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Unchecked solution by nhat: find all a,b in positive integer such that: $\frac {a^{2}*b+a+b}{a*b^2+b+7}$ in positive integer I want the another solution which is different to the solution of IMO here my solution for this case a=b we easy to have the solution if a<b then $a*b*(a-b)+a-7$<0 (contradition) hence a>b let a in $(m*b,(m+1)*b)$ (m i positive integer) then we easy to prove that : m<$\frac {a^{2}*b+a+b}{a*b^2+b+7}$<(m+1) (contradition) so a divisible to b give $a=k*b$ (k in positive integer) then $\frac{k^2*b^3+k*b+b}{k*b^3+b+7}$=$\frac{b*{k^2*b^2+k+1}}{k*b^3+b+7}$ if $(a,7)=1$ then $(b,k*b^3+b+7)=1$ so that $\frac {k^2*b^2+k+1}{k*b^3+b+7}$ in positive integer and we esay to prove that k>b so easy to prove that \[k\equiv 0 (mod b)\] so $k=q*b$ (b >= 3and q i positive integer) then $\frac {k^2*b^2+k+1}{k*b^3+b+7}$=${\{q^2*b^4+q*b+1}{q*b^4+b+1}$=$q$+$\frac {1-q}{q*b^4+b+1}$ (contradition) if b divisible to 7 so that $b=7*u$ (u in positive integer) and easy to prove that u=k (it's similar for this case two) so we have the solution $a=7*k^2$,$b=7*k$ my solution hasor hasn't correct

The solution given isn't quite correct... there are at least two missing solutions: $(11,1)$ and $(49,1)$. $11^2+11+1 = 133$, which is divisible by $11+1+7 = 19$. $49^2+49+1 = 2451$, which is divisible by $49+1+7 = 57$.

My full solution... If $(xy^2+y+7)|(x^2y+x+y)$, then there exists a positive integer $k$ such that $k(xy^2+y+7) = x^2y+x+y$ , or $7k-y = (xy+1)(x-ky)$ . First we claim that all solutions have both sides nonnegative. Suppose $7k-y < 0$ . But then we have $|7k-y| < |y| < |xy+1| < |xy+1||x-ky|$ , which is a contradiction. So both sides are nonnegative. Now consider the two cases: (1) both sides are zero, and (2) both sides are positive. Case 1: Both sides are zero. Then we have $7k-y = 0 \Rightarrow y = 7k$ . And also $(xy+1)(x-ky) = 0$, but since $xy+1 > 0$ we know $x-ky = 0 \Rightarrow x = ky = 7k^2$ . So we have all solutions of the form $(x,y) = (7k^2, 7k)$ . Case 2: Both sides are positive. Then $x > ky$ so the RHS is positive. So we have $7k > 7k-y = (xy+1)(x-ky) > (xy+1) > y^2k+1$. Hence $y < 3$ . So we check $y = 1, 2$ . For $y = 1$ , we have $(x+8)|(x^2+x+1)$. Since $\frac{x^2+x+1}{x+8} = x-7+\frac{57}{x+8}$ , we have $(x+8)|57 \Rightarrow x = 11, 49$ , giving the solutions $(11,1)$ and $(49,1)$ . For $y = 2$ , we have $(4x+9)|(2x^2+x+2) \Rightarrow (4x+9)|(4x^2+2x+4)$ . Since $\frac{4x^2+2x+4}{4x+9} = x-\frac{7x-4}{4x+9}$ , so $(4x+9)|(7x-4)$ . But since $2(4x+9) > 7x-4$ , we must have $4x+9 = 7x-4$ , which does not have an integer solution. Hence our only solutions are $(x,y) = (11,1); (49,1); (7k^2, 7k)$ for positive integers $k$ .

