http://www.artofproblemsolving.com/Forum/viewtopic.php?f=462&t=150372 helps.

yucky problem. $\left \lfloor \frac{(a-b)^2 - 1}{ab} \right \rfloor = -2 + \left \lfloor \frac{a^2 + b^2 - 1}{ab} \right \rfloor$ $\left \lfloor \frac{(a-b)^2 - 1}{ab-1} \right \rfloor = -2 + \left \lfloor \frac{a^2 + b^2 + 1}{ab-1} \right \rfloor$ Let $\left \lfloor \frac{a^2 + b^2 - 1}{ab} \right \rfloor = q$. Then $qab \le a^2 + b^2 - 1 < qab + ab$ It suffices to show $qab - q \le a^2 + b^2 + 1 < (q+1)(ab-1) = qab + ab - q - 1$ Suppose now, so $a^2 + b^2 + 1 \ge qab + ab - q - 1$. That means $qab + ab - q \le a^2 + b^2 - 2 < qab + ab - 2$ $2 < |a^2 + b^2 - (q+1)ab - 2| \le q < q+1$ Then by a result here (which is just easy vieta jumping to prove), we get $a^2 + b^2 - (q+1)ab$ is a perfect square. However, $a^2 + b^2 - (q+1)ab$ is negative so this is a contradiction. Hence $a^2 + b^2 + 1 < qab + ab - q - 1 \implies \left \lfloor \frac{a^2 + b^2 + 1}{ab-1} \right \rfloor = q$, and hence we are done. To prove the easy result, suppose $|a^2 + b^2 - abc-2| \le c-1$ with $a+b$ minimal and $a^2 + b^2 - abc$ is not a perfect square and $a,b,c \in \mathbb{Z}^+$. WLOG $a > b$ ($a=b$ case is trivial to deal with), and let $a^2 + b^2 - abc = q$. Consider the quadratic $x^2 - xbc + b^2 - q = 0$. We know $a$ is a root, and so is $bc - a = \frac{b^2-q}{a}$ by Vieta's. Remark that $\frac{b^2-q}{a} \le \frac{b^2+c+1}{a} \le \frac{b^2 + a + 1}{a} \le \frac{(a-1)^2 + a + 1}{a} < a$. Hence $b + (bc-a) \le a+b$. Remark this chain of inequalities fails only when $b=1$ as then we don't necessarily have $c \le a$, but this case is easy to check. However, then $(a,b,c) = (bc-a, b, c)$ satisfies the inequality. This gives a contradiction unless $bc - a = 0$, in which case $q = b^2$, contradiction. Hence $a,b$ does not exist and thus we're done.

I think something wrong. $ \left\lfloor \frac {(a-b)^2-1} {ab-1} \right\rfloor = -2 + \left\lfloor \frac {a^2+b^2-3} {ab-1} \right\rfloor . $

math154 wrote: Prove that if $a$ and $b$ are positive integers and $ab>1$, then \[\left\lfloor\frac{(a-b)^2-1}{ab}\right\rfloor=\left\lfloor\frac{(a-b)^2-1}{ab-1}\right\rfloor.\]Here $\lfloor x\rfloor$ denotes the greatest integer not exceeding $x$. Calvin Deng. if $|a-b|=1$ then we are done Note that $\frac{(a-b)^2-1}{ab-1}> \frac{(a-b)^2-1}{ab}$ so $[\frac{(a-b)^2-1}{ab-1}]\ge [\frac{(a-b)^2-1}{ab}]$ Assume that the integer parts are not equal. So there exist a positive integer $c$ such that $\frac{(a-b)^2-1}{ab-1}\ge c>\frac{(a-b)^2-1}{ab}$ We will prove that this inequality does not hold for any $a,b,c\in N$. Assume it holds, then w.l.o.g $a\ge b$ and we can assume that $a+b+c$ is the smallest sum from all $(a,b,c)$ triples that satisfy that inequality. İf inequality holds then we have $c\ge abc-(a-b)^2+1>0$ let $abc-(a-b)^2+1=t$ a fixed positive integer, then the quadratic trinomial $f(x)=x^2-(bc+2b)x+b^2+t-1=(x-x_1)(x-x_2)$ has a solution $a$ and let $a_1$ be the second solution. Then by Viet's theorem we have $a+a_1=bc+2b$ and $aa_1=b^2+t-1$ so from this we have that $a_1$ is also a positive integer. So the triple $(a_1,b,c)$ also satisfies our inequality. So we must have $a_1\ge a$ since $a+b+c$ is a minimal sum of solutions and $a_1b>1$. As we proved we must have $a-b\ge2$ so $a_1-b\ge a-b\ge 2$ so $f(b)=(b-a)(b-a_1)=(a-b)(a_1-b)\ge 4$ which is equivalent to $$b^2-(bc+2b)b+b^2+t-1\ge 4$$$$t-5\ge b^2(c-4)\ge t-4$$which is impossible, contradiction. So such triple does not exist.

Prove that if $a$ and $b$ are distinct positive integers and $ab>1$, then \[\left\lfloor\frac{(a-b)^2}{ab}\right\rfloor=\left\lfloor\frac{(a-b)^2-1}{ab-1}\right\rfloor.\]Here $\lfloor x\rfloor$ denotes the greatest integer not exceeding $x$.

any solution?

rightways wrote: Prove that if $a$ and $b$ are distinct positive integers and $ab>1$, then \[\left\lfloor\frac{(a-b)^2}{ab}\right\rfloor=\left\lfloor\frac{(a-b)^2-1}{ab-1}\right\rfloor.\]Here $\lfloor x\rfloor$ denotes the greatest integer not exceeding $x$. According to the previous problem we know that $\left\lfloor\frac{(a-b)^2-1}{ab}\right\rfloor=\left\lfloor\frac{(a-b)^2-1}{ab-1}\right\rfloor$

So this problem is equivalent to showing that $\left\lfloor\frac{(a-b)^2-1}{ab}\right\rfloor= \left\lfloor\frac{(a-b)^2}{ab}\right\rfloor$ Which in turn requires us to show that $ab$ does not divide $(a-b)^2$ when $a,b$ are distinct. proof

so we are done. the problem can also be shown directly through the vieta jumping in the previous post with a few minor tweaks.