0-yvy_0)\sqrt{y+1}]=t.$$Therefore $(v(2y+1)-2(y+1)u,(2y+1)u-2yv)$ is also the solution of the equation. We only need $\frac vu\in (\frac{2(y+1)}{2y+1},\frac{2y+1}{2y})$ to get a smallest solution, contradiction! When $u>2y+1$ it is obviously correct. When $u\le 2y+1,$ only need to consider $v\ge u+1.$ But now $yv^2-(y+1)u^2>t,$ contradiction! $\Box$

Btw, we can easily prove that crucial equation $ax^2-(a+1)y^2=t$ has no solutions in positive integers $x,y$ for $0<t<a$ applying Legendre criterium for convergents. Indeed, $$0<x-y\sqrt{\frac{a+1}{a}} =\frac{t}{a}\left(x+y\sqrt{\frac{a+1}{a}}\right)^{-1}<\frac{1}{2x},$$so $$0<\frac{x}{y}-\sqrt{\frac{a+1}{a}}<\frac{1}{2xy}<\frac{1}{2y^2}.$$Hence, $x/y$ should be on of convergents to $$ \sqrt{\frac{a+1}{a}} = [1;\overline{2a,2}].$$WLOG, we can assume that $x$ and $y$ are coprime, so $ax^2-(a+1)y^2$ should be one of $ap_i^2-(a+1)q_i^2$ for $i\leq 2$. But $$a\cdot 1^2-(a+1) \cdot 1^2 = -1, $$$$a(2a+1)^2-(a+1) (2a)^2 = a, $$$$a(4a+3)^2-(a+1) (4a+1)^2 = -1, $$but none of theese numbers belongs to $(0,a)$.