Answers:[/i][/u][/b] $(m,n)={(1,2),(2,1),(1,3),(3,1),(2,2),(2,5),(5,2),(3,5),(5,3)}$ Solution: We know that $n^3$ and $mn-1$ are relatively primes, thus from $n^3(m^3+1)=(m{}^n{}^3+1)+(n^3+1)$ and we see that $n^3+1$ is divisible by $mn-1$ if and only if $m^3+1$ is divisible by $mn-1$. Therefore, if the pairs $(m,n)$ holds then $(n,m)$ also holds. Now in the future we will say that $m \geq n$. $1.$ If $m=n$, $\frac{n^3+1}{n^2-1}=n+\frac{1}{n-1}$ which gives (2,2) $2.$ If $m>n=1$, we have $\frac{n^3+1}{mn-1}=\frac{2}{m-1}$ we get (2,1) and (3,1) $3.$ If $m>n>1$, since $(m,n)$ holds, then natural $\frac{n^3+1}{mn-1}+1$ is divisible by $n$. $\frac{n^3+1}{mn-1}=kn-1$ for some $k$. Thus we get inequality $kn-1<\frac{n^3+1}{n^2-1}=n+\frac{1}{n-1}$, or $(k-1)n<1+\frac{1}{n-1}$. (*) Since $n>1$, (*) holds only when $k=1$. By the way we observe $n^3+1=(n-1)(mn-1)$ gives us $m=\frac{n^2+1}{n-1}=n+1+\frac{2}{n-1}$ which gives us (5,2) and (5,3). davron

here we go- suppose $ \frac{n^3+1}{mn-1}$ is a positive integer, call it k: $ \frac{n^3+1}{mn-1}=k => n^{3}+1=kmn-k => 1+k \equiv 0\mod{n}$ therefore, exists integer t such that 1+k=nt: $ n^{3}+1=(nt-1)(nm-1)=mtn^{2}-n(t+m)+1 =>(*) n^{2}-(mt)n+(t+m)=0$ if we solve this as an equation in n, we must have $ (mt)^2-4(t+m)$ be a perfect square: $ (mt)^2-4(t+m)<(mt)^2$ $ (mt)^2-4(t+m)>(mt-1)^2 \leftrightarrow -4t-4m>-2tm+1 \leftrightarrow 2(t-2)(m-2)>9 \leftrightarrow (t-2)(m-2)>4$ so, if we want a solution, we need $ (t-2)(m-2) \leq 4$. there are few possible cases: m=1: $ \frac{n^3+1}{n-1}=n^{2}+n+1+\frac{2}{n-1}: n=2,3->$(2,1) (3,1) when m>1: t=6: m=2 or m=3. the solutions are not integers. t=5: m=2 or m=3. no integer solutions. t=4: m=2,3, or 4. no integer solutions. t=3: m=2,...,6. m=2: n=1,5: (1,2) (5,2). m=3,4,5,6: no integer solutions. t=2: $ m^2-m-2$ should be a perfect square. for m>3: $ m^2>m^2-m-2>(m-1)^{2}$. for m=2: we get 0 ,and then n=2. (2,2). for m=3: we get the solutions (1,3),(5,3) t=1: $ m^2-4m-4$ should be a square. for m>6: $ (m-2)^2>m^2-4m-4>(m-3)^2$. for 1<m<7: only m=5 gives a square, and the solutions are (2,5),(3,5). m=2: the same as t=2 ($ (*)$ is symmetrical with m and t) - $ t \leq 3$, and we already found the solutions for t=1,2,3. All the solutions are: $ (2,1) (3,1) (1,2) (5,2) (2,2) (1,3) (5,3) (2,5) (3,5)$

I don't understand this part of Davron's solution: Davron wrote: We know that $ n^3$ and $ mn - 1$ are relatively primes, thus from $ n^3(m^3 + 1) = (m{}^n{}^3 + 1) + (n^3 + 1)$

n and mn-1 are relatively prime, because mn-1 is one less than a multiple of n. It follows that $ n^3$ and mn-1 are relatively prime.

all of you right,it's an other question(it's very nice) p is a prime number and m,n are integers.find all(m,n,p): p=m^2+n^2 and m^3+n^3-4\p is integer