Here my solution I hope there are no mistakes Let $ab^2+b+7=n$ then $ab^2\cong -b-7 (mod n) \rightarrow a^2b^2\cong -ab-7a (mod n)$ and $a^2b\cong -a-b (mod n) \rightarrow a^2b^2\cong -ab-b^2(mod n)$ from here $\rightarrow 7a \cong b^2(mod n)$ so $n$ divides both $(ab^2+b+7)$ and $a(b^2-7a)$ so $(ab^2+b+7)-a(b^2-7a)=b+7+7a^2$ then n divides $b+7+7a^2$ , too. so $b+7+7a^2\geq ab^2+b+7 \rightarrow 7a \geq b^2$. Lets share it in two parts 1) $7a-b^2=0 \rightarrow (a,b)=(7k^2,7k)$ 2) $7a-b^2>0$ Since we have found that $n|(7a-b^2)$ then $7a-b^2\geq ab^2+b+7$ from here we will have $(7-b^2)(a+1)\geq b+14$. $b+14>0$ then $(7-b^2)(a+1)>0$ also $(a+1)>0$ then we have to have $(7-b^2)>0$ here $b$ is positive integer then $b= 1$ or $2$ otherwise inequality will lead to contradiction If $b=1$ then the problem will become as $\frac{a^2+a+1}{a+8}$ Let $k=a+8$ then $a^2\cong -a-1 (mod k) (i)$ $a\cong -8(mod k)\rightarrow a^2\cong 64 (mod k) (ii)$ from $(i)$ and $(ii)$ we will have $-a-1\cong 64 (mod k)$ and $a+65$ is divisible by $k$ then $\frac{a+65}{a+8}=\frac{a+8}{a+8} + \frac{57}{a+8}$ so $\frac{57}{a+8} (iii)$ is an integer, too. so there are just two solutions for $(iii)$ those are $a=11$ or $49$ ( think didisors of $57$) If $b=2$ by simliar of previous we will show that there is no solution for $b=2$ So the solutions are $(a,b) = (11,1), (49,1), (7k^2,7k)$ We are done

If $ x < y$ then it is easy to see there are no solutions. By the division algorithm, we can let $ x = ay + b$, where $ 0 \le b < y$. Then we have $ a^2y^3 + 2aby^2 + (b^2 + a + 1)y + b$ is divisible by $ ay^3 + by^2 + y + 7$ Let the ratio between these two numbers be $ a + k$, then $ a^2y^3 + 2aby^2 + (b^2 + a + 1)y + b = (a + k)(ay^3 + by^2 + y + 7)$ $ aby^2 + b^2y + y + b = aky^3 + kby^2 + ky + 7a + 7k$ If $ k = 0$, then $ aby^2 + b^2y + y + b = 7a$ Case 1:$ b = 0$ Then $ y = 7a$ and we have the solution $ (7a^2,7a)$ Case 2: $ b \ge 0$ Then either $ y = 1$ or $ 2$. If $ y = 1$, then $ b$ can't be more than $ 7$, we try out all values from $ 1$ to $ 7$ for $ b$. $ b = 4$ and $ b = 6$ works, this gives the solutions $ (11,1)$ and $ (49,1)$. If $ y = 2$ then $ b = 1$ and we have no solution. If $ k > 0$, then since $ b < y$, we must have $ k = 1$ since if $ k > 1$ then the left side is less than the right side (we would have $ aky^3 > aby^2 + b$, $ kby^2 > b^2y$, $ ky > y$). But if $ k = 1$, then $ b = 7a + 7 + (y - b)(ay^2 + by)$. But $ y - b \ge 1$ since $ b < y$, a contradiction. So $ k < 0$. Then we must have $ y = 1$ or the left side would be greater than the right. Since $ 0 \le b < y$, we must have $ b = 0$. Thus, $ 1 = ak + 7a + 8k$, or $ (a + 8)(k + 7) = 57$. We find $ (a,k) = (11, - 3)$ and $ (49, - 6)$ work, giving the solutions $ (11,1)$ and $ (49,1)$. So the three solutions are $ (7a^2,7a)$, $ (11,1)$, and $ (49,1)$

I thought to myself this morning - "I haven't really tried any IMO problems". I decided to remedy that by solving this one. First, suppose that $ gcd(y,7)=1$. Then the problem can be restated as follows. $ xy^2+y+7 \, | \, x^2y+x+y$ $ \Leftrightarrow xy^2+y+7 \, | \, y(x^2y+x+y)-x(xy^2+y+7)$ $ \Leftrightarrow xy^2+y+7 \, | \, y^2-7x$ It is easy to see that since $ x,y$ are positive, then $ xy^2+y+7 > y^2-7x$. Then we must have either $ y^2=7x$ or $ y^2-7x < 0$. The first case cannot arise since $ gcd(7,y)=1$ so we must have $ y^2-7x < 0$. Then our condition is equivalent to: $ xy^2+y+7 | 7x - y^2$ and we must have $ 7x - y^2 \ge xy^2 + y + 7$ The second condition implies that $ y < 3$ since otherwise we have: $ (7x - y^2) - (xy^2+x+y) \le 7x - 9 -9x -x-3 = -3x - 12 < 0$. Hence $ y=1$ or $ y=2$. Case 1: $ y=1$. Our problem restates as $ x+8 | 7x-1$ and so we write: $ \dfrac{7x-1}{x+8} = k$. It is clear that $ k$ must be positive. We rewrite this as: $ x=\dfrac{8k+1}{7-k}$ and since $ x$ is positive, the denominator must be positive and so $ k < 7$. Substituting the values $ k=1,2,3,4,5,6$ into this expression in turn shows that only $ k=4$ and $ k=6$ give integer values for $ x$. These are $ x=11$ and $ x=49$. So our two solutions are $ (x,y)=(49,1), (11,1)$. Checking in the original problem reveals that they work. Case 2: $ y=2$. Starting in the same way as above we get $ x= \dfrac{9k+4}{7-4k}$. In a similar way to above we must have $ k < 2$, but substituting $ k=1$ does not give integer $ x$. Hence, no solutions in this case. Now we must consider when $ gcd(7,y)=7$. This case is easy. Let $ y=7t$. Any pair $ (x,y)$ which satisfies the problem must also satisfy: $ xy^2+y+7 | y(x^2y+x+y)-x(xy^2+y+7)$ $ xy^2+y+7 | y^2-7x$ We cannot have $ y=1$ or $ y=2$ as above since $ gcd(y,7)=7$. Thus we must have the other possibility, $ y^2=7x$. Then we have $ 7x=y^2=49t^2$ So $ x=7t^2$ Then our solution is $ (x,y)=(7t^2,7t)$. Checking this in the original problem: $ \dfrac{x^2y+x+y}{xy^2+y+7}$ $ =\dfrac{343t^5+7t^2+7t}{343t^4+7t+7}$ $ =t$ which is an integer. Hence all solutions are given by: $ (x,y) = (11,1), (49,1), (7t^2,7t)$

