First note that we must have 1=< 2ab^2-b^3+1=b^2(2a-b)+1 and so, 1<=b<=2a. Now, we'll prove that the pairs that solve this problem are (i) (a,b)=(n,1), (ii) (a,b)=(n,2n), and (iii) (a,b)=(8n^4-n,2n) where n is a positive Indeed, if b=1, we get the pairs in (i), while if b=2a we get the pairs in (ii). Now assume that 1<b<2a and set r=a^2/( 2ab^2-b^3+1) Then the quadratic equation (*) x^2-2xrb^2+r(b^3-1)=0 has two solution that are positive integers, since one of them is a and their product is r(b^3-1)>0. Evaluating the discriminant D we have D=4(r^2b^4-rb^3+r). Thus r^2b^4-rb^3+r=m^2 for some natural number and the above equation has solution rb^2+m and rb^2-m. Now note that m^2 = r^2b^4-rb^3+r > r^2b^4-2rb^3+b^2=(rb^2-b)^2, with rb^2-b>0. Hence, b>rb^2-m. However, since a^2> b^2(2a-b>=b^2, we have a>b. Therefore, a=rb^2+m > b and the other (also positve) root, say c, is c=rb^2-m < b. Furthermore, if 2c>b, then we have r=c^2/(b^2(2c-b)+1 <= c^2/b^2 <1, a contradiction. Thus b=2c and r=c^2. Since ac=r(b^3-1), we get a=c(b^3-1)=8c^4-c. It is easy to verify that if c>=1, then the integers b=2c and a=c(b^3-1) solve the problem, since a^2/(2ab^2-b^3+1) = c^2(b^3-1)^2 / [(b^3-1)(2cb^2-1)]=c^2. The proof is complete.

if b = 1, a is an even number. so case(i) may be (2k,1)

why the denominator has to be positive? couldnt it be negative?

Oh, yes...when b=1, then a must be even so case (i) becomes (n,1) with n even. Sorry about that! The denominator must be positive since the numerator is positive and the ratio must be a positive integer.

sorry I didnt know that the ratio has to be positie integer. all i know was it has to be integer...

Be careful next time. a ² >=0 so DENOminator >=0

Who can help me with Dmitri Shiryayevs elegant solution : [russian team member imo 2003] Let $n=2ab^2-b^3+1$ than we have $2ab^2 \equiv b^3-1 (modn)$ , also we have $2a^2b^2 \equiv ab^3-a (modn)$ but $a^2 \equiv 0 (modn)$. Thus we have $ab^3-a \equiv 0 (modn)$, $2ab^3-2a \equiv 0 (modn)$ also $2ab^3 -b^4+b \equiv 0 (modn)$ since n divides n. So we have $2a \equiv b^4 -b (modn)$. That gives: $m=b^4-b-2a \equiv 0 (modn)$. And, $n=(2a-b)b^2+1>0$ since $\frac{a^2}{2ab^2-b^3+1}$ is positive and $a^2$ postitive, too. So that means denominator is also positive. By the way we have $2a-b \geq 0$ 1. $2a=b$ gives $(a,b)=(k,2k)$ 2. $2a-b>0$ ... $2a-b \geq 1$. It follows that $a^2$ divisible by $n$. $a^2 \geq n = (2a-b)b^2+1$ ... $a^2 \geq b^2+1$ than we have $a>b^2$ $(*)$ And $n=(2a-b)b^2+1>(2b^2-b)b^2+1=2b^4-b^3+1>b^4>b^4-b-2a=m$ From here we have $0 \geq m$. a-) $b^4-b-2a=0$ gives series $(a,b)=(8k^4-k,2k)$ b-)If $b^4-b-2a<0$ so we have $-(b^4-b-2a) \geq n=2ab^2-b^3+1$ than $2a(1-b^2) \geq b^4-b-b+1=(b-1)(b-1)^3$ So $b=1$ which gives $(2k,1)$. So we are done. Please explain me : $(*)$ davron