We see that \[ \frac{n^3+1}{mn-1} = \frac{m^2n^2+mn+1+\frac{m^3+1}{mn-1}}{m^3},\]which means that $n^3+1$ is divisible by $mn-1$ iff $m^3+1$ is divisible by $mn-1$ as well. This means that if $(m,n)$ works, $(n,m)$ does as well. From now on, we can therefore assume that $m \ge n$. If $n=1$, then $\frac{n^3+1}{mn-1} = \frac{2}{m-1}$, which means that $m=2,3$. Therefore, $(m,n)=(2,1),(3,1),(1,2),(1,3)$. If $n=2$, then $\frac{n^3+1}{mn-1} = \frac{9}{2m-1}$, which means that $m=5,2,1$. Therefore, $(m,n)=(2,5),(2,2),(2,1),(5,2),(1,2)$. We now know that $m \ge n > 2$. We now bring everything in terms of $n$. Since $m \ge n>1$, we have \[ \frac{n^3+1}{mn-1} \le \frac{n^3+1}{n^2-1} = n+\frac{1}{n-1}. \]Now, note that $\frac{n^3+1}{mn-1} + 1 = \frac{n^3+mn}{mn-1}$, which divides $n$. Since $\frac{n^3+1}{mn-1}+1 \le n+1+\frac{1}{n-1}$ and $n-1>1$, we must have $\frac{n^3+1}{mn-1}+1=n$. From this, we get $m=\frac{n^2+1}{n-1} = n+1+\frac{2}{n-1}$. This means that $(m,n)=(5,2),(5,3),(2,5),(3,5)$. Taking the union of all the solutions, we have \[ (m,n)=(2,1),(3,1),(1,2),(1,3),(2,5),(2,2),(5,2),(5,3),(3,5). \]

Instead of $n^3+1$ and $m^3+1$, I prefer using the smaller values $n^2+m$ and $m^2+n$.

Case bashy solution: \[mn - 1\mid m(n^3+1) - n^2(mn-1) = n^2+m.\]Let $\frac{n^2+m}{mn-1} = k\in \mathbb{Z}^{+}$. \[n^2+m = k(mn-1) \implies n = \frac{km \pm \sqrt{k^2m^2-4(m+k)}}{2}.\]Let $k^2m^2 - 4(m+k) = l^2$. As $4(m+k)>0$, \[4(m+k) = (km)^2-l^2 \geq (km)^2-(km-1)^2=2km-1.\]Therefore $4\geq (k-2)(m-2)$. It can be checked easily that none of the unordered pairs $(k,m)=(3,3),(3,4),(3,5),(3,6),(4,4)$ yield a solution. Thus $k=1$ or $k=2$ or $m=1$ or $m=2$. As the equation $(km)^2-l^2 = 4(m+k)$ is symmetric, we can only solve for $k=1$ and $k=2$. Case. $k=1$ $m^2-4m-4 - l^2 = 0$ or $(m-2-l)(m-2+l)= 8$. This gives $m=5,l=1$. Case. $k=2$ $4m^2 - l^2 = 4(m+2)$ or $(2m-1)^2 - l^2 = 9$. Thus $(2m-1-l)(2m-1+l)=9$. This gives $m=2,l=0$ and $m=3,l=4$. For $m=5,l=1,k=1$, $n = \frac{5\pm 1}{2} = 2,3$ and for $m=1,l=1,k=5$, $n = 2,3$. For $m=2,l=0,k=2$, $n = \frac{4}{2} = 2$. For $m=3,l=4,k=2$, $n = \frac{6\pm 4}{2} = 1,5$ and for $m=2,l=4,k=3$, $n=1,5$. Thus we have the solutions: \[\boxed{(m,n) = (2,5),(5,2),(3,5),(5,3),(3,1),(1,3),(1,2),(2,1),(2,2)}.\]