KrazyFK wrote: First, suppose that $ gcd(y,7)=1$. Then the problem can be restated as follows. $ xy^2+y+7 \, | \, x^2y+x+y$ $ \Leftrightarrow xy^2+y+7 \, | \, y(x^2y+x+y)-x(xy^2+y+7)$ $ \Leftrightarrow xy^2+y+7 \, | \, y^2-7x$ What is the importance of the statement $\gcd(y,7)=1$? Can you not perform the same manipulations even if $\gcd(y,7)=7$? Also, my solution involved using the division algorithm to get $\frac{a^2b+a+b}{ab^2+b+7}=\frac{a}{b}+\frac{b^2-7a}{ab^3+b^2+7b}$, can this only be done if $a\geq b$? From there I get $b^2-7a=0\implies b^2=7a\implies a=7k,b=7k^2$. Why can't I use the division algorithm to get the same result if $b<a$?

The statement of $\gcd(7, y)=1$ is not actually used for the algebraic manipulations, but to set aside a case that forces $y^2-7x$ to be positive (in other words, it's used right after your quote).

Oh, so from $ \frac{a^{2}b+a+b}{ab^{2}+b+7}=\frac{a}{b}+\frac{b^{2}-7a}{ab^{3}+b^{2}+7b} $, we must split it into cases. So I forgot the case where $b^2-7a<0$, so $7a-b^2>\frac{ab^{3}+b^{2}+7b}{b}$.

Yep. The $b^2-7a<0$ case leads to the two extra solutions, $(49, 1); (11, 1)$.

Thanks, but just from the division algorithm, what is the concrete rational that we must have $ 7a-b^{2}>\frac{ab^{3}+b^{2}+7b}{b} $ (division by $b$ on the RHS)? Is it just that if we factor out the $\frac1b$ on the RHS of $ \frac{a^{2}b+a+b}{ab^{2}+b+7}=\frac{a}{b}+\frac{b^{2}-7a}{ab^{3}+b^{2}+7b} $, we just want $a+\frac{b^2-7a}{ab^2+b+7}$ to be an integer?

That's right. From your division algorithm, if we want the LHS to be an integer, then $b$ has to divide $a+\frac{b^2-7a}{ab^2+b+7}$, so it must be an integer, and so $ab^2+b+7$ must divide $7a-b^2$.

Define $S=\left \{(a,b)\in \mathbb{Z}^{+}|\displaystyle \frac{a^2b+a+b}{ab^2+b+7}\in \mathbb{Z}^{+}\right \}$ We note that (i) $ab^2+b+7|a^2b+a+b\Rightarrow ab^2+b+7|a(b(a^2b+a+b)-a(ab^2+b+7))$ then $ab^2+b+7|ab^2-7a^2\Rightarrow ab^2+b+7|ab^2-7a^2-(ab^2+b+7)$ now we have that $ab^2+b+7|-7a^2-b-7\Rightarrow ab^2+b+7|7a^2+b+7$ note that $ab^2+b+7>0$ and $7a^2+b+7>0$ therefore $ab^2+b+7\leq 7a^2+b+7\Rightarrow b^2\leq 7a$ if $b^2=7a$ then $a=7t^2$ and $b=7t$ for all $t\in \mathbb{Z}^{+}$ therefore $(7t^2,7t)\in S$ now $b^2<7a$ By (i) we have that $ab^2+b+7|b^2-7a\Rightarrow ab^2+b+7|7a-b^2$ because $b^2<7a$ now because $b^2<7a$ we have $7a-b^2>0$ and note that $ab^2+b+7>0$ therefore $ab^2+b+7\leq 7a-b^2$ that is impossible for $b\ge 3$ therefore $b<3$ We have two cases: Case 1: $b=1$ we have $a+8|a^2+a+1\Rightarrow a+8|a^2+a+1-a(a+8)+7(a+8)$ therefore $a+8|57=1\cdot 3\cdot 19$ if we prove all the divisors we have that $(49,1)$ and $(11,1)$ are solutions. therefore $(11,1),(49,1)\in S$ Case 2: $b=2$ we have $4a+9|2a^2+a+2\Rightarrow 4a+9|4(2(2a^2+a+2)-a(4a+9))+7(4a+9)=79$ therefore $4a+9|79$ no solutions here Finally we have that $S=\{(11,1),(49,1),(7t^2,7t)\}$ Thanks

