orl wrote: Find all positive integers $ a$ and $ b$ for which \[ \left \lfloor \frac {a^2}{b} \right \rfloor + \left \lfloor \frac {b^2}{a} \right \rfloor = \left \lfloor \frac {a^2 + b^2}{ab} \right \rfloor + ab. \] I only remember the result $ (n,n^2+1)$ and $ (n^2+1,n)$

" wrote: we must have $ a\neq b$,so WLOG assume that $ a > b$ and let: $ \lfloor\frac {a^2}b\rfloor = \alpha,\lfloor\frac {b^2}a\rfloor = \beta$. so we must have: $ \begin{eqnarray*}\alpha + \beta & = & \lfloor\frac {a^2 + b^2}{ab}\rfloor + ab \\ & = & \lfloor\frac {a^2 + b^2}{ab}\rfloor + \frac {a^2}b\cdot\frac {b^2}a \\ & \geq & 2 + \alpha\beta\Rightarrow a\lpha + \beta\geq 2 + \alpha\beta \\ & \Rightarrow & (\alpha - 1)(\beta - 1)\leq - 1$ but we have $ \alpha\geq 1$ so $ \beta = 0$ which means that $ a > b^2$.so assume that $ a = b^2 + c$ in which $ c$ is a natural number,so if we plug in $ b^2 + c$ instead of $ a$ in the given equation,we get that: $ b^3 + 2bc + \lfloor\frac {c^2}b\rfloor = \lfloor\frac {b^4 + 2b^2c + c^2 + b^2}{b^3 + bc}\rfloor + b^3 + bc$ $ \Rightarrow (c - 1)b + \lfloor\frac {c^2}b\rfloor = \lfloor\frac {b^2(c + 1) + c^2}{b^3 + bc}\rfloor$ (1) now the last equation obviously holds for $ c = 1$ so $ (a,b) = (n^2 + 1,n)$ is a solution,now let $ c\geq 2$,from (1) we have: $ (c - 1)b\leq\frac {b^2(c + 1) + c^2}{b^3 + bc}$ which is equivalent to: ${ c^2(b^2 - 1) + b^2\{c(b^2 - 2) - (b^2 + 1)}\leq 0$ (2) $ \Rightarrow c(b^2 - 2) - (b^2 + 1)\leq 0$ $ \Rightarrow 2(b^2 - 2) - (b^2 + 1)\leq 0$ $ \Rightarrow b = 1\textrm{ or }2$ for $ b = 1$ we get $ a = 2$,also for $ b = 2$,using (2) we get that: $ 3c^2 + 4(2c - 5)\leq 0$ so we must have $ c < 2$,contradiction. so the only solutions are $ (a,b) = (n,n^2 + 1)$ and $ (a,b) = (n^2 + 1,n)$ in which $ n$ runs over the naturals.

We claim that all ordered pairs of integers $(a, b)$ are of the form $(k, k^2+1)$ or $(k^2+1, k)$ for $k\in\mathbb{Z}^{+}$. Let $a=b^2+1$. If $b>1$, then we have \[\text{LHS}=\left[b^3+2b+\frac{1}{b}\right]+\left[\frac{b^2}{b^2+1}\right]=b^3+2b,\] and \[\text{RHS}=\left[b+\frac{1}{b}+\frac{b}{b^2+1}\right]+b^3+b=b^3+2b,\] so we have a solution. It is easy to confirm that $b=1$ also yields a solution. There is an obvious symmetry between $a$ and $b$, so $b=a^2+1$ is also a solution. Now we must show there are no other solutions. WLOG, asume $b\ge a$. Let $a$ be fixed and let $f(b)=\frac{a^2}{b}+\frac{b^2}{a}-\left(\frac{a}{b}+\frac{b}{a}\right)-ab$ and $g(b)=\frac{a^2}{b}+\frac{b^2}{a}-\frac{a}{b}-\frac{b}{a}-ab$. From this we have $|g(a)-f(a)|<3$. Observe that we have \[g(a)=\frac{1}{a}\left(b-\frac{a^2+1}{2}\right)^2=\frac{1}{4m}\left(\frac{a^2+1}{2}\right)^2=\frac{a^2-a}{b}.\] So if $b\ge a^2+2$, then \[g(b)\ge a+\frac{2}{a}+\frac{a^2-a}{a^2+2}>a.\] Hence $f(b)>a-3$. So for $m>2$ there are no solutions for $b>a^2+1$. Also if $a\le b\le a^2-1$, then $\left(b-\frac{a^2+1}{2}\right)^2\le\frac{(a^2-3)^2}{4}$ and $\frac{a^2-a}{b}\le a-1$, so $g(b)\le -a+\frac{2}{a}-1<0$. Hence $f(b)<0$ for $b>2$. So for $b>2$ there are also no solutions for $b<a^2$. It is also easy to check that $f(a^2)=-(a-1)<0$. Now, it remains to check $a=1$ and $a=2$. For $a=1$, we have $f(1)=-1$. If $b>1$ then we have $f(b)=0+b^2-2b=b(b-2)$. So for $a=1$ the only solution is $b=2$. Finally take $a=2$, and consider $b\ge 2$. It is easy to check that $f(2)=-2$, $f(3)=-3$, and $f(4)=-1$. For $b\ge 6$ we have $f(b)=0+\frac{b^2}{2}-\frac{b}{2}-2b\ge\frac{b^2}{2}-\frac{1}{2}-\frac{5b}{2}>0$. Hence, $(a, b)= (k, k^2+1)$ or $(k^2+1, k)$ for $k\in\mathbb{Z}^{+}$. $\Box$