Standard problem using Vieta Jumping (finally I know how this method is called - big thx to the german team ). Standard manipulations of $ 4ab-1\mid (4a^{2}-1)^{2}$ shows that this implies $ 4ab-1\mid (a-b)^{2}$. Let's assume now that there exist some postive integers $ a\neq b$ so that $ 4ab-1\mid (a-b)^{2}$. Let $ k=(a-b)^{2}/(4ab-1)$. Let $ S=\{ (a,b): a,b\in \mathbb{Z}^{+}, (a-b)^{2}/(4ab-1)=k\}$. Since $ \mid S\mid \geq 1$, there is a pair $ (A,B)\in S$ which minimizes $ a+b$ over all $ (a,b)\in S$. wlog assume that $ A > B$. Consider now the quadratic equation \[ \frac{(x-B)^{2}}{4xB-1}=k \ \ \ \Leftrightarrow x^{2}-x(2B+4kB)+B^{2}+k=0 \] which has roots $ x_{1}=A$ (obviously) and $ x_{2}=2B+4kB-A=(B^{2}+k)/A$ (from Vieta). This implies that $ x_{2}\in \mathbb{Z}^{+}$, so $ (x_{2}, B)\in S$. By the minimality of $ A+B$, we however have $ x_{2}\geq A$, ie $ (B^{2}+k)/A\geq A$ and therefore $ k\geq A^{2}-B^{2}$. Thus, $ (A-B)^{2}/(4AB-1)=k\geq A^{2}-B^{2}$ and hence $ A-B\geq (A+B)(4AB-1)$, clearly impossible for $ A,B\in\mathbb{Z}^{+}$. $ \Box$

Let $ a-b=k\Longleftrightarrow b=a-k>0$ $ \therefore k<a$. $ 4ab-1=4a(a-k)-1=4a^{2}-4ka-1$, Dividing $ (4a^{2}-1)^{2}=16a^{4}-8a^{2}+1$ by $ 4ab-1=4a^{2}-4ka-1$ gives the quotient of $ 4a^{2}+4ka+4k^{2}-1$ and residue of $ 4k^{2}(4ka+1)$. $ \therefore 4ab-1|(4a^{2}-1)^{2}\Longleftrightarrow k=0$ or $ 4ka=-1$. Since $ k,\ a$ are integers, yielding $ k=0\Longleftrightarrow a=b$. Q.E.D.

If $ 4ab-1|4a^{2}-1$ it follows that $ 4a^{2}-1=4abk-k$ and $ kb=\frac{4a^{2}+k-1}{4a}.$ If $ k=1$ we get $ a=b.$ If $ k>1$ we have that $ 4a|k-1$ and $ k-1>0$ therefore $ k\geq 4a+1.$ But now we have $ 4a^{2}-1=k(4ab-1)\geq 16a^{2}b+4ab-4a-1\geq 16a^{2}-1$ which is a contradiction. Let's suppose that there is some $ p-$ prime divisor of $ 4ab-1$ such that if $ p^{\alpha}\parallel 4a^{2}-1$ and $ p^{\beta}\parallel 4ab-1$ then $ \beta>\alpha.$ Since $ p^{\beta}|(4a^{2}-1)^{2}$ it follows that $ b\leq 2\alpha.$ Since $ GCD(2a-1,2a+1)=1$ we have that either $ p^{\alpha}\parallel 2a-1$ or $ p^{\alpha}\parallel 2a+1.$ In both cases using that $ p^{\alpha}|4ab-1$ we get that either $ p^{\alpha}|2b-1$ or $ p^{\alpha}|2b+1.$ Since $ \beta\leq 2\alpha$ and the above we conclude that $ 4ab-1|(4a^{2}-1)(4b^{2}-1)=16a^{2}b^{2}-1-4a^{2}-4b^{2}+2$ and thus \[ 4ab-1|2a^{2}+2b^{2}-1.\] Therefore there exists some $ m\in\mathbb{N}$ such that \[ (*)\ \ \ \ 2b^{2}-4abm+2a^{2}+m-1=0.\] Suppose that there is at least one couple $ (a,b)$ with $ a<b.$ We choose such couple for which $ a-min$ and $ b$ is the maximal for this $ a.$ Using Viete we get that there is some $ b'\in\mathbb{N}$ which is solution with $ a.$ If $ b'<a$ we get a contradiction for $ a.$ If $ b'<b$ we get a contradiction for $ b.$ thus $ b'=a$ or $ b'=b.$ If $ b'=a$ we get $ m=1$ and $ b=a.$ If $ b'=b$ it follows that the descriminant of (*) should be 0, or $ 4a^{2}m^{2}-4a^{2}-2m+2=0$ or $ 4a^{2}=\frac{2m-2}{m^{2}}$ and therefore $ m=1$ and again $ a=b.$ Hope it's correct.