Achilleas Sinefakopoulos wrote: First note that we must have 1=< 2ab^2-b^3+1=b^2(2a-b)+1 and so, 1<=b<=2a. Now, we'll prove that the pairs that solve this problem are (i) $(a,b)=(n,1),$ (ii) $(a,b)=(n,2n)$, and (iii) $(a,b)=(8n^4-n,2n)$ where $n$ is a positive Indeed, if $b=1$, we get the pairs in (i), while if b=2a we get the pairs in (ii). Now assume that $1<b<2a$ and set $r=a^2/( 2ab^2-b^3+1)$ Then the quadratic equation (*) $x^2-2xrb^2+r(b^3-1)=0$ has two solution that are positive integers, since one of them is a and their product is $r(b^3-1)>0$. Evaluating the discriminant D we have $D=4(r^2b^4-rb^3+r)$. Thus $r^2b^4-rb^3+r=m^2$ for some natural number and the above equation has solution $rb^2+m$ and $rb^2-m$. Now note that $m^2 = r^2b^4-rb^3+r > r^2b^4-2rb^3+b^2=(rb^2-b)^2,$ with $rb^2-b>0$. Hence, $b>rb^2-m$. However, since $a^2> b^2(2a-b\geq b^2$, we have $a>b$. Therefore, $a=rb^2+m > b$ and the other (also positve) root, say c, is $c=rb^2-m < b$. Furthermore, if 2c>b, then we have $r=c^2/(b^2(2c-b)+1 \leq c^2/b^2 <1$, a contradiction. Thus $b=2c$ and $r=c^2$. Since $ac=r(b^3-1)$, we get $a=c(b^3-1)=8c^4-c$. It is easy to verify that if $c\geq 1$, then the integers b=2c and a=c(b^3-1) solve the problem, since $a^2/(2ab^2-b^3+1) = c^2(b^3-1)^2 / [(b^3-1)(2cb^2-1)]=c^2$. The proof is complete.

Davron wrote: $a^2 \geq b^2+1$ than we have $a>b^2$ I cant understand this...

what do you think about that, but it might be true since i read that from KVANT magazine [ i hope that someone will help] davron

bilarev wrote: Davron wrote: $a^2 \geq b^2+1$ than we have $a>b^2$ I cant understand this... ok now i got this point bilarev look it is really nice we know that $a^2\geq 2ab^2-b^3+1=b^2(2a-b)+1>b^2a$ thus we have $a^2>b^2a$ implies to $a>b^2$ nice right ? davron

Davron wrote: bilarev wrote: Davron wrote: $a^2 \geq b^2+1$ than we have $a>b^2$ I cant understand this... ok now i got this point bilarev look it is really nice we know that $a^2\geq 2ab^2-b^3+1=b^2(2a-b)+1>b^2a$ thus we have $a^2>b^2a$ implies to $a>b^2$ nice right ? davron I don't understand how from \[ a^2\geq b^2(2a-b)+1 \] you get \[ a^2>b^2a \]

ok i will explain this more clearly : we know that $2ab^2-b^3+1$ is positive since $a^2\geq 0$ Also we know that $a^2 \geq b^2(2b-b)+1>b^2+1$ from here we can easily get that $a>b$. ok ? Then we again rewwrite the inequality $a^2 \geq b^2(2a-b)+1=b^2[a+(a-b)]+1>b^2a+1$ since $a>b$ Thus we can easily see that $a^2>b^2a$ which leasds to the desired $a>b^2$ [please if you understand this let me know ] davron

i understand... thank you

This problem is very similar with 2007kmo first round,,, So I v got it

I was looking Davron's solution and I didn't understand the part the says that "From here $ m\le0$" . Why?

When they ask for positive prove it to be positive!

The problem is more interesting if the problem is to find all positive integer pairs $ (a,b) $ such that \[ \frac{a^2}{2ab^2-b^3+1} \] is an integer