The answer is the set of $9$ pairs \[ (1,2), \, (1,3),\, (5,2),\, (5,3),\, (2,2),\, (2,1),\,(3,1),\,(2,5),\,\text{and }(3,5). \]This can be seen to work; it remains to prove these are the only solutions. Let $\tfrac{n^3+1}{mn-1} = k$, so that $n^3 + 1 = k(mn-1)$. Taking this equality modulo $k$ yields $1\equiv -k\pmod n$, so $k\equiv -1\pmod n$. This means there exists an integer $p$ such that $k = pn-1$, which means we now have the symmetric-looking equation \[ n^3 + 1 = (mn-1)(pn-1). \]Expanding and cancelling out the $1$ yields $n^3 = mpn^2 - (m+p)n$; as $n\neq 0$, we may divide by $n$, which leaves us with the quadratic \[ n^2 - mpn + (m+p) = 0.\qquad(*) \]The discriminant of this quadratic is $\Delta = (mp)^2 - 4(m+p)$; it remains to determine which ordered pairs $(m,p)$ make this a perfect square. We case on the values of $m$ and $p$. If $m = 1$, then the discriminant is $p^2 - 4p - 4 = (p-2)^2 - 8$; this is a perfect square precisely when $p-2 = 3$, or $p = 5$. Similarly, $p = 1$ produces $m = 5$ as the lone solution. If $m = 2$, then the discriminant is $4p^2 - 4p - 8 = (2p-1)^2 - 9$; this is a perfect square precisely when $2p-1$ is either $3$ or $5$, which yields the two solutions $p=2$ and $p=3$. The case $p=2$ similarly yields $m=2$ and $m=3$. Finally, if both $m$ and $p$ are greater than $2$, then the inequality $(m-1)(p-1) > 2$ rewrites as $mp > m + p + 1$, and so \[ (mp)^2 > (mp)^2 - 4(m+p) > (mp)^2 - 4(mp-1) = (mp-2)^2. \]This is a contradiction, as $(mp)^2 - 4(m+p)$ is a perfect square with the same parity as $(mp)^2$ and thus must be at most $(mp-2)^2$. The case analysis above produced three possibilities for the unordered set $\{m,p\}$: $\{1,5\}$, $\{2,2\}$, and $\{2,3\}$. Solving for $n$ in $(*)$ for each of these three possibilities yields the nine ordered pairs shown above.

First, let $m,n \geq 4$. $mn-1 \mid n^3 +1 => mn-1 \mid n(n^2 +m) => mn-1 \mid n^2 +m$. As $mn-1 \mid n^2 +m => mn-1 \mid n^2m+m^2 => mn-1 \mid mn^2+m^2 -n(mn-1) => mn-1 \mid m^2 +n$. Now, we need $m+n^2 \geq mn-1 => n^2 +1 \geq m(n-1) => \frac{(n+1)(n-1) +2}{n-1} \geq m => n+1 + \frac2{n-1} \geq m => n+1 \geq m$. Similarly, we get $m+1 \geq n$ (Note: till here we've assumed $m,n \geq 4$). Thus, $m=n+1,n,n-1$. Now it's just a game of substituting- put $m=n-1$,$m=n$,$m=n+1$ and you'll get no solution for $m,n \leq 4$- here's an example: If $m=n-1 => n^2 -n -1 \mid n^3 +1 => n^2 -n-1 \mid n^2 +n+1 => n^2 -n-1 \mid 2n+2 => 2n+2 \geq n^2 -n-1 => n^2 -3n-3 \leq 0$ which of course has no solutions for $n \geq 4$. What we have left is cases when either $m$ or $n$ is lesser than $4$- we have no choice but to put in values and see which values work, and the math part is over...

Directly from the divisibility condition we get $mn-1|m^2+n, n^2+m, m^3+1$, so $WLOG m \geq n$. We have $mn-1 \leq n^2+m$, so $m(n-1) \leq n^2-1+2$ and therefore $m \leq n+1+\frac{2}{n-1}$. Assume $n \leq 3$. Substituting in the expression we get one-variable polynomial divisibilities and obtain the solutions $(1, 2), (2, 1), (1, 3), (3, 1), (2, 2), (2, 5), (5, 2), (3, 5), (5, 3),$ counting the ones with $n > m$ too. Now suppose $n \geq 4$. This means $m \leq n+1+\frac{2}{n-1} < n+2$. We have the only possibilities $m=n$ and $m=n+1,$ both of which are simple one-variable polynomial divisibilities and bring no new solutions, therefore we are done $\blacksquare$.