LEMMA 1 Call $ (a,b)$ a "good" pair if it satisfies $ 4ab-1|(4a^{2}-1)^{2}$. If $ (a,b)$ is good, then so does $ (b,a)$ LEMMA 2 Call an integer $ m$ "great" if it satisfies i) $ m|(4a^{2}-1)^{2}$ ii) $ m \equiv-1 (mod 4a)$ If $ m$ is great, then so does $ \frac{(4a^{2}-1)^{2}}{m}$ [Karena bukti kedua lemma tidak sulit, maka bukti diserahkan pada pembaca] We'll use induction on $ a$ so that if $ (a,b)$ is good, then $ a = b$. If $ a = 1$, then $ 4b-1|9$ and it's clear that $ 4b-1 = 3 \Leftrightarrow b = 1 = a$. Assume it's true for any $ a < n$. Consider a good pair $ (n,b)$ (This pair exists since setting $ n = b$ works well). If $ n < b$, then let $ R = 4nb-1$. By Lemma 2, we have $ S = \frac{(4n^{2}-1)^{2}}{R}$ satisfies $ S \equiv-1 (mod 4a)$. Set $ b' = \frac{S+1}{4a}$. Since $ b > n$, then $ R > 4n^{2}-1$ and thus $ S < 4n^{2}-1 \Leftrightarrow b' < n$. Hence, $ (n,b')$ is good, too. In particular, any good pair $ (n,b)$ with $ n < b$ correspondences to a good pair $ (n,b')$ with $ n > b'$. Now, consider the case if $ n > b$. By Lemma 1, we have $ (b,n)$ is a good pair, too. But, by induction hypothesis, all good pair $ (x,y)$ with $ x < a$ satisfies $ x = y$. Since $ b < a$, then $ b = n$, contradicting the condition $ b < n$. Now, we are left for the case $ n = b$. Indeed, this is true since $ 4nb-1 = 4n^{2}-1$ divides $ (4n^{2}-1)^{2}$. This completes our induction and our proof.

kunny, how can you be sure that $ k=0$ or $ 4ka =-1$? As far as I can see, you only get that $ 4ab-1$ divides $ 4k^{2}(4ka+1)$, not $ 4k^{2}(4ka+1) = 0$...

gabrielsebastian wrote: LEMMA 1 Call $ (a,b)$ a "good" pair if it satisfies $ 4ab-1|(4a^{2}-1)^{2}$. If $ (a,b)$ is good, then so does $ (b,a)$ LEMMA 2 Call an integer $ m$ "great" if it satisfies i) $ m|(4a^{2}-1)^{2}$ ii) $ m \equiv-1 (mod 4a)$ If $ m$ is great, then so does $ \frac{(4a^{2}-1)^{2}}{m}$ [Karena bukti kedua lemma tidak sulit, maka bukti diserahkan pada pembaca] We'll use induction on $ a$ so that if $ (a,b)$ is good, then $ a = b$. If $ a = 1$, then $ 4b-1|9$ and it's clear that $ 4b-1 = 3 \Leftrightarrow b = 1 = a$. Assume it's true for any $ a < n$. Consider a good pair $ (n,b)$ (This pair exists since setting $ n = b$ works well). If $ n < b$, then let $ R = 4nb-1$. By Lemma 2, we have $ S = \frac{(4n^{2}-1)^{2}}{R}$ satisfies $ S \equiv-1 (mod 4a)$. Set $ b' = \frac{S+1}{4a}$. Since $ b > n$, then $ R > 4n^{2}-1$ and thus $ S < 4n^{2}-1 \Leftrightarrow b' < n$. Hence, $ (n,b')$ is good, too. In particular, any good pair $ (n,b)$ with $ n < b$ correspondences to a good pair $ (n,b')$ with $ n > b'$. Now, consider the case if $ n > b$. By Lemma 1, we have $ (b,n)$ is a good pair, too. But, by induction hypothesis, all good pair $ (x,y)$ with $ x < a$ satisfies $ x = y$. Since $ b < a$, then $ b = n$, contradicting the condition $ b < n$. Now, we are left for the case $ n = b$. Indeed, this is true since $ 4nb-1 = 4n^{2}-1$ divides $ (4n^{2}-1)^{2}$. This completes our induction and our proof. What does "Karena bukti kedua lemma tidak sulit, maka bukti diserahkan pada pembaca" mean?