Let $k\in\mathbb N$ be equal to the given expression. Consider a solution $(a,b)$. If $b=1$ then $k=\dfrac a2$ so $a$ is even. We can verify that all solutions of the form $(a,b)=(2n,1)$ for any $n\in\mathbb N$ work. Assume that $b\ge2$. We have $a^2-2kb^2a+kb^3-k=0$, a quadratic in $a$. Now $\exists a_0$ which is another solution to this equation. By Vieta’s: $a_0+a=2kb^2$ $a_0\cdot a=kb^3-k$ Therefore $\underbrace{a_0}_{\mathbb N}=\underbrace{2kb^2-a}_{\mathbb Z}=\underbrace{\dfrac{kb^3-k}a}_{>0}$. Since $k\in\mathbb N$, it must be true that $a^2\ge2ab^2-b^3+1>0\implies a^2-1\ge b^2(2a-b)>-1$. If $2a-b=0$ then we have that $(n,2n)$ is a solution $\forall n\in\mathbb N$ (which works). If $2a-b<0$ then $b^2(2a-b)<0$, contradiction. Now we must have $2a-b\ge1$, so $a^2\ge b^2+1>b^2$ and $a>b$. Since for fixed $b$ and $k$, $a$ has two solutions (namely $a$ and $a_0$), another solution is $a_0=n$ and $b=2n\forall n\in\mathbb N$. Then: $a_0=n=8n^2\dfrac{a^2}{8an^2-8n^3+1}-a=\dfrac{8an^3-a}{8an^2-8n^3+1}$ $\implies a=8n^4-n$ And we see that $(a,b)=(8n^4-n,2n)\forall n\in\mathbb N$ is a solution. I claim that the only solutions are $\boxed{(n,2n)}$, $\boxed{(8n^4-n,2n)}$, and $\boxed{(2n,1)}$ for arbitrary $n\in\mathbb N$. Assume that $b\ge2$. Here, a contradiction proves our claim. WLOG $a+b$ is minimal. $a_0<a$ $\iff kb^2<a$ $\iff k<\dfrac a{b^2}$ Now, we have that: $k=\dfrac{a^2}{2ab^2-b^3+1}=\dfrac{a^2}{ab^2+\left(b^2(a-b)+1\right)}$ Since $b^2(a-b)+1>0$, we see that: $k<\dfrac{a^2}{ab^2}=\dfrac a{b^2}\bot$, which is a contradiction, thus proving our claim. $\square$

jgnr wrote:

that's wonderful

The solutions are $(x, 2x), (8x^4 -x, 2x), (2x, 1)$ for all positive integers $x$, which can be easily verified to work. Let $a^2 = k(2ab^2-b^3+1)$ for some positive integer $k$. We can solve to get: $$a^2 = k(2ab^2-b^3+1) \implies a^2-a(2kb^2)+k(b^3-1) = 0 \implies a = \frac{2kb^2\pm\sqrt{(2kb^2)^2-4k(b^3-1)}}{2},$$so the discriminant is $4k^2b^4-4kb^3+4k.$ For $a$ to be an integer this must be a square, so set this equal to $n^2$ for some positive integer $n$. We have that $$4k^2b^4-4kb^3+4k = (2kb^2-b)^2-(b^2-4k) = n^2.$$We have three cases: $b^2-4k > 0$, $b^2-4k < 0$, or $b^2-4k = 0$. Case 1: $b^2-4k > 0.$ We have $$b^2-4k = (2kb^2-b)^2-n^2 \ge (2kb^2-b)^2-(2kb^2-b-1)^2 = 4kb^2-2b-1 > b^2-4k,$$a contradiction. Case 2: $b^2-4k < 0$. We have $$4k-b^2 = n^2-(2kb^2-b)^2 \ge (2kb^2-b+1)^2-(2kb^2-b)^2 = 4kb^2-2b+1.$$Notice that if $b \ge 2$, then $4kb^2-2b+1 > 4k-b^2$, a contradiction. Only if $b = 1$, then $4kb^2-2b+1 \le 4k-b^2$, and we actually achieve equality when $n = 2kb^2-b+1$. Plugging in $b = 1$ gives $a = 2x$ for all positive integers $x$, and we have our first set of solutions: $(2x, 1)$. Case 3: $b^2-4k = 0$. This means $n = 2kb^2-b$ and $4k = b^2$, so solving for $a$, we get $$a = \frac{2kb^2 \pm (2kb^2-b)}{2},$$so $a = \frac{b}{2}$, so we get our second set of solutions $(x, 2x)$, or $a = \frac{4kb^2-b}{2} = \frac{b^4-b}{2}$ for even $b$. Substituting $b = 2c$ gives $a = 8c^4-c$ for all positive integers $c$, which gives our final set of solutions $(8x^4-x, 2x)$. $\blacksquare$ Remarks: Nice discriminant bounding problem. This problem was quite straight forward and wasn't as contrived as I thought it would be.