ISL 1994 N2 wrote: Find all ordered pairs $ (m,n)$ where $ m$ and $ n$ are positive integers such that $ \frac {n^3 + 1}{mn - 1}$ is an integer. Solution: Note $\gcd(mn-1,n)=1$ $$\implies \frac{n^3+1}{mn-1}=\left(\frac{n^3+1+mn-1}{mn-1}\right) -1 \implies \frac{n^3+mn}{mn-1}=\frac{n(n^2+m)}{mn-1} \in \mathbb{Z} \implies mn-1 \mid n^2+~m $$Now, $$\frac{n^2+m}{mn-1} \in \mathbb{Z} \implies \frac{n(mn-1)+m^2+n}{mn-1} \in \mathbb{Z} \implies mn-1 \mid m^2+~n \implies \begin{cases} \tfrac{m^2+1}{m-1} \geq n \\ \tfrac{n^2+1}{n-1} \geq m \end{cases}$$Hence, $$\begin{cases} m \le \frac{n^2+1}{n-1}=n+1 +\frac{2}{n-1} < n+2 ~ ~ \qquad [ \text{for } n>3] \\n \le \frac{m^2+1}{m-1}=m+1+\frac{2}{m-1} < m+2 ~ ~ \qquad [ \text{for } m>3] \end{cases}$$Hence, $n<m+2<n+4 \implies m=n-1,n,n+1$ $\implies$ Yields No solution. Manually checking values when $m,n \le 3$, we get $\boxed{(m,n) = (1,2), (1,3), (2,1), (2,2), (2,5), (3,1), (3,5), (5,2), (5,3) } \qquad \blacksquare$

Here is a strange solution idea inspired by this. Clearly the quotient is congruent to $-1 \pmod{n}$, so there exists $m_0$ with $n^3 + 1 = (mn - 1)(m_0n - 1)$, which becomes $n^2 = mm_0n - (m + m_0)$ or $n^2 - (m + m_0)n + mm_0$. Thus the quantities $mm_0 - n$ and $\tfrac{m + m_0}{n}$ are equal to some positive integer $n_0$. Thus $mm_0 = n + n_0$ and $nn_0 = m + m_0$, so \[(m - 1)(m_0 - 1) + (n - 1)(n_0 - 1) = 2.\]Now it's easy to finish.

Let $$\frac{n^3 + 1}{mn - 1} = k \Longleftrightarrow n^3+ 1 = mnk - k$$Clearly, taking modulo $n$ from both sides, we see that $k \equiv -1 \mod n$. Therefore, $k$ can be written as $an - 1$ for some $a \in \mathbb{Z}$. We can re-write what we have as: $$\frac{n^3 + 1}{mn - 1} = k \Longleftrightarrow \frac{n^3 + 1}{mn - 1} = an - 1 \Longleftrightarrow n^3 + 1=amn^2 - n(a+m) + 1 \Longleftrightarrow n^2 = amn - (a+m)$$Solving, we see that the discriminant is $(pm)^2 - 4(p+m)$, which we seek to make a perfect square. Claim: $(am)^2 - 4(a+m) > (am)^2$ Assume by sake of contradiction that $(am)^2 - 4(a+m) \le (pm)^2$. That would imply that $a+m$ is nonnegative which clearly isn’t true. Case 1: $(am)^2 - 4(a+m) = (am-k)^2$, where $k$ is odd. If $am$ is odd, then the LHS is odd while the RHS is even. On the other hand, if $am$ is even, the LHS is even, while the RHS is odd. This is a contradiction. Case 2: $(am)^2 - 4(a+m) = (am - 2)^2$ This is the same as saying that $(am)^2 - 4(a+m) = (am)^2 - 4am + 4 \Longleftrightarrow a+m = am - 1 \Longleftrightarrow (a-1)(m-1) = 2 \Longleftrightarrow (a,m) = (3,2),(2,3)$. Case 3: $(am)^2 - 4(a+m) = (am-k)^2$ where $k \ge 4$ Assume by sake of contradiction that $(am)^2 - 4(a+m) = (am-k)^2 \le (am-4)^2 \Longleftrightarrow (2a-1)(2m-1) \le 9$. Here the only solutions are of the form $(a, m) = (4,1),(1,4),(5,1),(1,5),(1,3),(3,1),(2,1),$ and $(1,2),(2,2)$. Checking, we see that the only $(a,m)$ that actually produce a perfect square are $(5,1)$ and $(1,5)$ and $(2,2)$. Putting this all together and included what we found in the previous cases, we see that the final answer is $$\boxed{(1,2), \, (1,3),\, (5,2),\, (5,3),\, (2,2),\, (2,1),\,(3,1),\,(2,5),\,\text{and }(3,5)}.$$

