Edited . Sorry , I didn't check one of the possible cases and miss a family of solutions . I'll try to complete it

Can you show your proof mahanmath?

First we have: a and b have the same prime divisors $p_1,p_2,...,p_k$ and we write: $a=\prod ({p_i}^{a_i})$ and $b=\prod ({p_i}^{b_i})$ from the 2 first conditions, we have $\frac{b_i}{2}\le a_i\le 2b_i$ (*) denote: $min (a_i,b_i)=c_i$ +, $a+1|b^2+1$ then $a+1|a^2+b^2=\prod ({p_i}^{2c_i})*(\prod {p_x}^{2|a_x-b_x|}+\prod {p_y}^{2|a_y-b_y|})$ then we must have $a+1=\prod ({p_i}^{a_i})+1|(\prod {p_x}^{2|a_x-b_x|}+\prod {p_y}^{2|a_y-b_y|})\le (\prod {p_x}^{2a_x})+(\prod {p_y}^{2a_y})$ (from *) ($mn+1\le m+n$ then m or n=1) then we have 2 cases :$a|b$ or $b|a$ +,b|a then a=kb and k|a and we have $kb+1|b^2+1$ or $kb+1|b(b-k)$ then $kb+1|b-k$ so b=k or $a=b^2$ similarly, the second case -->no answer *)answers are $b=a^2,a\in N$

Seems mahanmath made a typo/ has not all and tuandokim has made lots of mistakes: $ a+1=\prod ({p_{i}}^{a_{i}})+1|(\prod{p_{x}}^{2|a_{x}-b_{x}|}+\prod{p_{y}}^{2|a_{y}-b_{y}|})\le (\prod{p_{x}}^{2a_{x}})+(\prod{p_{y}}^{2a_{y}}) $ is nothing else that conclusion $a+1\le a^2+b^2$ again. So sth as $mn+1\le m+n$ couldn't be used and your notation with primes isn't nice to read. Sth as $a=xy,b=xz,y,z|x,gcd(y,z)=1$ is easier to conclude $xy+1|y^2+z^2$ , where we have to prove one $y,z$ is $1$ to conclude $a|b$ or $b|a$ and we can easily finish. SO after this, the problem is solved. Also $xy+1|y^2+1$ gives $x=y$ is easy and $x+1|z^2+1$ giving $z^2=x$ would have to be possible to prove. If $b|a$ then $a=b^2$ was the only correct conclusion. If $a|b=ka$ we get similar $a+1|k^2+1$ with $k|a$ and so we can easily get $k^2=a$ Hence the solutions are of form $(r^2,r),(r^2,r^3)$ which gives $a+1|a^3+1$ and trivial things, so they are all correct.

I think my solution is still right( except case a|b where we have $(r^2,r^3)$ as our solution) When I take ${p_i}^{c_i}$ out, I almost consider the set $A$ of x in $\prod{p_{x}}^{2|a_{x}-b_{x}|}$ ($\prod $ of only $x\in A$) is the set of number that satisfy +,$1\le x\le k$ +,$a_x>b_x$ and the set $B$ of y in $\prod{p_{y}}^{2|a_{y}-b_{y}|}$ is $1\le y\le k$ and $a_y<b_y$ then $(A\cup B)\subset {1,2,...,k}$ hence $(\prod{p_{x}}^{2a_{x}})*(\prod{p_{y}}^{2a_{y}})\le \prod ({p_i}^{a_i}) $

tuandokim wrote: I think my solution is still right( except case a|b where we have $(r^2,r^3)$ as our solution) When I take ${p_i}^{c_i}$ out, I almost consider the set $A$ of x in $\prod{p_{x}}^{2|a_{x}-b_{x}|}$ ($\prod $ of only $x\in A$) is the set of number that satisfy +,$1\le x\le k$ +,$a_x>b_x$ and the set $B$ of y in $\prod{p_{y}}^{2|a_{y}-b_{y}|}$ is $1\le y\le k$ and $a_y<b_y$ then $(A\cup B)\subset {1,2,...,k}$ hence $(\prod{p_{x}}^{2a_{x}})*(\prod{p_{y}}^{2a_{y}})\le \prod ({p_i}^{a_i}) $ Suppose $(a,b) = d$, $a = dx, b = dy$, $(x,y) = 1$. $a\left| {{b^2}} \right. \Rightarrow x\left| d \right.$, $b\left| {{a^2}} \right. \Rightarrow y\left| d \right.$，so $xy\left| d \right.$. Because $a + 1\left| {{a^2} + 1} \right.$, we have $a + 1\left| {{b^2} + {a^2}} \right.$, since $(d,a + 1) = 1$, we have $dx + 1\left| {{y^2} + {x^2}} \right.$. From this we have ${x^2} + {y^2} \geqslant dx + 1 \geqslant y{x^2} + 1$, not ${x^2} + {y^2} \geqslant dx + 1 \geqslant {y^2}{x^2} + 1$. So the solution is not end.