When in doubt, do some bounds. Let $a^2=k(2ab^2)-b^3k+k$ then $a^2-(2b^2k)a+b^3k-k,$ so by the quadratic formula $a=b^2k\pm \sqrt{b^4k^2-b^3k+k}$ For $a$ to be an integer, $b^4k^2-b^3k+k$ must be a perfect square. If $b=1$ then clearly all even $a$ work, so let $b>1.$ Let $r=b^2k-\frac{b}{2}+c$ then $r^2=b^4k^2-b^3k+\frac{b^2}{4}+2b^2kc-bc+c^2.$ If $b^4k^2-b^3k+k=r^2$ then $\frac{b^2}{4}+2b^2kc+c^2=bc+k.$ If $c$ is positive, then $b^2kc\ge bc$ and $b^2kc\ge k$ so $\frac{b^2}{k}+c^2\le 0$ implying $c=0$ a contradiction. If $c$ is negative, then let $c=-c_0$ note that $\frac{b^2}{4}-bc+c^2=k(1-2b^2c)$ so $1+2b^2c_0\le \frac{b^2}{4}+bc_0+c_0^2.$ Note that for $c_0\ge 1$ we have $2c_0< \frac{1}{4}$ while $1\le bc_0+c_0^2$ so we have another contradiction. Thus, $c=0.$ This implies that $k=\frac{b^2}{4}.$ We can plug in for $a.$ Thus, we have the following solutions: $(2k,1),(k,2k),(8k^4-k,2k)$ for positive integer $k$ which all work.

#2 in $\LaTeX$. First note that we must have $$1\leq 2ab^2-b^3+1=b^2(2a-b)+1$$And so, $$1\leq b \leq 2a.$$Now, we'll prove that the pairs that solve this problem are $$(a,b)=(n,1),(n,2n),(8n^4-n,2n),$$where $n$ is positive. Indeed, if $b=1,$ we get the first pair, while if $b=2a$ we get the second pair. Now assume that $1<b<2a$ and set $r=\frac{a^2}{ 2ab^2-b^3+1}.$ Then the quadratic equation $$x^2-2xrb^2+r(b^3-1)=0$$ has two solution that are positive integers, since one of them is $a$ and their product is $r(b^3-1)>0.$ Evaluating the discriminant $D$ we have $D=4(r^2b^4-rb^3+r).$ Thus $r^2b^4-rb^3+r=m^2$ for some natural number and the above equation has solution $rb^2+m$ and $rb^2-m.$ Now note that $$m^2 = r^2b^4-rb^3+r > r^2b^4-2rb^3+b^2=(rb^2-b)^2,$$ with $rb^2-b>0.$ Hence, $b>rb^2-m.$ However, since $a^2> b^2(2a-b) \geq b^2$, we have $a>b.$ Therefore, $a=rb^2+m > b$ and the other (also positive) root, say $c,$ is $c=rb^2-m < b.$ Furthermore, if $2c>b$, then we have $$r=c^2/(b^2(2c-b)+1 \leq c^2/b^2 <1,$$ a contradiction. Thus $b=2c$ and $r=c^2$. Since $ac=r(b^3-1)$, we get $a=c(b^3-1)=8c^4-c$. It is easy to verify that if $c\geq 1$, then the integers $b=2c$ and $a=c(b^3-1)$ solve the problem, since $$\frac{a^2}{2ab^2-b^3+1}= \frac{c^2(b^3-1)^2}{(b^3-1)(2cb^2-1)}=c^2.$$ The proof is complete.