Nice problem. This one will go into PEN soon. A little easy for a number 5 though - although not as easy as Kunny pretends it to be.

$ 4ab-1|b^{2}(4a^{2}-1)^{2}$ implies $ 4ab-1|(a-b)^{2}$. $ (a-b)^{2}=(4ab-1)k$ so that $ a^{2}-(2b+4kb)a+b^{2}+k=0$. assume there is $ (a,b)$ satisfying this condition and $ a \neq b$. wlog, let $ a>b$. then by Vieta's theorem, $ (b, \frac{b^{2}+k}{a})$ is a solution too, since $ \frac{b^{2}+k}{a}=2b+4kb$ is a positive integer. we can check that $ \frac{b^{2}+k}{a}<b$ bu the direct substitution of $ k$. therefore, if there is a solution $ (a,b)$ with minimum $ b$, we can get a solution with smaller minimum, contradiction.

iwan_no2000 wrote: gabrielsebastian wrote: LEMMA 1 Call $ (a,b)$ a "good" pair if it satisfies $ 4ab-1|(4a^{2}-1)^{2}$. If $ (a,b)$ is good, then so does $ (b,a)$ LEMMA 2 Call an integer $ m$ "great" if it satisfies i) $ m|(4a^{2}-1)^{2}$ ii) $ m \equiv-1 (mod 4a)$ If $ m$ is great, then so does $ \frac{(4a^{2}-1)^{2}}{m}$ [Karena bukti kedua lemma tidak sulit, maka bukti diserahkan pada pembaca] We'll use induction on $ a$ so that if $ (a,b)$ is good, then $ a = b$. If $ a = 1$, then $ 4b-1|9$ and it's clear that $ 4b-1 = 3 \Leftrightarrow b = 1 = a$. Assume it's true for any $ a < n$. Consider a good pair $ (n,b)$ (This pair exists since setting $ n = b$ works well). If $ n < b$, then let $ R = 4nb-1$. By Lemma 2, we have $ S = \frac{(4n^{2}-1)^{2}}{R}$ satisfies $ S \equiv-1 (mod 4a)$. Set $ b' = \frac{S+1}{4a}$. Since $ b > n$, then $ R > 4n^{2}-1$ and thus $ S < 4n^{2}-1 \Leftrightarrow b' < n$. Hence, $ (n,b')$ is good, too. In particular, any good pair $ (n,b)$ with $ n < b$ correspondences to a good pair $ (n,b')$ with $ n > b'$. Now, consider the case if $ n > b$. By Lemma 1, we have $ (b,n)$ is a good pair, too. But, by induction hypothesis, all good pair $ (x,y)$ with $ x < a$ satisfies $ x = y$. Since $ b < a$, then $ b = n$, contradicting the condition $ b < n$. Now, we are left for the case $ n = b$. Indeed, this is true since $ 4nb-1 = 4n^{2}-1$ divides $ (4n^{2}-1)^{2}$. This completes our induction and our proof. What does "Karena bukti kedua lemma tidak sulit, maka bukti diserahkan pada pembaca" mean? I'm sorry, he just copied the solution from Indonesian forum. It means that because the proof for both lemmas aren't too hard, so it's given to the reader..

milo wrote: kunny, how can you be sure that $ k=0$ or $ 4ka =-1$? As far as I can see, you only get that $ 4ab-1$ divides $ 4k^{2}(4ka+1)$, not $ 4k^{2}(4ka+1) = 0$... kunny wrote: Let $ a-b=k\Longleftrightarrow b=a-k>0$ $ \therefore k<a$. $ 4ab-1=4a(a-k)-1=4a^{2}-4ka-1$, Dividing $ (4a^{2}-1)^{2}=16a^{4}-8a^{2}+1$ by $ 4ab-1=4a^{2}-4ka-1$ gives the quotient of $ 4a^{2}+4ka+4k^{2}-1$ and residue of $ 4k^{2}(4ka+1)$. $ \therefore (4a^{2}-1)^{2}=(4ab-1)\{4a^{2}+4(a-b)a+4(a-b)^{2}-1\}+4(a-b)^{2}\{4(a-b)+1\}$ $ =(4ab-1)(8a^{2}-4ab-1)+16a^{2}(a-b)^{2}.$ Since we have $ 8a^{2}-4ab=4a(2a-b)\neq 1$ for $ a,\ b\in \mathbb{N}$, $ \therefore 4ab-1|(4a^{2}-1)^{2}\Longleftrightarrow 16a^{2}(a-b)^{2}=0$, yielding $ a=b$. Q.E.D. How about this time? kunny