Let $gcd(a,b)=d$ then $a=d.x,b=d.y$ and $(x,y)=1$ $ a \mid b^{2}, b \mid a^{2}$ then $d=k.x.y$. we have $a=kx^{2}y, b=kxy^{2}$ case1: $a \geq b$ $b^{2}=ka$ then $a+1 \mid k-1$ then $a^{2}\geq b^{2} \geq (a+2)a >(a-1)^{2}$(*) or $k=1$ with k=1 then $a=b^{2}$, do with (*) that is easy. case 2: $ x<y$ then we have $(kx^{2}.y+1) \mid (kxy)^{2}(x^{2}+y^{2})$ hence $ (kx^{2}y+1) \mid (x^{2}+y^{2})$ 1)If $x^{2} \geq y$ then $x^{2}+y^{2}\geq k.x^{2}.y+1 > k.y^{2}$ then $x^{2}> (k-1)y^{2}>(k-1)x^{2}$ then k=1. And do next not difficult. 2) if $ x^{2}<y$ then $x^{2}+y^{2}=(l.y+x^{2})(k.x^{2}.y+1)$( do with modulo y. then l=0 hence $y=k.x^{4}$ , (x,y)=1 then $x=1$ we have $(a,b)=(k^{2}, k^{3})$

vntbqpqh234 wrote: 1)If $x^{2} \geq y$ then $x^{2}+y^{2}\geq k.x^{2}.y \geq k.y^{2}$ then $x^{2}\geq (k-1)y^{2}>(k-1)x^{2}$ then k=1. And do next not difficult. Well done, but in this case, we can get an other solution: $\left( {{k^2}({k^2} - 1),k{{({k^2} - 1)}^2}} \right),k \geqslant 2$ So the all solutions are: $(a,b) = \left( {{{(k - 1)}^2},k - 1} \right),\left( {{{(k - 1)}^2},{{(k - 1)}^3}} \right),\left( {{k^2}({k^2} - 1),k{{({k^2} - 1)}^2}} \right),k \geqslant 2$ Or $(a,b) = \left( {{k^2},k} \right),\left( {{k^2},{k^3}} \right),\left( {{{(k + 1)}^2}k(k + 2),(k + 1){k^2}{{(k + 2)}^2}} \right),k \in {N^ + }$. For example $(a,b) =(12,18)$ is a solution.

yunxiu wrote: vntbqpqh234 wrote: 1)If $x^{2} \geq y$ then $x^{2}+y^{2}\geq k.x^{2}.y \geq k.y^{2}$ then $x^{2}\geq (k-1)y^{2}>(k-1)x^{2}$ then k=1. And do next not difficult. Well done, but in this case, we can get an other solution: $\left( {{k^2}({k^2} - 1),k{{({k^2} - 1)}^2}} \right),k \geqslant 2$ So the all solutions are: $(a,b) = \left( {{{(k - 1)}^2},k - 1} \right),\left( {{{(k - 1)}^2},{{(k - 1)}^3}} \right),\left( {{k^2}({k^2} - 1),k{{({k^2} - 1)}^2}} \right),k \geqslant 2$ Or $(a,b) = \left( {{k^2},k} \right),\left( {{k^2},{k^3}} \right),\left( {{{(k + 1)}^2}k(k + 2),(k + 1){k^2}{{(k + 2)}^2}} \right),k \in {N^ + }$. For example $(a,b) =(12,18)$ is a solution. Yes, you are right. that case for us : $ x^{2}y+1=x^{2}+y^{2}$ then have your roots.