Nice and straightforward solution $xy^2+y+7 \mid x^2y+x+y$ $\Rightarrow xy^2+y+7 \mid y(x^2y+x+y)-x(xy^2+y+7)$ $\Rightarrow xy^2+y+7 \mid y^2-7x$ Clearly $y^2-7x <xy^2+y+7$ so $y^2-7x=0$ which gives $(x,y)=(7t^2,7t)$ ot $y^2-7x <0$.This gives $7x-y^2 \ge xy^2+y+7 \Rightarrow (x+1)y^2+y+7-7x \le 0 \Rightarrow y \le 2$.$y=2 \Rightarrow 4x+9 \mid 2x^2+x+2 \Rightarrow 4x+9 \mid 4x^2+2x+4-4x^2-9x \Rightarrow 4x+9 \mid 7x-4 \Rightarrow 4x+9 \mid 7(4x+9)-4(7x-4)=79 \Rightarrow$ no solution. For $y=1$ we have $x+8 \mid x^2+x+1 \Rightarrow x+8 \mid x^2+x+1 -x^2-8x \Rightarrow x+8 \mid 7x-1 \Rightarrow x+8 \mid 7(x+1)-(7x-1) \Rightarrow x+8 \mid 57 \Rightarrow \boxed{x=11,49}$. Solutions $(x,y)=(7t^2,7t),(11,1),(49,1)$.

Nasty problem. The desired ordered pairs are $(x,y) = (11,1)$, $(49,1)$, and $(7k^2,7k)$ for all positive integers $k$. These pairs can be easily checked to work, and so it remains to prove that they are the only ones. First suppose $y\geq 3$. The condition that $xy^{2}+y+7$ divides $x^2y+x+y$ implies \[ \mathbb Z\ni \frac{x^2y^2 + xy + 7x}{x^2y+x+y} = x + \frac{y^2 - 7x}{x^2y+x+y}.\qquad(\dagger) \]We now claim the bounds \[ -1 < \frac{y^2-7x}{x^2y+x+y} < \frac 1x. \]First, since $x$ and $x^2y+x+y$ are both positive, the upper bound is equivalent to $xy^2 - 7x^2 < xy^2 + y + 7$, which is trivially true since $y$ is positive and $-7x^2$ is negative. In a similar fashion, the lower bound is equivalent to \[ -xy^2 - y - 7 < y^2 - 7x, \]which in turn is equivalent to \[ x(7-y^2) < y^2 + y + 7. \]Since $y\geq 3$, the left hand side is negative while the right hand side is positive, so this inequality is true as well. It follows that since $\tfrac{y^2-7x}{x^2y+x+y}$ is an integer, it must be equal to $0$; hence $y^2 = 7x$, and the pairs $(x,y)$ which satisfy this are precisely those of the form $(7k^2,7k)$. (Indeed, the condition implies $7\mid y$, so $y = 7k$ for some positive integer $k$; simplification thus yields $x = 7k^2$.) It remains to handle the $y=1$ and $y=2$ cases. In the former case, we may write \[ \frac{x^2y+x+y}{xy^2+y+7} = \frac{x^2+x+1}{x+8} = x-7 + \frac{57}{x+8}; \]this is an integer precisely when $x+8$ is a divisor of $57 = 3\cdot 19$, and from this we get the two solutions $x=11$ and $x=49$. In the latter case, we recall by $(\dagger)$ that \[ \frac{y^2-7x}{xy^2+y+7} = \frac{4-7x}{4x+9} \]must be an integer, say $A$. But then \[ 4A + 7 = \frac{16-28x}{4x+9} + 7 = \frac{79}{4x+9} \]must be an integer as well, which is not possible since $79$ is a prime congruent to $3$ modulo $4$.

