Igor wrote: Find all ordered pairs $(a,b)$ of positive integers for which the numbers $\dfrac{a^3b-1}{a+1}$ and $\dfrac{b^3a+1}{b-1}$ are both positive integers. We have $a+1|a^3b-1$ implies $a+1|b(a^3+1)-(b+1)$. Therefore $a+1|b+1$. Similarly we have $b-1|a+1$. Hence $b-1|b+1$. It follows that $b-1|2$. From here it is easy to find $a,b$.

\[ a+1|a^3+1 , a+1|a^3b-1 \Longrightarrow a+1|a^3+a^3b = a^3(b+1) \Longrightarrow a+1|b+1 \]\[ b-1|b^3-1 , b-1|b^3a+1 \Longrightarrow b-1|b^3+b^3a = b^3(a+1) %Error. "Longrigharriw" is a bad command. b-1|a+1 \] So we get: \[ b-1|a+1|b+1 \Longrightarrow b-1|b+1 , b-1|b-1 \Longrightarrow b-1|2 \Longrightarrow b=2,3 \] Case 1: \[ b=2 \Longrightarrow a+1|3 \Longrightarrow a=2 \] Which is a solution. Case 2: \[ b=3 \Longrightarrow a+1|4 \Longrightarrow a=1,3 \] Which are solutions. So the solutions are: \[ (a,b)=(2,2)=(1,3)=(3,3) \]

You just correctly got $b$ equal to $2$ or $3$, and then you try $b=1$ and $b=2$. Two more solutions for $b=3$ ($a=1$ and $a=3$). And, since we discuss your effort, $b=1$ leading to $a=1$, so $a^3b-1 = 0$, is not the contradiction, rather it is since $b=1$ is disallowed, as being a denominator.

Sorry for my mistake , edited . Thanks

$ b-1 | b^3a+1 $ , $ b-1 | b-1 $ $\implies$ $ b-1 | b(b^2a+1) $, but $ (b, b-1)=1$ $\implies$ $ b-1 | b^2a+1 $ Doing twice again the same thing, we'll obtain that $ b-1 | a+1 $ $(1)$ $ a+1 | a^3b-1 $ , $ a+1 | a+1$ $\implies$ $ a+1 | a^2b+1 $. From $(1)$ $\implies$ $ b-1 | a^2b+1 $, but $ b-1|b-1 $ $\implies$ $b-1 | a^2+1$ $(2)$. From $(1)$ and $(2)$, we get $ b-1 | 2 $ $\implies$ $b \in \{2,3\}$. Now we just have to find $a$ and we can do it in the same way as War-Hammer did.

$a+1|a^3b-1$ and $a+1|a^3+1$ => $a+1|b+1$. $b-1|b^3a+1$ and $b-1|b^3-1$ => $b-1|a+1$=> $a+1|b+1$ and $b-1|a+1$ => $b-1|b+1$ => $b-1|2$ => $b=2;3$ => $a=2;1,3$=> $(a,b)=(2,2)(1,3)(3,3)$!

Same as post #3.

Such a problem doesn't teach one anything, it clearly shows what one has to do ... First problem from Romanian District Round is harder than this one ...

The issue is not how difficult it is; problem 1 should be affordable by a large range of differently trained juniors. The issue is with the lack of relevance and the banal technique needed; it has no moment when one needs the slightest jump of insight. Look, for example at this, now classical, type of question. Find all pairs $(a,b)$ of positive integers for which $\dfrac {a^2+1} {b}$ and $\dfrac {b^2+1} {a}$ are simultaneously integer. There are some easy steps, like $\bullet$ Notice $\gcd(a,b) = 1$. $\bullet$ Solve for $a=b$, whence therefore $a=b=1$. There is a simple step, but which requires that "jump of insight". $\bullet$ If we write the relations as $a\mid b^2+1$ and $b\mid a^2+1$, we derive $a, b \mid a^2+b^2+1$. This seems inconsequential, but now, based on $\gcd(a,b) = 1$, we can also derive $ab \mid a^2+b^2+1$. This is creative, and almost essential for solving the problem in the sequel, where the hard techniques arrive. Finally, the technically hard part, known in our circles as Vieta Jumping (of course, too difficult and ill-suited for a problem 1). $\bullet$ Demand $\dfrac {a^2+b^2+1} {ab} = k\in \mathbb{N}^*$, and write $a^2 - kba + b^2 + 1 = 0$ for a solution $(a,b)$ with $a > b$. The other root $a'$ obeys $a+a' = kb$ and $aa' = b^2+1$. Thus $a' = kb - a$ is an integer, and since $a>0$, $aa' = b^2+1>0$, it follows $a' > 0$ and $a' \leq b$. But $(a,b)$ being a solution, so is $(a', b)$, therefore also $(b,a')$. Now, if $b > a'$, we can repeat the procedure, until by the principle of infinite descent we must fall into the case when the two variables are equal, shown to only possibly be $(1,1)$. But then we must have $k=3$, and reversing the procedure allows the creation of the unique family of solutions $(1,1)\mapsto (2,1) \mapsto (5,2)\mapsto (13,5)\mapsto \cdots$, in general $(a,b)\mapsto (3a-b, a)$. We may then notice the relation to Fibonacci numbers, get a closed form, etc.