kunny wrote: milo wrote: kunny, how can you be sure that $ k=0$ or $ 4ka =-1$? As far as I can see, you only get that $ 4ab-1$ divides $ 4k^{2}(4ka+1)$, not $ 4k^{2}(4ka+1) = 0$... kunny wrote: Let $ a-b=k\Longleftrightarrow b=a-k>0$ $ \therefore k<a$. $ 4ab-1=4a(a-k)-1=4a^{2}-4ka-1$, Dividing $ (4a^{2}-1)^{2}=16a^{4}-8a^{2}+1$ by $ 4ab-1=4a^{2}-4ka-1$ gives the quotient of $ 4a^{2}+4ka+4k^{2}-1$ and residue of $ 4k^{2}(4ka+1)$. $ \therefore (4a^{2}-1)^{2}=(4ab-1)\{4a^{2}+4(a-b)a+4(a-b)^{2}-1\}+4(a-b)^{2}\{4(a-b)+1\}$ $ =(4ab-1)(8a^{2}-4ab-1)+16a^{2}(a-b)^{2}.$ Since we have $ 8a^{2}-4ab=4a(2a-b)\neq 1$ for $ a,\ b\in \mathbb{N}$, $ \therefore 4ab-1|(4a^{2}-1)^{2}\Longleftrightarrow 16a^{2}(a-b)^{2}=0$, yielding $ a=b$. Q.E.D. How about this time? kunny the same mistake as last one,how can you sure that $ 16a^{2}(a-b)^{2}=0$?

If $ 16a^{2}(a-b)^{2}\neq 0$, then $ 4ab-1$ is not a divisor of $ (4a^{2}-1)^{2}$, so $ 16a^{2}(a-b)^{2}=0$, yielding $ a=b$. Is this a right reasoning, or wrong the whole of my solution basically?

kunny wrote: If $ 16a^{2}(a-b)^{2}\neq 0$, then $ 4ab-1$ is not a divisor of $ (4a^{2}-1)^{2}$, so $ 16a^{2}(a-b)^{2}=0$, yielding $ a=b$. Is this a right reasoning, or wrong the whole of my solution basically? You have only shown that $ 4ab-1$ divides $ 16a^{2}(a-b)^{2}$. Since $ 4ab-1$ and $ 16a^{2}$ are relative prime, this implies that $ 4ab-1$ divides $ (a-b)^{2}$. The main part of the argument is still missing.

I can understand now, thank you. kunny

Here is another solution, Is this also wrong? Let $ a+b=x,\ a-b=y$, we have $ 4ab-1=(a+b)^{2}-(a-b)^{2}-1=x^{2}-(y^{2}+1),$ $ (4a^{2}-1)^{2}=\{(2a)^{2}-1\}^{2}=\{(x+y)^{2}-1\}^{2}$ $ =x^{4}+4yx^{3}+2(3y^{2}-1)x^{2}+4y(y^{2}-1)x+(y^{2}-1)^{2}$. $ \therefore x^{4}+4yx^{3}+2(3y^{2}-1)x^{2}+4y(y^{2}-1)x+(y^{2}-1)^{2}$ $ =\{x^{2}-(y^{2}+1)\}\{x^{2}+4yx+(7y^{2}-1)\}+8y^{3}x+8y^{4}+4y^{2}$. $ \therefore 4ab-1|(4a^{2}-1)^{2}\Longleftrightarrow 4y^{2}\{2y(x+y)+1\}=0$, yielding $ y=0\Longleftrightarrow a=b$. $ \because 2y(y+x)\ is \ even$.

gabrielsebastian wrote: LEMMA 1 Call $ (a,b)$ a "good" pair if it satisfies $ 4ab-1|(4a^{2}-1)^{2}$. If $ (a,b)$ is good, then so does $ (b,a)$ Proving this alone completes the problem, because if $ 4ab-1|(4a^{2}-1)^{2}$, then the quotient is also $ -1 \mod{4a}$ and so there is some $ c$ such that $ (4ab-1)(4ac-1)=(4a^{2}-1)^{2}$, and so one of $ b, c$ is less than or equal to $ a$, so you can assume WLOG that $ b \le a$. Now invoke the inductive hypothesis and you're done.