Let $(a,b)$ satisfy $$\frac{a^2}{2ab^2-b^3+1}=m, ~~~ m\in \mathbb{N}.$$It can be simplified to $$(2(a-b^2m))^2=(2b^2m-b)^2+4m-b^2. ~~~~~(1)$$ Case 1: $4m-b^2>0.$ By (1) we have \begin{align*}4m-b^2=(2(a-b^2m))^2-(2b^2m-b)^2 \\ \geq (2b^2m-b+1)^2-(2b^2m-b)^2=2(2b^2m-b)+1 \end{align*} This is true because we can set $$M=2(a-b^2m), N=2b^2m-b, ~~~ M,N \in \mathbb{N}.$$Due to $M^2-N^2>0$ and $N>0,$ \begin{align*} |M| \geq N+1 \\ \implies M^2-N^2 \geq (N+1)^2 -N^2=2N+1 \\ \implies 2(2b^2m-b)+1+b^2-4m \leq 0 \\ \implies 4m(b^2-1)+(b-1)^2 \leq 0. \end{align*}So it must be that $b=1,$ giving $\boxed{(a,b)=(2m,1)}$ as a satisfying pair. Case 2: $4m-b^2<0.$ By (1) we have, \begin{align*} 4m-b^2&= (2(a-b^2m))^2-(2b^2m-b)^2 \\ \leq (2b^2m-b-1)^2 -(2b^2m-b)^2=-2(2b^2m-b)+1 \\ \implies (4m-1)b^2-2b+(4m-1)\leq 0~~~~~ (2) \\ \implies D=4(1-(4m-1)^2)< 0, \end{align*}where $D$ is the discriminant of the quadratic polynomial of $b$. Therefore because $4m-1>0,$ (2) is not true. And thus there is no pair $(a,b)$ satisfying the property in this case. Case 3: $4m-b^2=0.$ So, $$(2(a-b^2m))^2=(2b^2m-b)^2 \implies 2a-2b^2m=\pm (2b^2k-b).$$ Subcase 1: $2a-2b^2m=-2b^2m+b$ implies that $b=2a.$ So we get the pair $\boxed{(a,b)=(k,2k)}$ where $k$ is any natural number, as a satisfying pair. Subcase 2: $$2a-2b^2m=2b^2m-b \implies 4b^2m-b=2a.~~~~~(3)$$However $4m-b^2=0$ implies $b$ is even. So there exists a $c$ such that $b=2c$ and $m=c^2.$ By (3) we get $a=8c^4-c.$ So another pair is $\boxed{(a,b)=(8c^4-c, 2c)}$ and we are done satisfying all cases.

Solved with pi271828 a while back. The answer is $\boxed{(2n,1), (n,2n), (8n^4-n,2n)}$ for any positive integer $n$. These work. We now prove they are the only solution. Let $\frac{a^2}{2ab^2-b^3+1}=k$. Clearly if $b=1$, then $k=\frac{a}{2}$, so $a$ must be even. This falls in the solution set ($(2n,1)$). Then we have \[a^2=2ab^2k - kb^3 + k = a^2 - 2kb^2 \cdot a + (kb^3 - k) = 0\] Claim: Either $b=2a$ or $a>b$. Proof: Suppose $a\le b$. Note that $a^2\ge 2ab^2 - b^3 + 1 > 0$. So $a^2 > b^2 (2a-b)>-1$. This implies $2a-b\ge 0$. If $2a-b>0$, then we must have $a^2>b^2$, contradiction. So $b=2a$. $\square$ Now, consider the quadratic \[X^2-2kb^2 X + (kb^3 - k) = 0\] Let the two roots of this equation be $a_0$ and $a_1$, with $a_0\ge a_1$. We have \begin{align*} a_0+a_1 = 2kb^2 \\ a_0\cdot a_1 = kb^3 - k = k(b^3-1) \\ \end{align*} If $a_0= a_1$, then $a_0=a_1=kb^2$, so \[(kb^2)^2 = k(b^3-1)\implies kb^4 = b^3 - 1,\]not true. So we have $a_0>a_1$. This implies $a_0>kb^2$. If $b=2a_0$, then $b\ne 2a_1$, which implies $a_1>b$, however $b=2a_0>2a_1$. So $b\ne 2a_0$. Case 1: $b=2a_1$. Then we have $a_0 = 2kb^2 - \frac{b}{2}$, so $\left(2kb^2 - \frac{b}{2}\right)\cdot \frac{b}{2} = kb^3 - k$, which implies $kb^3 - \frac{b^2}{4} = kb^3 - k$., so $k=\frac{b^2}{4}$. So $a_0 = \frac{b^4-b}{2}$. Setting $b=2a_1$ gives the solutions $(a_0, b) = (8a_1^4 - a_1, 2a_1)$ and $(a_1,b) = (a_1, 2a_1)$, both of which fall into the solution set described above ($(8n^4-n,2n)$ and $(n,2n)$). Case 2: $b\ne 2a_1$. In this case, $a_0>a_1>b$. Since $a_0>kb^2$, we have $a_0\cdot a_1 > kb^2 \cdot b = kb^3$, which implies $a_0\cdot a_1\ne k(b^3-1)$, contradiction.