ISL 1998 N1 wrote: Determine all pairs $(x,y)$ of positive integers such that $x^{2}y+x+y$ is divisible by $xy^{2}+y+7$. $x^2y+x+y \equiv x^2y^2+xy+y^2 \equiv (-y-7)x+xy+y^2 \equiv 7x-y^2 \equiv 0 \pmod{xy^2+y+7}$ For $7x=y^2$ we have $(x,y)=(7k^2,7k)$ for a positive integer $k$. $7x-y^2 \equiv 7x^2-xy^2 \equiv 7x^2+y+7 \equiv 0 \pmod{xy^2+y+7} \implies 7x^2+y+7 \ge xy^2+y+7 \implies 7x-y^2 \ge 0$ $\mathbb{N} \ni \frac{7x-y^2}{xy^2+y+7} \leq \frac{7x}{xy^2} = \frac{7}{y^2}$ This gives $y=1,2$. Bashing out gives that $(x,y)=(11,1),(16,1)$

Storage Also, unexpectedly clean 1998 N1 wrote: Determine all pairs $(x,y)$ of positive integers such that $x^{2}y+x+y$ is divisible by $xy^{2}+y+7$. Note that $xy^2+y+7 \mid x(xy^2+y+7)-y(x^2y+x+y)=7x-y^2$. Now if the latter is zero, end of discussion: we can check $(7t^2, 7t)$ is a solution for all $t \ge 1$. Else we get that $xy^2+y+7 \le |7x-y^2| \le 7x+y^2$ so $xy^2-y^2 \le 7x$ (yes, bounds this silly work!) so $y^2(x-1) \le 7x$ hence $y \le 3$ or $x=1$ but we can check that that gives no solutions. Now for $y=1$ we check that $x+8 \mid 7x-1 \implies x+8 \mid 57$ which sadly isn't prime (hi Grothendieck!) but factors as $3 \cdot 19$ and so $x=11, x=49$ are the only possible choices here. We can quickly verify that $y=2, 3$ give nothing, so these are the only solutions. $\blacksquare$

We get that $\frac{x^2y+x+y}{xy^2+y+7} = k.$ Clearing the denominator and rearranging, we get $y=xy(ky-x) + (ky-x)+7k$ and factoring we get $y=(xy+1)(ky-x)+7k$. Clearly, if $ky-x>0$, then the right hand side is larger than the left hand side. If $ky-x=0$, then $y=7k$ meaning $x=7k^2$. So all pairs of the form $(7k^2, 7k)$ work for positive integers $k$. Now if $ky-x<0$, then $x>ky$. Let $x=ky+m$. Then $7k-mky^2-m^2y-m=y$. Since $y$ is positive, we get $7k>m(ky^2+my+1).$ If $y\geq 3$, then we have $9k$ added up which won't work. So $y=1,2$. If $y=1$, $7k-mk-m^2-m=1$. This rearranges to $k = \frac{m^2+m+1}{7-m}$. This means $1\leq m \leq 6$. Testing each one, the ordered pairs we get are $(x,y)=(49, 1), (11,1).$ Now if $y=2$, we get $7k-4mk-4m^2-m=2$. The only possible value is $m=1$ which gives $3k=7$ which is not an integer. So our final answer is the ordered pairs $\boxed{(7k^2, 7k), (49,1), (11,1)}.$

I claim the solutions are $(49,1)$, $(11,1)$ and $(7n^2,7n)$ for any positive integer $n$. First rewrite the divisibility as $$xy^2+y+7 \mid y(x^2y+x+y)-x(xy^2+y+7) = y^2-7x$$ Now we have two cases. Case 1: $y^2-7x \geq 0$. Note that since $x,y \geq 1$ we clearly have $xy^2+y+7+7x> y^2$ so we must have $y^2- 7x = 0$ or $y^2 = 7x$. If we let $x = 7n^2$ where $n$ is any positive integer then $y = 7n$ and we have our first pair $(7n^2,7n)$. Case 2: $y^2 - 7x < 0$. In this case, we are essentially looking at $$xy^2+y+7 \mid 7x-y^2$$Thus we must have $xy^2+y+7 \leq 7x-y^2$ which rearranges to $xy^2+y^2+y+7 \leq 7x$. Note that in order for this to be true, we must have $y \leq 2$ or else $xy^2 > 7x$. Now we simply just check the cases where $y=1$ and $y=2$. If $y = 1$, then $x+8 \mid 7x-1$ which is equivalent to $x+8 \mid 7(x+8)-(7x-1) = 57$ from which we extract $x = 11$ or $x = 49$. This gives us the pairs $(49,1)$ and $(11,1)$. If $y = 2$, then $4x+9 \mid 7x-4$ which is equivalent to $4x+9 \mid 7(4x+9)-4(7x-4) = 79$ from which we see there are no positive integer solutions for $x$. We've exhausted all cases and we're done.