vntbqpqh234 wrote: Let $gcd(a,b)=d$ then $a=d.x,b=d.y$ and $(x,y)=1$ $ a \mid b^{2}, b \mid a^{2}$ then $d=k.x.y$. we have $a=kx^{2}y, b=kxy^{2}$ case1: $a \geq b$ $b^{2}=ka$ then $a+1 \mid k-1$ $ b^2 = ka $ ? Are you sure?? I think $ b^2 = ky^3 a $

yunxiu wrote: vntbqpqh234 wrote: 1)If $x^{2} \geq y$ then $x^{2}+y^{2}\geq k.x^{2}.y \geq k.y^{2}$ then $x^{2}\geq (k-1)y^{2}>(k-1)x^{2}$ then k=1. And do next not difficult. Well done, but in this case, we can get an other solution: $\left( {{k^2}({k^2} - 1),k{{({k^2} - 1)}^2}} \right),k \geqslant 2$ Indeed, correct remark, but how to show that is the only other solution? To syk... : I think he used an other $k$ there and that case was already solved correctly.

Complete solution： If $a \geqslant b$, suppose ${b^2} = ta$, from $a + 1\left| {ta + 1} \right.$ we have $a + 1\left| {t - 1} \right.$. Because $ta = {b^2} \leqslant {a^2}$,so $0 \leqslant t - 1 < t \leqslant a < a + 1$, hence $t - 1 = 0$, $a = {b^2}$, which is a solution. Next we suppose $a < b$, let $(a,b) = d$, $a = dx, b = dy$, $(x,y) = 1$, from $x\left| d \right.$ and $y\left| d \right.$, we have $xy\left| d \right.$, suppose $d = kxy$, then $a = k{x^2}y, b = kx{y^2}$, from $a < b$ we have $x < y$. Because $a + 1\left| {{a^2} - 1} \right.$, so $a + 1\left| {{b^2} + {a^2}} \right.$, as $(d,a + 1) = 1$, we have $dx + 1\left| {{y^2} + {x^2}} \right.$, so$k{x^2}y + 1\left| {{y^2} + {x^2}} \right.$. 1)If ${x^2} \geqslant y$, then $2{y^2} > {x^2} + {y^2} > k{x^2}y \geqslant k{y^2}$, so $k = 1$, and we have ${x^2}y + 1\left| {{y^2} + {x^2}} \right.$, so $2{y^2} > {x^2} + {y^2} \geqslant {x^2}y + 1 > {y^2}$, so $\frac{{{x^2} + {y^2}}}{{{x^2}y + 1}} = 1$, ${x^2} + {y^2} = {x^2}y + 1$, because $y - 1 > 0$, we have $y + 1 = {x^2}$, and get an other solution $(a,b) = \left( {{r^2}({r^2} - 1),r{{({r^2} - 1)}^2}} \right),r \geqslant 2$. 2)If ${x^2} < y$, let ${x^2} + {y^2} = s\left( {k{x^2}y + 1} \right)$, then we have $s \equiv {x^2}(\bmod y)$,so we can suppose$s = {x^2} + ly,l \geqslant 0$, and we have ${x^2} + {y^2} = s\left( {k{x^2}y + 1} \right) = k{x^4}y + {x^2} + kl{x^2}{y^2} + ly$. Because ${x^2} + {y^2} \leqslant {x^2}{y^2}$, we have $l = 0$, so $y = k{x^4}$, hence $x = 1$, $y = k$, we get an other solution $(a,b) = ({k^2},{k^3})$. Above all, the all solutions are $(a,b) = \left( {{r^2},r} \right),\left( {{r^2},{r^3}} \right),\left( {{{(r + 1)}^2}r(r + 2),(r + 1){r^2}{{(r + 2)}^2}} \right),r \in {N^ + }$.