Vieta jumping. We claim the solutions are $(1,2),(1,3),(2,2),(2,5),(3,5)$ and permutations. We first take care of the case $m=n.$ Note that this implies $n^2-1\mid n^3+1,$ or $n^2-1\mid n+1,$ or $n-1\mid 1,$ implying $n=2.$ For the rest of the proof, we assume $m\neq n.$ Claim (Symmetry): For $m,n,$ $\frac{n^3+1}{mn-1}$ is integer if and only if $\frac{m^3+1}{mn-1}$ is integer. Proof: Obviously this only needs to be proven in one direction. Note that \[mn-1\mid n^3+1\implies\]\[mn-1\mid mn^3+m\implies\]\[mn-1\mid n^2+m\implies\]\[mn-1\mid mn^2+m^2\implies\]\[mn-1\mid n+m^2\implies\]\[mn-1\mid nm+m^3\implies\]\[mn-1\mid m^3+1.\] Now take $mn-1$ and $n^3+1$ mod $n,$ and note that $\frac{n^3+1}{mn-1}\equiv -1\pmod{n},$ so we can express it as $m'n-1.$ Claim (Bounding): We claim that for $n\geq 4,$ both $m$ and $m'$ must be less than $n.$ \end{fact} Proof: Note that $(mn-1)(m'n-1)=n^3+1.$ We cannot have $m=1$ since $n-1\mid n^3+1\implies n-1\mid 2,$ and $n-1\geq 3.$ Since $(2n-1)(n^2-1)>n^3+1$ for all $n\geq 4,$ we must have $m<n.$ Claim: We cannot have $m,n\geq 4.$ Proof: Put Symmetry and Bounding together. Note that Bounding implies $n<m,$ but Symmetry implies that $(n,m)$ being a valid pair implies $(m,n)$ is a valid pair, which requires $m<n,$ contradiction. Now we only need to manually check $n=1,n=2,n=3,$ and then we can permute.

Given $mn - 1 | n^3 + 1 \implies mn-1|n(n^2 + m) \implies mn-1|n^2 + m$ (since $mn-1, n$ are coprime). We have obtained $mn - 1 | n^2 + m \qquad (1)$. Continuing this, we get $mn-1|m^2(n^2 + m) - m^2n^2 + mn \implies mn - 1 | m^3 + mn \implies mn-1|m(m^2 + n) \implies mn-1|m^2 + n \qquad (2)$. From $(1)$ and $(2)$, we have both $mn - 1 \leq m^2 + n \implies n(m - 1) \leq m^2 + 1$ and $m(n-1) \leq n^2 + 1$. Suppose $m < n$. Then we have from $n(m-1) \leq m^2 + 1$ that $n=m+1$ or $n = m+2$. If $n = m+2$, then $m \leq 3$ gives the solutions $(1, 3), (3, 5), (1, 2), (2, 5)$. Suppose $n = m+1$. Then we require $m^2 + m - 1 | m^3 + 3m^2 + 3m + 2$ in the original question. This gives $$m^2 + m - 1 | m^3 + 3m^2 + 3m + 2 - m(m^2 + m - 1) \implies m^2 + m - 1 | 2m^2 + 4m + 2$$But $m^2 + m - 1$ is always odd, so $m^2 + m - 1 | m^2 + 2m + 1 \implies m^2 + m - 1 | m+2 \implies m = 1$ which gives the solutions $(1, 2), (1, 3)$ which we already had. Now suppose $m > n$. From $m(n-1) \leq n^2 + 1$, we get that $m = n+1$ or $m = n+2$. If $m = n+2$, we easily see $n \leq 3$ which gives the new solutions $(5, 3), (3, 1), (2, 1)$. Therefore, $m = n+1$. Plugging it in the original question gives us $n^2 + n - 1 | n^3 + 1 \implies n^2 + n - 1 | n^3 + 1 - n(n^2 + n - 1) \implies n^2 + n - 1 | n^2 - n -1$. This is only possible if $n = 1, m = 2$ which gives a solution that we already have. Finally, if $m = n$, it is trivial to see that $m = n = 2$. So, our obtained solutions are $\boxed{(m, n) = (1, 3), (3, 5), (1, 2), (2, 5), (5, 3), (3, 1), (2, 1), (2, 2)}$ and we are done.