If $ \frac{a^3b-1}{a+1} $ and $ \frac{b^3a+1}{b-1} $ are both integers, then $a+1|a^3b-1$ and $b-1|b^3a+1$. But $a+1|a^3+1\implies a+1|a^3b+b$ so $a+1|(a^3b+b)-(a^3b-1)=b+1$. Similarly, $b-1|b^3-1\implies b-1|b^3a-a$ so $b-1|(b^3a+1)-(b^3a-a)=a+1$. So $b-1|a+1|b+1$. Thus $b-1|b+1$. Thus we have either $b-1=b+1$ which is impossible or $b+1\geq 2(b-1)\implies b\leq 3$ so $b=\{1, 2, 3\}$. Trying all three possibilities we have $b=1$ is impossible since that would make the denominator of $ \frac{b^3a+1}{b-1} $ to be zero, $b=2$ yields $a=2$ and $b=3$ yields $a=\{1, 3\}$ so we have $(a,b)=\boxed{\{(1, 3),(2, 2),(3,3)\}}$.

Do anyone have open question for this problem ??

Yes, I did mention (if you bother to look) that these solutions are related to the Fibonacci numbers. But just noticing they are solutions (which is not difficult to see and prove), is only half the way; you need also show there are no others!

from the 1st condition , we have , $a+1$ divides $b+1$. from the second condition , we get that , $b-1$ divides $a+1$ . hence $b-1$ divides $b+1$ ,i.e. , $b-1$ divides $2$ and now it's a straightforward case chase

$ a+1|b+1 $ and $ b-1|a+1, a\le\ b $ and $ b-1\le\ a+1 $ or $ a+2\ge\ b\ge\ a $ Now, because $ b $ is positive integer, we need to check the cases $ b=a+2, b=a+1 $ and $ b=a $

Using mods: \[a^3b -1 \equiv 0 \pmod{a+1} \Rightarrow b \equiv -1 \pmod{a+1} \\ b^3a + 1 \equiv 0 \pmod{b-1} \Rightarrow a \equiv -1 \pmod{b-1}\] Therefore, we have $b-1 | a+1$ and $a+1 | b+1$, so $b-1|b+1$. But if $b-1 | b+1$, then $b-1 | (b+1)-(b-1) = 2$. Therefore, $b = 2$ or $b = 3$. If $b = 2$, then we have $2 \equiv -1 \pmod{a+1}$, so $a+1 | 3$. This means $a = 2$. If $b = 3$, then we have $3 \equiv -1 \pmod{a+1}$, so $a+1 | 4$, which gives $a = 3$ or $a = 1$. Therefore, our solutions are $(1, 3); (2, 2); (3, 3)$.

a+1|a^3*b-1=b(a^3+1)-b-1=b(a+1)(a^2-a+1)-b-1a+1|b+1(1) b-1|b^3*a+1=a(b^3-1)+a+1=a(b-1)(b^2+b+1)+a+1b-1|a+1 (2) an observation (n|m and n,m€N→n<=m) so a+1<=b+1 and b-1<=a+1→b-1<=a+1<=b+1 we have 3 cases 1)a+1=b-1 so b-1|b+1 and b-1|b-1→b-1|2→b-1€{1,2}→b€{2,3} and a+1€{1,2}→a€{0,1} a>0→the firs solution is (a,b)€(1,3) 2)a+1=b→b|b+1 and b|b→b|1→b=1→a+1=1→a=0(F) a>0 3)a+1=b+1→a=b→b-1|b+1 and b-1|b-1→b-1|2→b-1€{1,2}→b€{2,3} and a€{2,3} So the solution are (a,b)€{(1,3),(2,2),(3,3)}

Possible pairs are (1,3),(2,2),(3,3). This follows directly from the fact that (b-1)|(b+1) So, b+1 = m(b-1) for some m>1 Thus, (b+1)-(b-1) = (m-1)(b-1), or, (m-1)(b-1) = 2 Some simple case chasing yields the values.

$a+1|a^3b–1$,$b–1|b^3a+1$ $a+1\le a^3b–1$,$b–1\le b^3a+1$ $\implies a+1+b–1\le ab(a^2+b^2)$ $\implies a+b\le (a^2+b^2)ab$ If} $a|b\implies a\le |b|$ $a+b=ab$ $\implies (a–1)(b–1)=1$ $(a, b) =(2,2)$ case 2 $a+b=ab(a^2+b^2)$,$a+b=(a^2+b^2)$ good case $(a+b)(ab(a+b)–1)=2a^2b^2$ Case $ab=3,ab(a+b)=3$ ( case) $(a, b) =(1,3)$ And also $a+b\neq 1$,$a, b\in\mathbb{Z^+}$ Also $(3,3)$ is a solution Clearly all fits in solution $a+b\neq 2$ .also I left other cases as it will not be possible can show it by $mod$ .