This is a pretty nice Vieta jumping problem. The answers are $(2n, 1), (n, 2n), (8n^2-4, 2n)$ for all positive integers $k$. They obviously work. Assume $b>1$ in what follows. Fix $k$ to be the value of the expression, and assume it is an integer. Then, the condition is equivalent to $$a^2-2ab^2k+b^3k - k = 0.$$In other words, if $(a, b)$ is a solution, then so is $\left(\frac{b^3k-k}a, b\right).$ Claim. We will have $b^3k-k < a^2$ unless $2a=b$ or $a \leq b$. Proof. Note that $b^3k-k < b^3k <\frac{a^2b}{2a-b}$, so the condition simplifies to $a > b$. $\blacksquare$ On the other hand, if $a\leq b$ in the original equation, then we have $a^2 > b^2(2a-b) > -1$, which implies that $b=2a$ as otherwise $a^2>b^2$. As a result, this means that the minimal solution $(a_0, b_0)$ to this equation for fixed $k$ takes the form $(n, 2n)$ for some positive integer $n$. Plugging back into the recursive sequence of solutions, we notice that the solutions alternate between $(n, 2n)$ and $(n, 8n^4-n)$. Thus, these yield the entire curve of solutions.

Nothing new of a concept here; imo these problems should be less frequent because they're similar to a computational problem where you literally know the approach, you just need to follow through, whereas in problems like sequences or a combo game you need to be creative. Note that $b=2a$ works, and $(a,b)=(2m,1)$ both work. We claim that the only other solution is $(a,b)=(8m^2-4,2m)$; henceforth assume the first two are not true. Then $$\frac{a^2}{(2a-b)b^2+1}:=k,b^3k-k < b^3k <\frac{a^2b}{2a-b}<a^2\iff a>b;a^2-2ab^2k+b^3k-k=0\implies(a,b)\Leftrightarrow\left(\frac{b^3k-k}a,b\right)$$yields a smaller solution as long as $a>b$. If $$a\le b,b\ne 2a,a^2\ge2ab^2-b^3=1>0,\text{ then }a^2>b^2(2a-b)\ge0\stackrel{\implies2a-b\ge0}{a\not>b\implies}2a-b=0,$$which we've already accounted for. Plugging back in, we find that the solution set alternates between $(m,2m)\leftrightarrow(m,8m^4-m)$, which is all the solutions by our Vieta jumping, as desired.

What is a quadratic The answer is $(n,2n)$, $(8n^4-n,2n)$, $(2n,1)$ only for any $n \in \mathbb{Z}^+$, which clearly work. Obviously if $b=1$ then $a$ must be even, so suppose $b>1$. Rewrite the denominator as $b^2(2a-b)+1$. For the quantity to be positive, we need $2a-b \geq 0$. If $2a=b$ then we get the first curve of solutions. Otherwise, I claim that $a>b^2$. Indeed, if $b \geq a$ then the denominator is at least $b^2+1$ which is too big, and if $\sqrt{a} \leq b \leq a$ then the denominator is at least $(\sqrt{a})^2a+1$ which is also too big. Now, if $2ab^2-b^3+1 \mid a^2$ then $2ab^2-b^3+1 \mid a^2(2b^2)^2$, and by the Euclidean algorithm it follows that $2ab^2-b^3+1 \mid (b^3-1)^2$. Obviously, $2ab^2-b^3+1$ is $1 \pmod{b^2}$, and since $(b^3-1)^2 \equiv 1 \pmod{b^2}$ as well it follows that $\tfrac{(b^3-1)^2}{2ab^2-b^3+1}$ is a $1 \pmod{b^2}$ positive integer as well. On the other hand, since $a>b^2$ we have $$\frac{(b^3-1)^2}{2ab^2-b^3+1}<\frac{(b^3-1)^2}{2b^4-b^3+1}\leq b^2 \impliedby b^6-2b^3+1\leq 2b^6-b^5+b^2 \iff b^6+2b^3+b^2 \geq b^5+1,$$which is obvious since $b$ is positive integer. Hence we need $$\tfrac{(b^3-1)^2}{2ab^2-b^3+1}=1 \implies b^3(b^3-1)=2ab^2 \implies a=\frac{1}{2}b(b^3-1).$$Plugging this back into the original, for this curve of solutions we would need $$\frac{\frac{1}{4}b^2(b^3-1)^2}{b^3(b^3-1)-b^3+1}=\frac{b^2(b^3-1)^2}{4(b^3-1)^2}=\frac{b^2}{4} \in \mathbb{Z},$$which clearly implies $b=2n$ for some $n \in \mathbb{Z}$ and $a=8n^4-n$. $\blacksquare$