We have: $$mn-1\mid n^3+1\Rightarrow mn-1\mid m(n^3+1)-n^2(mn-1)=n^2+m$$and $$mn-1\mid m(n^2+m)-n(mn-1)=m^2+n.$$We will find all solutions to these two divisibility conditions. WLOG $m\ge n$, and suppose $n\ge4$. Let $f(x)=\frac{x^2+1}{x-1}$. Note that $x+2-f(x)=\frac{x-3}{x-1}$, so $f(x)<x+2$ for $x\ge4$. Since $mn-1\le n^2+m$, we have $m\le f(n)<n+2$, so $m\le n+1$. Then either $m=n$ or $m=n+1$. If $m=n$, then $n^2-1\mid n^2+n$, so $n^2-1\mid n+1$. From $n^2-1\le n+1$, we derive $n\le2$ which is a contradiction. If $m=n+1$, then $n^2+n-1\mid n^2+3n+1$, so $n^2+n-1\mid 3n+2$. From $n^2+n-1\le3n+2$, we derive $n\le3$ which is a contradiction. Now, we check the cases $n\le3$. $n=1$: Then $m-1\mid m+1$, so $m-1\mid 2$. This gives $m=2,3$. $n=2$: Then $2m-1\mid m+4$, so $2m-1\le m+4$ or $m\le5$. Checking each of these gives that only $m=2,5$ work. $n=3$: Then $3m-1\mid m+9$, so $3m-1\le m+9$ or $m\le5$. Checking each of these gives that only $m=5$ works. This gives the only solutions: $$(m,n)=(1,2),(1,3),(2,2),(2,5),(3,5)$$along with their reverses: $$(m,n)=(2,1),(3,1),(5,2),(5,3).$$

If mn-1 is divisible by n^3+1 then mn-1 is divisible by m(n^3+1)-n^2(mn-1)=n^2+m as it is the same. simillarly m(n^2+m)-n(mn-1)=m^2+n. hence, $mn-1,$ is divisible by n^2+m, and m^2+n. then rest is just casework an modular arithmetic

$mn - 1 \mid n^3 + 1 \implies mn - 1 \mid n^3 + 1 + (mn - 1) = (n^2 + m)n \implies mn - 1 \mid n^2 + m$. Let $k = \frac{n^2 + m}{mn - 1} \implies$ $n^2 + m = mnk - k \implies n^2 - mnk + m + k = 0.$ Supose $m \neq n$, take $n_1$ the other root so $\begin{cases} n_1 + n = mk \\ n_1\cdot n = m + k \end{cases} \implies n_1 \in \mathbb{Z_+}$. If $(m, n)$ has the smallest sum then $n_1 \geq n \implies 2n_1 \geq mk \implies m + k \geq \frac{mk}{2}n \implies \frac{2}{k} + \frac{2}{m} \geq n \implies 4 \geq n$ $\bullet$ $n = 1$ $m - 1 \mid m + 1 \implies m \in \{2, 3\}$ $i)$ $m = 2$ $\implies k = 3 \implies (m, n) = (2, 1) \to (2, 5)$ $ii)$ $m =3$ $\implies k = 2 \implies (m, n) = (3, 1) \to (3, 5)$ $\bullet$ $n = 2$ $2m - 1 \mid m + 4 \implies m \in \{1, 2, 5\}$ $i)$ $m = 1$ $\implies k = 5 \implies (m, n) = (1, 2) \to (1, 3)$ $ii)$ $m = 2$ $(m, n) = (2, 2)$ $iii)$ $m = 5$ $\implies k = 1 \implies (m, n) = (5, 2) \to (5, 3)$ $\bullet$ $n = 3$ $3m - 1 \mid m + 9 \implies m \in \{1, 5\}$ $i)$ $m = 1$ we did this case before $ii)$ $m = 5$ we did this case before $\bullet$ $n = 4$ $4m - 1 \mid m + 16 \implies 4m - 1 \mid 65$ but all divisors of $65$ are $\equiv 1 \pmod{4}$, so this case doesn't have solution. $\boxed{(m, n) = (1, 2), (1, 3), (2, 1), (2, 2), (2, 5), (3, 1), (3, 5), (5, 2), (5, 3)}$