A more ugly approach: Let $ 4abk-k=a^{2}-2ab+b^{2}$, then $ a^{2}-a(2b+4bk)+b^{2}+k=0$ Assume $ (a,b)$ is the minimal pair with $ a>b$, So, $ a=b+2bk-\sqrt{4b^{2}k^{2}+4b^{2}k-k}>b$ (since this value is positive) $ 2bk>\sqrt{4b^{2}k^{2}+4b^{2}k-k}$, but $ 4b^{2}k>k$, Contradiction! I wonder if $ 4b^{2}k^{2}+4b^{2}k-k$ can never be a perfect square cuz I couldn't prove it independently (without assuming the problem is true).

never mind

Tomaths wrote: nealth wrote: I wonder if $ 4b^{2}k^{2}+4b^{2}k-k$ can never be a perfect square cuz I couldn't prove it independently (without assuming the problem is true).[/b] Its false set $ b=1$ and then we have $ \sqrt{k(4k+3)}$ then assume they are coprime you could construct solutions using Pell's i.e $ 4k+3=m^{2}$ and $ k=n^{2}\implies 4n^{2}-m^{2}=3$ Unfortunately, $ 4k+3$ cannot be a square. Actually, this must be true (otherwise the original problem is false), but i want to know if we can prove this independently.

Another solution by Vieta Jumping. Thank's the hint by #2 $4ab-1\mid(4a^2-1)^2b^2=(4a^2b-b)^2\Rightarrow\dfrac{(a-b)^2}{4ab-1}\in\mathbb{Z}$ Let $S=\{(a,b)\in\mathbb{N}^2:a<b,\dfrac{(a-b)^2}{4ab-1}\in\mathbb{N}\}$ Let $(x,y)\in S$ s.t. $\forall\,(x',y')\in S,x<x'\lor(x=x'\land y\leq y')$ Consider the equation $f(b)=\dfrac{(x-b)^2}{4xb-1}=k$ That is $b^2-2(2k+1)xb+(x^2+k)=(b-y)(b-y')=0$ ,where $b=y,y'$ with $x<y$ $\Rightarrow\begin{cases}y+y'=2(2k+1)x\Rightarrow y'\in\mathbb{Z}\\yy'=x^2+k\Rightarrow y'>0\end{cases}\therefore y'\in\mathbb{N}$ By assuming: $y\leq y'$ $\Rightarrow 0<(x-y)(x-y')=x^2-2(2k+1)x^2+(x^2+k)=k(1-4x^2),\to\leftarrow$

Nothing new but here is my solution. Notice that we have $(4ab-1, 4ab) = 1$, so $$4ab-1\mid (4a^2-1)^2 \implies 4ab-1 \mid (4a^2-4ab)^2 = (4a)^2(a-b)^2 \implies 4ab-1 \mid (a-b)^2.$$Assume FTSOC that $(A, B)$ is an integer solution to $k = \frac{(a-b)^2}{4ab-1}$ with $a \ne b$, such that $A>B$ and $A$ is as small as possible. We consider the quadratic $x^2-x(2B+4Bk)+(B^2+k)$ which has $A$ as a root by our hypothesis. Notice that the other root $r = 2B+4Bk-A = \frac{B^2+k}{A}$, so $(r, B)$ is also a solution which means $A < r$. However, $$k = \frac{(A-B)^2}{4AB-1} < (A-B)^2 \le A^2-AB \le A^2-B^2$$yet $r > A \implies \frac{B^2+k}{A} > A \implies k > A^2-B^2$, a contradiction. Therefore, this shows that $a = b$ as desired. $\blacksquare$