Set the expression equal to $k$. Our equation can be rewritten as \[a^2-(2b^2k)a+(b^3-1)k=0.\] The discriminant of this quadratic is \[\ell^2 := 4b^4k^2-4b^3k+4k = (2b^2k-b)^2 - (b^2-4k).\] This value is at almost always strictly between $(2b^k-b-1)^2$ and $(2b^k-b+1)^2$, and would produce no solutions unless: $b=1$: Substituting into the original expression, we have the solution $\boxed{(2c,1)}$. $\ell=\tfrac{b^2}{4}$: Then $n = \pm (2b^2k-b) = \pm \left(\tfrac{b^4}{4} - \tfrac b2\right)$, so quadratic formula gives the solutions $\boxed{(c,2c), (8c^2-1,2c)}$. $\blacksquare$

The solutions are $(a, b) = (k, 2k), (k, 1), (8k^{4}-k, 2k)$ for positive integers $k.$ Obviously these work. We will show that these are all of the solutions. Take a pair $(a, b)$ which is not of one of the first two forms. Claim: Such pairs must satisfy $a > b^{2}.$ Assume for the sake of contradiction that $a \le b^{2}.$ We have that $$\frac{a^{2}}{2ab^{2}-b^{3}+1} < \frac{1}{b^{2}} \left(\frac{a^{2}}{2a-b} \right) = \frac{1}{b^{2}} \left(\frac{a}{2}+\frac{b}{4}+\frac{b^{2}}{4(2a-b)} \right). $$Since $2a-b \neq 0,$ we have that $$\frac{b^{2}}{4(2a-b)} \le \frac{b^{2}}{4}, $$and $b < b^{2},$ since $b > 1.$ So, we have $$\frac{1}{b^{2}} \left(\frac{a}{2}+\frac{b}{4}+\frac{b^{2}}{4(2a-b)} \right) < \frac{1}{b^{2}} \left( \frac{b^{2}}{2}+\frac{b^{2}}{4}+\frac{b^{2}}{4} \right) = 1,$$so this number cannot be a positive integer, as it is smaller than $1.$ Notice that if $$\frac{a^{2}}{2ab^{2}-b^{3}+1} = k, $$then $$a^{2}-2akb^{2}+(b^{3}-1)k = 0.$$So, now, if $a'$ is the other solution to this equation, $a'$ is an integer (since $a+a^{\prime} = 2kb^{2} $ by Vieta's), and is positive, since $aa^{\prime} = (b^{3}-1)k ,$ again by Vieta's. Claim: $2a' = b.$ Assume for the sake of contradiction that this is not true, i.e. $(a^{\prime}, b) $ is not of one of the first two forms. (Notice that $b \neq 1,$ so it cannot be the second form.) We get $$a^{\prime} = \frac{(b^{3}-1)k}{a} < \frac{b^{3}k}{a} = \frac{b^{3}a}{2ab^{2}-b^{3}+1} < \frac{ab}{2a-b} < \frac{b}{2} + \frac{\frac{b^{2}}{2}}{2b^{2}-b} \le \frac{b}{2} + \frac{1}{3} < b. $$But, $(a^{\prime}, b)$ is a valid solution set, so $b^{2} < a^{\prime}, $ a contradiction. Now, if we take $b = 2 a^{\prime}$ and $$a = \frac{(b^{3}-1)k}{a^{\prime}} = \frac{(8(a^{\prime})^{3}-1)k}{a^{\prime}},$$and replace $\ell = a^{\prime}, $ we get that $$a = \frac{(8\ell^{3}-1) \ell^{2}}{\ell} = 8\ell^{4}-\ell,$$giving us our third solution.