Here is my solution. Let $ gcd (a,b)=d $ and $ a = dx, b = dy $ where $ gcd(x,y)=1$ From $ a | b^2 $, $ b | a^2 $ we get $ x | d $, $ y | d $. Since $ gcd(x,y)=1$, $ xy | d $. Let $ d = kxy $ $ \implies a = kx^2 y $, $ b = kx y^2 $ $ a +1 | b^2 + 1 \implies kx^2 y +1 | k^2 x^2 y^4 + 1 $ $ \implies kx^2 y +1 | (k^2 x^2 y^4 + 1) - (kx^2 y +1)(kx^2 y -1) y^2 = x^2 + y^2 $ $ \therefore \frac{x^2 + y^2}{kx^2 y +1} \in \mathbb{N} $ (Claim) $ k, x, y \in \mathbb{N} $ , $ gcd(x,y)=1 $ , $ \frac{x^2 + y^2}{kx^2 y +1} \in \mathbb{N} $ then $ \implies \frac{x^2 + y^2}{kx^2 y +1} = 1 $ (proof of Claim) Suppose that there exist $ n \ge 2 $ such that there exist $ x, y \in \mathbb{N} $ which $ \frac{x^2 + y^2}{kx^2 y +1} = n $ and $ gcd(x,y)=1 $. Let $ A = \{ x | \frac{x^2 + y^2}{kx^2 y +1} = n , \ x, y \in \mathbb{N} , gcd(x,y)=1 \} $ Now we will prove that $ A \subset \{ \sqrt{n} \} $ Suppose there exist $ x_{0} \in A - \{ \sqrt{n} \} $ and choose the minimum $ y_0 $ such that $ \frac{x_{0} ^2 + y_{0} ^2}{kx_{0} ^2 y_{0} +1} = n $ and $ gcd (x_0 , y_0 )=1 $. Case 1. $ x_0 \ge y_0 $ $ 2 x_0 ^2 \ge x_0 ^2 + y_0 ^2 \ge kx_0 ^2 y_0 +1 > kx_0 ^2 y_0 \implies 2 > ky_0 \implies k=y_0 = 1 $ $ \therefore n = \frac{x_{0} ^2 + y_{0} ^2}{kx_{0} ^2 y_{0} +1} = \frac{ x_0 ^2 + 1}{x_0 ^2 + 1} = 1 $ which is contradiction. Case 2. $ x_0 < y_0 $ $ \frac{x_{0} ^2 + y_{0} ^2}{kx_{0} ^2 y_{0} +1} = n \iff y_0 ^2 - nkx_0 ^2 y_0 + (x_0 ^2 - n ) = 0$ Let $ y_1 = nkx_0 ^2 - y_0 \in \mathbb{Z} $. $ \implies gcd(y_1 , n )= 1$ Then $ y_0 , y_1 $ are the roots of $ X^2 - nkx_0 ^2 X + (x_0 ^2 -n)=0 $ $ \therefore \frac{x_{0} ^2 + y_{1} ^2}{kx_{0} ^2 y_{1} +1} = \frac{x_{0} ^2 + y_{0} ^2}{kx_{0} ^2 y_{0} +1} = n $ $ kx_{0} ^2 y_{1} +1 = \frac{x_{0} ^2 + y_{1} ^2}{n} >0 \implies y_1 \ge 0 $ and $ y_0 y_1 = x_0 ^2 -n \ne 0 \implies y_1 \in \mathbb{N} $ $ y_1 = \frac{x_0 ^2 -n}{y_0} < \frac{y_0 ^2 -n}{y_0} < y_0 $ $ \therefore \frac{x_{0} ^2 + y_{1} ^2}{kx_{0} ^2 y_{1} +1} = n $, $ gcd(y_1 , n)=1 $, $ 0< y_1 < y_0 $ $ \implies $ contradiction that $ y_0 $ is minimum. $\therefore A \subset \{ \sqrt{n} \} $ since $ A \ne \emptyset \implies A = \{ \sqrt{n} \} $ $ x= \sqrt{n} , \frac{x^2 + y^2}{kx^2 y +1} = n \implies y=kn^2 $ Since $ n \ge 2 $, $ gcd(x,y) >1$ which is a contradiction. (Claim proof end) $ \therefore \frac{x^2 + y^2}{kx^2 y +1} = 1 $ (i) $ k = 1 $ $ x^2 + y^2 = x^2 y + 1 $ , $ ( x^2 - y - 1 ) (y-1) = 0 $ $ y = 1 $ : $ (k, x, y)=(1, m, 1) $ $ \implies $ $(a,b)=(m^2 , m) $ $ y = x^2 - 1 $ : $(k, x, y)=(1, m, m^2 -1 ) $ $ \implies $ $ (a,b)=( m^2 (m^2 -1) , m (m^2 -1 )^2 ) $ (ii) $ k \ge 2 $ If we use the same method above (Vieta Jumping) we get $ x=1 $ (Just suppose that there exist $ x_0 \ge 2 $ and choose the minimum $ y_0 $ ... ) $ \therefore x=1 $ , $ y^2 + 1 = ky + 1 \implies y=k $ $ \therefore (k, x, y)=(m, 1, m) $ $ \implies (a,b)=(m^2 , m^3 ) $