We claim the answer is $(7n^2,7n),(49,1),(11,1).$ These clearly all work. Note that $$x^2y+x+y\equiv 0\pmod{xy^2+y+7}$$$$x^2y^2+xy+y^2\equiv 0\pmod{xy^2+y+7}$$$$y^2-7x\equiv0\pmod{xy^2+y+7}.$$(at the second step we subtracted $x$ times the modulus.) The point of this is that now everything is linear in $x$. Case 1: $y^2-7x>0$. Then, we have $$y^2-7x<y^2\leq xy^2<xy^2+y+7,$$hence there are no solutions as the expression is positive and smaller than the modulus. f Case 2: $y^2-7x=0$. Then, clearly $(x,y)=(7n^2,7n)$, which clearly works. Case 3: $y^2-7x<0$. Then, we have $$7x-y^2=c(xy^2+y+7).$$We can rearrange this to $$x=\frac{y^2+cy+7c}{7-cy^2}.$$ Since $cy^2\leq 6$, there are not that many cases. If $y=1$, we have $$x=\frac{8c+1}{7-c}.$$For $c=4$ and $c=6$ this gives $x=11$ and $x=49$, and other values of $c$ up to 6 do not give integer $x$. If $y=2$, we have $$x=\frac{9c+4}{7-4c},$$here only $c=1$ is allowed which does not give an integer anyway. Finally, $y\geq3$ is not possible, hence done.

We have $$0\equiv x^2y+x+y\equiv x^2y^2+xy+y^2-(x(xy^2+y+7))\equiv y^2-7x\pmod{xy^2+y+7}.$$Due to size reasons $y^2-7x\le0$, if it equals 0 then the solution set is $(7n^2,7n)$. If it's <0 we must have $xy^2+y+7\le7x-y^2<7x\implies y\in\{1,2\}$; easy calculation reduces it into $\boxed{(x,y)\in\{(7n^2,7n),(11,1),(49,1)\}.}\blacksquare$ wait im not sure why but why are my sols so short lol on overleaf typing they take up like 10 lines

Note that this implies that $xy^2 + y + 7$ divides $y(x^2y + x + y) - x(xy^2 + y + 7) = y^2 - 7x$. This implies that either $y^2 = 7x$, or that $x \ge 2, y^2 \ge 16$ can not hold. In the first case, we get the solution set $(x, y) = \left(7k^2, 7k\right)$ which can be seen to hold. We now bash out the remaining cases. It can be seen that if $x = 1$, then there are no solutions. Then, if $y = 1$, it follows that $x + 8 \mid 1 - 7x$ so $x \in \{11, 49\}$. We can check that $(x, y) = (49, 1)$ and $(x, y) = (11, 1)$ works. If $y = 2$, then $4x + 9 \mid 4 - 7x$ which implies $4x + 9 \mid 3x - 13$ which has no solutions. If $y = 3$, then $9x + 10 \mid 9 - 7x$ which has no solutions.

Rewrite the condition as $$xy^2+y+7 \mid (x^2y^2+xy+7x) - (x^2y^2+xy+y^2) = 7x-y^2.$$Now we have a few cases: If $y^2 > 7x$, then $y^2 - 7x < y^2 < xy^2+y+7$, which is obviously impossible. If $y^2 < 7x$, then $$xy^2+(y+y^2) + 7 \leq 7x,$$which implies $y^2 \leq 7$ and $y \in \{1, 2\}$. This yields the solutions $(49, 1)$ and $(11, 1)$. If $y^2 = 7x$, then the entire curve of solutions $(7n^2, 7n)$ can be checked to work.

Bruh, the first time I solved the problem, I solved it for $x^y + x + y \mid xy^2 + y + 7$ ;-; . Also, these edge cases are so hard for me to find without making sillies. I claim that the answers are $(x,y)=(11,1)$, $(49,1)$ and $(7n^2,7n)$ for any $n\in\mathbb N$. Firstly, we have $xy^2 + y + 7 \mid x^2y + x + y \mid x^2y^2+xy+y^2\equiv (x^2y^2+xy+y^2)-x(xy^2 + y + 7) = y^2 - 7x$. Now if $y^2 > 7x$, then we get that $xy^2 + y + 7 \le y^2 - 7x$ which gives us $y^2 -y(1+x^2) -(7+7x) \ge 0$. But then the discriminant must be $\le 0$, that is $(1+x^2)^2 + 4(7+7x) \le 0$ which clearly has no solution. Now if $y^2 < 7x$, then we get that $xy^2 + y + 7 \ge y^2 - 7x$. Then we get that $x^2y -7x +(y^2+y+7) \le 0$. This means that the discriminant must be $\ge 0$, that is $7^2 -4y(y^2+y+7)\ge 0 \implies 4y(y^2+y+7)\le 49$. This gives only solution $y=1$ since the left side is strictly increasing and it exceeds $49$ for $y=2$. For $y=1$, from the problem statement we get that $x+1+7 \mid x^2 + x + 1 \equiv (x^2+x+1)-x(x+8) = -7x+1 \equiv (-7x+1)+7(x+8) = 57$. This gives us that $x\in \left\{11,49\right\}$ each of which are solutions. Now for the other case when $y^2 = 7x$, then clearly $7\mid y$ and so $49n^2 = 7x \implies x = 7n^2$ which is another solution.