Adding $1$ to both the given numbers we get: $\dfrac{a^3b-1}{a+1} + 1$ is also a positive integer so we have: $\dfrac{a^3b+a}{a+1}$ = $\dfrac{a(a^2b+1)}{a+1}$ is a positive integer $ => (a+1) | (a^2b+1)$ $ => (a+1) | (((a+1) - 1)^2b+1)$ $ => (a+1) | (b+1)$ Similarly, $\dfrac{b^3a+1}{b-1} + 1$ is also a positive integer so we have: $\dfrac{b^3a+b}{b-1}$ = $\dfrac{b(b^2a+1)}{b-1}$ is a positive integer $ => (b-1) | (b^2a+1)$ $ => (b-1) | (((b-1) + 1)^2a+1)$ $ => (b-1) | (a+1)$ Combining above $2$ results we get: $(b-1) | (b+1)$ $ => b=2,3 $ $Case 1: b=2$ $ => a+1|3 => a=2 $ which is a valid solution. $Case 2: b=3$ $ => a+1|4 => a=1,3 $ which are valid solutions. Thus, all solutions are: $(2,2), (1,3), (3,3)$

just put that a^3 is -1 mod (a+1) because a is -1 mod (a+1) . Do the same for b,and you get b-1 divides b+1. . This means b-1 divides 2 and just prove the cases.

Looks like vieta jumping

Пусть $O$ - центр окружности, описанной около треугольника $ABC$. Тогда $AO \perp BC$ и $AH \perp BC$, следовательно, $A, O, H$ лежат на одной прямой. Так как $AE$ является медианой треугольника $ABC$, то $AE = \frac{1}{2} AC = \frac{1}{2} AB$, а значит, треугольник $ABE$ является равнобедренным, и $BE \parallel OH$. Также заметим, что $CD = \frac{1}{3} BC$, а значит, точка $D$ делит отрезок $BC$ в отношении $1:3$. Тогда $BD = \frac{2}{3} BC$. Проведем высоту $HK$ треугольника $BDH$ из вершины $H$ на сторону $BD$. Так как $\angle BHD = 90^\circ$, то $HK$ является медианой треугольника $BDH$, а значит, $HK = \frac{1}{2} BD = \frac{1}{3} BC$. Теперь рассмотрим треугольник $BEC$. Так как он равнобедренный, то $BE = EC$. Кроме того, так как $AE$ является медианой, то $EH = \frac{1}{2} AH$. Но $AH = 2 \cdot HK$, поэтому $EH = \frac{1}{3} BC$. Таким образом, мы получили, что $BE = EC$ и $EH = \frac{1}{3} BC$, а значит, треугольник $BEH$ является прямоугольным и равнобедренным. Следовательно, $BE \perp HD$, что и требовалось доказать.

Notice that, $$a+1 |b(a^3+1)-(a^3b-1)=b+1$$$$b-1|(b^3a+1)-a(b^3-1)=a+1$$In particular, we have $$b-1 \le a+1 \le b+1$$If $a+1=b-1$, $a+1=b-1|b+1 \rightarrow (a,b) =(1, 3)$. If $a+1=b$, $a+1|b+1=a+2$. No solution. If $a+1=b+1$, $b-1|a+1=b+1 \rightarrow (a, b) =(2, 2),(3,3)$. Hence, all possible solutions are $(a, b) =(1, 3),(2,2),(3,3)$.

If $a+1|a^3b-1$ then $a+1$ should also perfectly divide $a²b+1$ and repeatedly $1-ab$ and lastly $b+1$ too. So $a+1|b+1$. Samely $b-1|a+1$. We know that $gcd(b+1,b-1)$ can be equal to 1 if they are odd except one of them equal to 1 in positive integer. Let we call this one case 1 and extra one case would be chosen for $b-1=1$ , notice that $b+1$ cannot be equal to $1$ in positive integers. Another case is that both $b+1 and b-1$ is even numbers and their gcd=2. => Case 1 $(b+1)= (a+1).k k≥1$ $(a+1)= (b-1).p p≥1$ $-->(b+1)=(b-1).t.k$ however in this case $gcd(b+1,b-1)= 1$ so k=t=1 but $b-1≠b+1$ so this case fails. ==>Case 2 If $(b-1)=1 , b+1=3 -> gcd(b+1,b-1)=3$ and it doesn't really matter whether $k=3$ or $t=3$ , the conclusion is the same $(a,b)= (2,2)$ ==>Case 3 Now , whether $t$ or $k$ is equal to $gcd(b+1,b-1)=2 as even integers$ matters. Both t and k cannot be equal to 2 because of greatest common divisors of this integers cannot be 4. If $t=2 , k=1$ : $(b+1)=(a+1)=2(b-1)$ -> $b=3$ and $a=3$. $(a,b)=(3,3)$ If $t=1 , k=2$ $(b+1)=2(a+1)=2(b-1)$ ->$b=3$ but notice that a is different in here , $a=1$ $(a,b)=(1,3)$ We are done. All $(a,b)$ pairs are $(2,2)$ , $(3,3)$ , $(1,3)$. H.Y.E