if the answer is $ (a,b) =\left({{k^2},k}\right),\left({{k^2},{k^3}}\right),\left({{{(k+1)}^2}k(k+2),(k+1){k^2}{{(k+2)}^2}}\right),k\in{N^+} $ the problem is enough confusion.

Let $(a,b)=d$, so $a=dx$ and $b=dy$. From $a|b^2$ we have $x|d$ and from $b|a^2$ we have $y|d$, so there is $z$ such that $d=xyz$. Then we have from $a+1|b^2+1$ that $x^2yz+1|x^2y^4z^2+1$. Also $x^2yz+1|x^4y^2z^2-1$, so $x^2yz+1|x^2+y^2$. Also $x^2yz+1|x^4yz+x^2$, so $x^2yz+1|x^4z-y$. Obviously $y\le x^2yz$ and $-x^4z<1$, so $y-x^4z<x^2yz+1$. Then either $x^4z-y=0$, or $x^4z-y\ge x^2yz+1$. If $x^4z=y$, then because $x$ and $y$ are coprime, we have $x=1$ and $y=z$. Then we get $a=y^2$ and $b=y^3$, which obviously works. Either $x^2z=1$, or $x^2z>1$. If $x^2z=1$, then $x=1$ and $z=1$, so $y+1|1-y$, so $y+1|2$, so $y=1$, which is part of previous solution. Let $x^2z>1$. Then from $x^4z-y\ge x^2yz+1$ we get $y\le \frac{x^4z-1}{x^2z-1}\le x^2+1$. Then let $x^2-y=k\ge-1$. From $x^2yz+1|x^2+y^2$ we get $x^2y+1\le x^2y+1\le x^2+y^2$, so $0\ge y(x^2-y)-x^2+1=yk-y-k+1=(y-1)(k-1)$. If $y=1$, then $x^2z+1|x^2+1$, so $z=1$ and we get $a=b^2$, which works. Now let $y>1$. Then $1\ge k\ge-1$. If $k=1$, then $y=x^2-1$. Then $x^4z-x^2z+1|x^4-x^2+1$, so $z=1$ and we get $a=x^2(x^2-1)$, $b=x(x^2-1)^2$, which works. If $k=0$, then $x^2=y$, so (because $x$ and $y$ are coprime), we have $x=y=1$, so $z+1|2$, so $x=y=z=1$ and $a=b=1$, which was here before. Finally if $k=-1$, then $y=x^2+1$. But we know that $y\le \frac{x^4z-1}{x^2z-1}\le x^2+1$, where equality in second inequality is exactly if $z=1$. Then $z=1$ and then $x^4+x^2+1|x^4+3x^2+1$, so $x^4+x^2+1|2x^2$. Then $(x^2-1)^2+x^2\le 0$ - contradiction. The solutions $(a,b)$ are $(t^2,t)$, $(t^2,t^3)$ and $(t^2(t^2-1),t(t^2-1)^2)$. Q.E.D.