Notice the LHs also divides \[y(x^2y+x+y)-x(xy^2+y+7) = y^2-7x.\] If $y^2-7x=0$, we have the solutions $\boxed{(7k^2,7k)}$. Otherwise, we notice \[|xy^2+y+7| > |y^2-7x|,\]implying there are no solutions, unless $y=1,2$ where we get the pairs $\boxed{(11,1),(49,1)}$. $\blacksquare$

I claim that all solutions are $\boxed{(a, b) \in \{(11, 1), (49, 1), (7t^2, 7t), \}}$, for $t \in \mathbb{Z}^+$. Observe that \[ab^2 + b + 7 \mid a^2b + a + b \implies ab^2 + b + 7 \mid b(a^2b + a + b) - a(ab^2 + b + 7) = b^2 - 7a.\]We now split into two separate cases: Case 1:$(b^2 - 7a = 0).$ Then $b^2 = 7a$, so that $b \equiv 0 \pmod 7$, which in turn gives us $b = 7t$, for some $t \in \mathbb{Z}^+$. Then we find the corresponding $a = 7t^2$. $\square$ Case 2:$(b^2 - 7a < 0).$ First observe that $b^2 - 7a \ngeq 0$, else $b^2 - 7a \geq ab^2 + b + 7 \implies b^2 \geq a(b^2 + 7) + b + 7$, which is a clear contradiction for $a, b > 0$. Hence $b^2 - 7a < 0$, so that we obtain \[ab^2 + b + 7 \leq 7a - b^2 \implies b^2(a + 1) + b + 7 \leq 7a \implies b^2 < 7.\]Hence we split into cases for $b = 1, 2$: Subcase 2.1: $(b = 1).$ Then \[\frac{7a - 1}{a + 8} \in \mathbb{Z} \implies 7 - \frac{57}{a + 8} \implies a + 8 = 1, 3, 19, 57 \implies \boxed{a = 11, 49}. \bigstar\] Subcase 2.2: $(b = 2).$ Then \[\frac{7a - 4}{4a + 9} \in \mathbb{Z} \implies 1 + \frac{3a - 13}{4a + 9} \implies a \leq -22, \]which is impossible. Therefore no solution exists in this subcase. $\bigstar$ The only solutions, therefore, are $\boxed{(a, b) \in \{(11, 1), (49, 1), (7t^2, 7t))\}}$, for arbitrary $t \in \mathbb{Z}^+$, as claimed. $\blacksquare$

The solutions are $(x,y) = (11,1)$, $(49,1)$, $(7k^2,7k)$ for all $k\in\mathbb N$. Note that, we are given that: \begin{align*} xy^2+ y+ 7\mid x^2y + x + y \implies xy^2+ y+7\mid y(x^2y + x + y) - x(xy^2+y+7)= y^2-7x \end{align*} Now we divide into cases based on the sign of $y^2-7x$ When $y^2-7x> 0$. The divisibility condition implies that $y^2-7x\geq xy^2+ y+ 7$ Clearly, $0<y^2-7x<xy^2+ y+ 7$, contradicting the divisibilty condition. When $y^2-7x=0$. in this case we get , $y^2=7x$ , let $y = 7k$ , so$ x = 7k^2$. Plugging this back in to the original equation reads: \begin{align*} 343k^4 + 7k + 7 \mid 343k^5 + 7k^2 + 7k \end{align*}which is always valid, hence these are always solutions. When $y^2-7x<0$. We get: \begin{align*} |y^2-7x|\geq xy^2+ y+ 7 \implies 7x-y^2\geq xy^2+y+7 \iff x(y^2-7)+y^2+y+7\le 0 \iff y \in \{1,2\}. \end{align*} When $y=1$ we get: \begin{align*} x+8 \mid 7x-1 \iff x+8 \mid 57 \end{align*}This gives $x=11$ and $x=49$. When $y=2$ \begin{align*} 4x+9 \mid 7x-4\iff 4x+9 \mid 79 \end{align*}which gives no solutions.