Let $\frac{a^2}{2ab^2 - b^3 + 1} = k$, where k is an integer. We get the quadratic $a^2 - (2b^2k)a + b^3k - k = 0$. We have that $D = 4b^4k^2 + 4(k - b^3k) = (2b^2k - b)^2 - b^2 + 4k$. We want a and b to be integers $\Rightarrow$ $D = n^2 = (2b^2k - b)^2 - b^2 + 4k$. Case 1: $b^2 - 4k = 0$ From the quadratic $a_1 = \frac{2b^2k + (2b^2k - b)}{2}$ and $a_2 = \frac{2b^2k - (2b^2k - b)}{2}$ $\Rightarrow$ $a_1 = \frac{4b^2k - b}{2} = \frac{b^4 - b}{2}$ for even b. Let b = 2c and from that we get the solutions $(a,b) = (8x^4-x,2x)$. Also $a_2 = \frac{b}{2}$ and from that we get the solutions $(a,b) = (x,2x)$. Case 2: $b^2 - 4k > 0$ We have that $b^2-4k = (2kb^2-b)^2-n^2 \ge (2kb^2-b)^2-(2kb^2-b-1)^2 = 4kb^2-2b-1 > b^2-4k$, which obviously is a contradiction $\Rightarrow$ we get no solutions from this case. Case 3: $b^2 - 4k < 0$ We have $4k-b^2 = n^2-(2kb^2-b)^2 \ge (2kb^2-b+1)^2-(2kb^2-b)^2 = 4kb^2-2b+1$. If $b \ge 2$, then $4kb^2-2b+1 > 4k-b^2$, which leads to contradiction. If b = 1, then $4kb^2-2b+1 \le 4k-b^2$, and we get $n = 2kb^2-b+1$. Plugging in $b = 1$ gives $a = 2x$, from which we get the solutions (a,b) = (2x,1) $\Rightarrow$ all solutions we get are: $(a,b) = (8x^4-x,2x); (x,2x); (2x,1)$.

We claim solutions are $(2n,1),(n,2n),(8n^4-n,2n).$ First note $2ab^2-b^3+1\mid a^2\implies 2ab^2-b^3+1\mid a(b^3-1),$ and letting $k(2ab^2-b^3+1)=a(b^3-1)$ rearranges to $2akb^2=(a+k)(b^3-1).$ If $b=1$ we get the first solution set $(2n,1)$ so assume $b>1.$ Then $b^2\mid a+k.$ Now we can rearrange our equation to \[\frac1a+\frac1k=\frac{2b^2}{b^3-1}.\]Since this is symmetric in $a,k$ we WLOG $a\le k.$ Then $\frac2a\ge\frac{2b^2}{b^3-1}$ so $a\le\frac{b^3-1}{b^2}\implies a\le b-1.$ In particular $a<\frac{b^2}{b+1},$ which rearranges to $b^2-a>ab.$ We also have $k\ge b^2-a,$ so $k>ab.$ This gives us the following estimates: \[\frac1a\left(1+\frac1b\right)>\frac1a+\frac1k=\frac{2b^2}{b^3-1}>\frac1a.\]The lower bound rearranges to $a>\frac{b^3-1}{2b^2}>\frac{b-1}2$ and the upper bound becomes $a<\frac{(b^3-1)(b+1)}{2b^3}<\frac{b+1}2.$ Thus $a=\frac b2$ which gives $k=\frac{b^4-b}2.$ Since we assumed $a\le k$ it follows that $a$ can take both of these, and setting $b=2n$ gives our other solution sets $(n,2n),(8n^4-n,2n).$ Checking, these work so we are done.

Let $\frac{a^2}{2ab^2 - b^3 + 1} = k \implies$ $a^2 - 2ab^2k + b^3k - k = 0$ Obvious $a \neq b$, so take $a_1$ the other root and $\begin{cases} a + a_1 = 2b^2k \\ a \cdot a_1 = b^3k - k \end{cases}$ If $a_1 = 0 \implies b = 1$ (cause $k \neq 0$) $\implies a = 2k \implies$ $(a, b) = (2k, 1)$ If $a_1 > 0$, let $(a, b)$ the pair with the smallest sum, so $a_1 \geq a \implies a_1 \geq b^2k$ $\implies (b^3 - 1)k \geq b^2ka \implies$ $b > a$ and $k > 0 \implies$ $2ab^2 - b^3 = (2a - b)b^2 \geq 0$ $\implies 2a \geq b$. Supose $2a > b \implies \frac{a^2}{2ab^2 - b^3 + 1} \leq$ $\frac{a^2}{b^2 + 1} <$ $\frac{a^2}{a^2 + 1} < 1$ $\implies k < 1$ ABS! Then $2a = b \implies k = a^2$ $\implies (a, b) = (a, 2a) \to (8a^4 - a, 2a)$ (by vieta) $\boxed{\therefore (a, b) = (2t, 1), (t, 2t), (8t^4 - t, 2t)}$