I don't know what others have written above to be really honest. Here is a well-written(that's why long) solution featuring Vieta Jumping which I just learned. I will give slightly rephrased problem here for easier referencing in the solution. Solved with proxima1681. Problem. Find all positive integers $(a,b)$ such that $b \mid a^2$ $a \mid b^2$ $a+1 \mid b^2 + 1$ Solution: The answers are $(a,b) = (q^2,q)$, $(q^2, q^3)$ for all positive integers $q$. $(a,b) = \left((q^2-1)q^2, q(q^2-1)^2\right)$ for all integral $q > 1$. All these solutions clearly work, so we now proceed towards the other direction. From $(b)$, we get that $b^2 = ak$ for some integer $k$. This eliminates one of the divisibility conditions. Actually putting this in $(c)$, we get \[a + 1 \mid ak + 1 \iff a + 1 \mid k-1\]Let $d = \gcd(a,k)$. Then we can actually say there exists integers $x,y$ such that $a = dx^2$ and $k = dy^2$ such that $\gcd(x,y) = 1$. Putting this in $b^2 = ak$, one would get $b = dxy$. Plug this in above derived divisibility relation to get \begin{align*} dx^2 + 1 & \mid dy^2 - 1 \\ dx^2 + 1 & \mid dy^2 - 1 + dx^2 + 1 \\ dx^2 + 1 & \mid d(x^2 + y^2) \\ dx^2 + 1 & \mid x^2 + y^2 \tag{1} \label{1} \end{align*}Putting $b = dxy$ and $a = dx^2$ in $(a)$, we would get $y \mid d$. Again now write $ym = d$. We have again eliminated this $(a)$ condition. Finally we're going to swing this in $(1)$ to get \[ymx^2 + 1 \mid x^2 + y^2. \tag{2}\]Now, we will now make two cases on $m = 1$ and $m > 1$. Also, do note that $y>x$ unless $y^3m =1$ by the trivial bound in $dx^2 + 1 \mid dy^2 - 1$. The first part of $m=1$ will be straight divisibility arguments, the latter part will be by Vieta Jumping. Now for $m = 1$, we have \begin{align*} yx^2 + 1 & \mid y(x^2 + y^2) -yx^3 - 1 \\ yx^2 + 1 & \mid y^3 - 1 \\ yx^2 + 1 & \mid (y-1)(y^2+y+1) \end{align*}Now note that if $y = 1$, we would get one of the claimed solutions upon suitable parametrization. Therefore assume $y > 1$. By the Euclidean algorithm, $\gcd(y-1, y^2 + y + 1) = 1,3$. If the gcd is $1$, then we would have two cases to explore. The case of $yx^2 + 1 \mid (y-1)$ would succumb to size. The latter case deserves a bit of discussion. \begin{align*} yx^2 + 1 & \mid y^2 + y + 1 \\ yx^2 + 1 & \mid x^2(y^2 + y + 1) - (y+1)(yx^2 + 1) \\ yx^2 + 1 & \mid y + 1 - x^2 \end{align*}By size reasons, we're forced that $y+1 = x^2$. Upon suitable parametrization, we get the solution in second bullet as advertised. The final part is dealing when $y \equiv 1 \pmod{3}$ where gcd evaluates to be $3$. Then cool thing is $\nu_3(y^2 + y + 1) = 1$ when you take $y \equiv 1 \pmod{3}$(one can see by putting $y = 3z+1$). This would make the case very similar to what we did right now. It would actually again amount to $x^2 = y+1$. But this time its impossible due to a modulo $3$ argument. These details can be figured out easily. We now move to the harder part when $m > 1$. Let $t = \dfrac{x^2 + y^2}{ymx^2 + 1}$ with obviously $t>0$. Fix $x$ and $t$ from here on. Let $(x,y) = (p,z)$ such that $p+z$ is minimal. Define \[P(X) = X^2 - Xmtp^2 + p^2 - t.\]Note that roots of $P$ are exactly the values of $y$ for a given $p$ and $t$. By Vieta's sum of roots relation, we clearly have the other root of $P$(call it $\beta$) will be clearly an integer. That root must be necessarily non-negative otherwise we will get that $t$ is negative(refer to fractional definition of $t$). By Vieta's relation we must have $z\beta = p^2-t$. One would achieve contradiction if we show that $z > \beta$ which would give contradict the minimality of $p + z$. Working backwards we get \[z > \frac{p^2-t}{z} \iff z^2 - p^2 > -t.\]Since $m \ne 1$, recall that we have $z>p$ which makes the desired inequality trivially true. This gives us the desired contradiction unless the other root $\beta = 0$. If $\beta = 0$, then we have $p^2 = t$. All the above arguments actually mean, that if we have any solution with $m>1$, then $t = x^2$ is forced. Plug this back into definition of $t$ to get that $y = mx^4$. But since $\gcd(x,y) = 1$, then $x = 1$. Finally upon parametrization, we get the last solution which we claimed and we're done. $\blacksquare$