Assume that $\gcd(y, 7)=1$. Then, $$\gcd(xy^2+y+7,x^2y+x+y)=\gcd(xy^2+y+7, y^2-7x)=\gcd(7x^2+y+7, 7x-y^2).$$ We want this to equal $xy^{2}+y+7$, so $7x^2+y+7\ge xy^2+y+7$ so $7x\ge y^2$. We also want $7x-y^2\ge xy^2+y+7$ or $7(x-1)\ge (x+1)y(y+1)$. This means that $y=1$ or $y=2$. When $y=1$, we get $(x+8)|(x^2+x+1)$ so $(x+8)|57$ so $x=11$ or $x=49$. When $y=2$, we get $(4x+9)|(2x^2+x+2)$ so $(4x+9)|(4x^2+2x+4)$ so $(4x+9)|(x+22)$ so there are no solutions to $x$. Now assume that $y=7b$. Then, $7|(xy^{2}+y+7)|(x^{2}y+x+y)$ so $x=7a$. Plugging this in means $(49ab^2+b+1)|(49a^2b+a+b)$. Note that $$\gcd(49ab^2+b+1, 49a^2b+a+b)=\gcd(49ab^2+b+1, b^2-a).$$ Note that $|b^2-a|< 49ab^2+b+1$ so $a=b^2$. Plugging $a=b^2$ back in, we get $(49b^4+b+1)|(49b^5+b^2+b)$, which is always true. Therefore, the solutions are $(11,1)$, $(49,1)$, and $(7k^2,7k)$ for all positive integer $k$.

Observe that we then have $xy^2 + y + 7 \mid x^2y^2 + xy + y^2$, so $xy^2 + y + 7 \mid y^2 - 7x$. We can then divide into cases, if $y^2 > 7x$, then clearly we have no solutions by size. If $y^2 = 7x$, then write $y = 7k, x = 7k^2$, we write $x^2y + x + y = 343k^5 + 7k^2 + 7k, xy^2 + y + 7 = 343k^4 + 7k + 7$, we can see all solutions of this form work. If $y^2 < 7x$, by size we still require $7x - y^2 \ge xy^2 + y + 7$, or equivalently $y^2 < 7$. We then check $y = 2$, we require $4x + 9 \mid 7x - 4$, equivalently $4x + 9 \mid 28x - 16$, equivalently $4x +9 \mid 79$, so there are no solutions for $x$ by divisor analysis. We now check $y = 1$, we require $x + 8\mid 7x - 1$, so we have $x + 8 \mid 57$, so we have $x = 11, 49$. We check $x = 11$ gives $x^2y + x + y = 133$ , $xy^2 + y + 7 = 19$, so this pair works. We then check $x = 49$, we get $x^2y + x + y = 2451, xy^2 + y + 7 = 57$, so this pair works as well. The answers are then $(7k^2, 7k), (49,1), (11,1)$.

\Rightarrow supremacy $ab^2+b+7|a^{2}b+a+b \Rightarrow b(ab+1)+7|ab(ab+1)+b^2 \Rightarrow ab^2+b+7|b^2-7a $ $ ab^2+b+7|ab^2-7a^2 \Rightarrow ab^2+b+7|7a^2+b+7$. Since both sides are positive: $ab^2 \leq 7a^2 \Rightarrow b^2\leq 7a$. So there is two cases to consider. First case $b^2=7a$: if we set $b=7k$, then $a=7k^2$, We can easily check that it is a solution with $k$ being a positive integer. Second case $b^2<7a$: Then from earlier, $ab^2+b+7<7a-b^2 \Rightarrow b^2+b+7<a(7-b^2)$ LHS is positive so RHS should follow. Meaning $7>b^2$. Then we can manually check and see that $(a,b)=(49,1); (11,1)$ is a solution.

\[xy^2+y+7 \mid y(x^2y+x+y)-x(xy^2+y+7)=y^2-7x \]which implies that $|y^2-7x| \geq xy^2+y+7$ which gives $x=1$ or $y^2\leq 7.$ or $y^2=7x\implies (x,y)=(7t^2,7t)$ which Convsersely always works. The edge cases give $(1,49)$ and $(1,11)$ as solutions.

A bit of long division leads us to $xy^2+y+7 \mid y^2-7x$. Clearly $y^2-7x>0$ isn't possible due to size. If $y^2-7x<0$ then $$7x>7x-y^2>xy^2+y+7>xy^2 \implies y=1,2.$$After checking we find the solution $(11,1),(49,1)$. If $y^2=7x$ then $(x,y)=(7k^2,7k)$ for some $k$, which works.