Note \[4ab-1|(4a^2-1)^2 \implies 4ab-1|(4a^2b-b)^2 \implies 4ab-1|(a-b)^2\]where we can perform the second step as $\gcd(4ab-1,b^2)=1$. Now we Vieta jump. Consider a pair $(a,b)$ that works and FTSOC assume $a\neq b$ and WLOG $a>b$. Let $k=\frac{(a-b)^2}{4ab-1}$. Then we have \[a^2-a(2b+4kb)+b^2k=0\], so if $(a,b)$ is a solution, then $(\frac{b^2+k}{a},b)$ is also a solution. Now consider the pair $(a,b)$ with $a>b$ WLOG and let it be the pair with the minimum maximum value of the pair (i.e. $a$, or the max value of the pair, is the minimum across all pairs). Then we also need to have $\frac{b^2+k}{a}\geq a$. We show this is impossible. Now first note that $a \geq k$. To prove this, note we need to show \[a(4ab-1) \geq (a-b)^2 \implies 4a^2b \geq a^2+b^2 \implies 2a^2+2a^2 \geq a^2+b^2\]which is obvious. Thus, we need to show that $\frac{b^2+a}{a} \geq a$ or $b^2+a \geq a^2$ is impossible. We do this by maximizing $b$ i.e. set it to $a-1$. This turns the inequality to $(a-1)^2+a \geq a^2 \implies a^2+1 \geq a^2+a$ which is a contradiction as $a>1$, so we are done.

Consider a solution $a,b$ with $a+b$ minimal. $4ab-1\mid(4a^2-1)^2$ implies $4ab-1\mid(a-b)^2$, so let $\frac{(a-b)^2}{4ab-1}=k$. Assume FTSOC that $a\ne b$, WLOG $a>b$. We have: $$a^2+(-4kb-2b)a+b^2+k=0.$$There is another solution $a_0$ with: $$a_0=\underbrace{4kb+2b-a}_{\in\mathbb Z}=\underbrace{\frac{b^2+k}a}_{>0}.$$Thus $a_0\in\mathbb N$. We have $(2ab-1)(b-a)<0$ and hence the inequalities: $$2ab^2+a-b<2a^2b$$$$2b^3+a-b<2a^2b$$$$2b(a-b)^2+4ab^2-b<4a^2b-a$$$$\frac{2b(a-b)^2}{4ab-1}+b<a$$$$4kb+2b-a<a$$$$a_0<a,$$contradiction.

See that: $$4ab-1\mid(4a^2-1)^2\implies4ab-1\mid(4a^2-1)^2-(4ab-1)(4a^3+a^2-2ab)=(a-b)^2.$$Suppose, FTSOC, that there exists a solutions $(a_0,b_0)$, with $a_0+b_0$ minimal and, WLOG, $a_0>b_0$. We have: $$\frac{(a_0-b_0)^2}{4a_0b_0-1}=k\implies a_0^2-a_0(2b_0+4b_0k)+b_0^2+k=0,$$and if $x$ is the other root of this quadratic, we have $x+a_0=2b_0+4b_0k$, which implies $x\in\mathbb{Z}$, and $x=\frac{b^2+k}{a_0}$. But since $a_0,b_0>0$, we have $k<(a_0-b_0)^2=(a^2-a_0b_0)+b_0(b_0-a_0)<a_0^2-b_0^2\implies x=\frac{b^2+k}{a_0}<\frac{a_0^2}{a_0}=a_0$, which is a contradiction if $x\neq b_0$. But if $x=b_0$, we have $a_0=b_0(4k+1)$ and $4b_0^2(4k+1)-1\mid16b_0^2k^2\implies4b_0^2(4k+1)-1\mid4b_0^2k-k$, which is false by its size. $\blacksquare$

Observe that $$4ab - 1\mid (4a^2-1)^2 \iff 4ab-1 \mid (4a^2b-b)^2 \iff 4ab-1 \mid (a-b)^2.$$Next let $a+b$ be minimal and suppose $b \neq a$. Then there exists a positive integer $c$ such that $$c(4ab-1) = (a-b)^2 \iff a^2-(2b+4bc)a + b^2+c=0.$$Thus, the positive integer solution $\left(\frac{b^2+c}a, b\right)$ works as well, and by minimality we must have $\frac{b^2+c}a \geq a$. So $$b^2+(a-b)^2 \geq b^2+c \geq a^2 \iff a^2-b^2 \leq a-b,$$which is impossible; thus we have a contradiction.

Note that since $\gcd(b,4ab-1)=1$, we have $4ab-1\mid b^2(4a^2-1)^2$. Expanding, we see that $b^2(4a^2-1)^2\equiv (a-b)^2 \pmod{4ab-1}$, so we get $4ab-1\mid (a-b)^2$. Now assume for contradiction that $a\neq b$, and WLOG, $a>b$. Fix $k$,and consider all positive integer values of $x$ such that $k=\tfrac{(x-b)^2}{4bx-1}$. This is $$x^2-2bx-4bkx+b^2+k=0.$$We know that $a$ is a solution, and there is another solution $x_1$. Since by Vieta's, we have $x_1=2b+4bk-a$, so $x_1$ is an integer, since $k,a,b$ are integers. Again by Vieta's, we have $x_1=\tfrac{b^2+k}{a}$, which is positive, so $x_1$ is a positive integer. We now prove that $x_1<a$. We want to show that $x_1=\tfrac{b^2+k}{a}<a$, which is equivalent to showing that $k=\tfrac{(a-b)^2}{4ab-1}<a^2-b^2$. Suppose for the sake of contradiction that $k>a^2-b^2$, then we get $\tfrac{a-b}{4ab-1}>a+b$, since $a-b>0$ by assumption. Multiplying by $4ab-1$ (which is positive since $a,b$ are positive integers), expanding, and simplifying, we get $a>2ab+4b^2$. This is clearly false, since $a,b$ are positive integers and $2ab>a$, so $k<a^2-b^2$, i.e. $x_1<a$. Thus if an ordered pair $(a,b)$ of distinct positive integers satisfies $k=\tfrac{(a-b)^2}{4ab-1}$, where $k$ is a positive integer, then a smaller ordered pair of positive integers also satisfies this relation, which is a contradiction due to infinite descent. Therefore $a\neq b$ is impossible, i.e. we must have $a=b$.

We have $$(4ab-1)|(16a^4-8a^2+1)$$$$(4ab-1)|(16a^4b-8a^2b+b)$$$$(4ab-1)|(4a^3-8a^2b+b)$$$$(4ab-1)|(4a^3-2a+b)$$$$(4ab-1)|(4a^3b-2ab+b^2)$$$$(4ab-1)|(a^2-2ab+b^2)$$$$(4ab-1)|(a-b)^2.$$ Note that by symmetry, if $(a,b)$ is a solution, then $(b,a)$ is also a solution. We will then use Vieta jumping. Claim: If $n=\frac{(a-b)^2}{4ab-1}$, then $\frac{b^2+n}{a}<a.$ Expanding this out, we get $n<a^2-b^2$, or $(a-b)^2<(a+b)(a-b)(4ab-1)$ so $(a-b)<(a+b)(4ab-1)$, which is obviously true since $a+b>a-b$ and $4ab-1>1.$ Letting $n=\frac{(a-b)^2}{4ab-1},$ we have that $$a^2+a(-2b-4bn)+b^2+n.$$Suppose FTOSC that there exists a solution with $a>b$ (symmetry can be used for $a<b$), and further suppose that $(a_1,b_1)$ is the solution with the smallest value of $a_1$ that has $a_1>b_1$. If $(a_1,b_1)$ is a solution, by Vieta jumping, setting $a$ equal to $2b_1+4b_1n-a_1=\frac{b_1^2+n}{a_1}$ is a solution, so $(\frac{b_1^2+n}{a_1},b_1)$ is also a solution. However, we have that $\frac{b_1^2+n}{a_1}<a_1$ by our claim, contradicting the minimality of $a$.

Firstly observe that $(a-b)^2\equiv (4a^2b-b)^2\equiv (4a^2-1)^2 \pmod {4ab-1}.$ Assume $a>b$ and denote by $(A,B)$ the solution of $\frac{(a-b)^2}{4ab-1}=k\in \mathbb{Z}$ which minimizes $a+b$. We have a Vieta jump $$(A,B)\rightarrow (A_0,B)=(4Bk+2B-A,B)=\left( \frac{B^2+k}{A},B\right),$$with $A_0\in \mathbb{Z}^+.$ But clearly $2Bk<A-B\implies A_0+B<A+B,$ the contradiction completes our proof.

If $4ab - 1$ divides $(4a^{2} - 1)^{2}$, then we have: $$\frac{(4a^2 - 1)^2}{4ab - 1} = k$$ where $k$ is a positive integer. Expanding the left side: $$\frac{16a^4 - 8a^2 + 1}{4ab - 1} = k$$ Multiplying both sides by $4ab - 1$: $$16a^4 - 8a^2 + 1 = k(4ab - 1)$$ Expanding the right side: $$16a^4 - 8a^2 + 1 = 4kab - k$$ Isolating $16a^4$: $$16a^4 = 4kab + 8a^2 + k - 1$$ Since $a$ and $b$ are positive integers, both sides must be positive, which implies that $k > 0$. Thus, $4kab + 8a^2 + k - 1 > 0$. Factoring the left side: $$(2a\sqrt{k})^2 + (2\sqrt{k})^2 + k - 1 > 0$$ Since $k > 0$, both $(2a\sqrt{k})^2$ and $(2\sqrt{k})^2$ are positive, so the left side is always positive. However, $k - 1 < 0$ for all $k > 0$, so we have a contradiction. Hence, it must be the case that $a = b$.

A neat Vieta Jumping Exercise, assume $a \neq b$ and $a > b$, now we take the minimal $a+b$ which satisfies this equation. Note that $4ab - 1 \mid (a-b)^{2}$. $$\frac{(a-b)^{2}}{4ab-1} = c \implies (a-b)^{2} = c(4ac-1)$$Just take $a' = 2b + 4bc - a = \frac{b^{2}+c}{a}$, which satisfies the above quadratic equation and contradicts the minimality of $a+b$ since $a' < a$, so we're done.

AFTSOC WLOG $a > b$ with minimal $a+b$ which satisfies this equation. Notice $$(4ab-1)\mid(4a^2b-b-a(4ab-1))^2=(a-b)^2\iff\frac{(a-b)^2}{4ab-1} = c \implies (a-b)^{2} = c(4ab-1)\implies a^2-(4bc+2b)a+(b^2+c)=0.$$But notice taking $a_0=\frac{b^2+c}{a}$ gives a smaller solution, which is easily seen because substituting and multiplying by $\frac{a^2}{b^2+c}$ returns to the original equation equaling 0; in particular, this is a contradiction because we assumed minimum, yet it's easy to check that $\frac{b^2+c}{a}<a$ making a smaller solution. As a result, a must equal b.

Yay! My first actual Vieta Jump. This was pretty standard. Something funny that I'm surprised no one noticed yet is that the names of both the problem proposers are birds. We first manipulate the divisibility constraint to make everything more symmetric. Note that, \begin{align*} 4ab-1 &\mid (4a^2-1)^2\\ 4ab-1 &\mid 16a^4 - 8a^2 +1\\ 4ab-1 &\mid 16a^4 -8a^2+4ab\\ 4ab-1 &\mid 16a^3 - 8a + 4b\\ 4ab-1 &\mid 16a^3 - 8a + 16ab^2\\ 4ab-1 &\mid 16a^2 - 8a + 16b^2\\ 4ab-1 &\mid 2a^2+2b^2-1 \end{align*}where much of the cancellation was done leveraging the fact that $4ab-1$ is coprime with each of $a$ , $b$ and $4$. Now, we can rewrite the divisibility condition as, \begin{align*} 2a^2 + 2b^2 -1 &= k(4ab-1)\\ 2a^2 - (4kb)a + 2b^2 +k-1 &=0 \end{align*}We wish to show that this equation has no solutions for $a\ne b$. It is easy to check that if $k=1$ then we must have $a=b$ and vice versa. In what proceeds, we consider $k>1$. Assuming for the sake of contradiction there exists solutions, consider a minimal solution pair $(a_0,b_0)$, where WLOG $a_0 > b_0$. Then, looking at $2a^2 - (4kb_0)a + 2b_0^2 +k-1 =0$ as a quadratic in $a$, consider the roots $a_0$ and $a_1$. We know, $a_1=4kb_0-a_0$ from which it follows that $a_1 \in \mathbb{Z}$. Further since, $a_1=\frac{2b^2+k-1}{a_0} > 0$, it follows that $a_1$ is actually a positive integer. Thus, $(b_0,a_1)$ is also a valid solution pair. Considering the minimality of the solution pair $(a_0,b_0)$ we have \[a_1 = 2kb_0 - a_0 \ge a_0\]from which it follows that $k \ge \frac{a_0}{b_0}$. Thus, \begin{align*} 2a_0^2 + 2b_0^2 -1 &\ge k(4a_0b_0-1)\\ 2a_0^2 + 2b_0^2 -1 &\ge \frac{a_0}{b_0}(4a_0b_0-1)\\ 2a_0^2 + 2b^2 -1 &\ge 4a_0^2 - \frac{a_0}{b_0}\\ 2a_0^2b_0 &\le 2b_0^3 + a_0-b_0\\ 2b_0(a_0-b_0)(a_0+b_0) & \le a_0-b_0\\ 2b_0(a_0+b_0) &\le 1 \end{align*}since $a_0 > b_0$. But this is a clear contradiction since $a_0$ and $b_0$ are both positive integers. Thus, there cannot exist such a minimal solution, so there indeed exists no solutions to the given equation for $k>1$, as desired.