zhoujef000 wrote:

replace $x$ and $y$ with $a$ and $b$ since i was given this with $a$ and $b$ and am too lazy to change everything

We claim the answer is $\boxed{(11,1)},$ $\boxed{(49,1)},$ and $\boxed{(7c^2, 7c)}$ for all positive integers $c.$ We proceed as follows.

Firstly, we show these work.

Observe that $11\cdot1^2+1+7=19\mid 19\cdot 7=133=11^2\cdot 1+11+1,$ and $49\cdot 1^2+1+7=57\mid 57\cdot 43=2451=49^2\cdot1+49+1,$ so $(11,1)$ and $(49,1)$ work. Also, for all positive integers $c,$ $7c^2\cdot(7c)^2+7c+7=343c^4+7c+7\mid (343c^2+7c+7)c=343c^5+7c^2+7c=(7c^2)^2(7c)+7c^2+7c,$ so $(7c^2,7c)$ works for all positive integers $c.$ We now show there are no more solutions.

Since $ab^2+b+7\mid a^2b+a+b,$ we have that $b(a^2b+a+b)-a(ab^2+b+7)=a^2b^2+ab+b^2-(a^2b^2+ab+7a)=b^2-7a\equiv 0\pmod{ab^2+b+7}.$ If $b^2-7a>0,$ then since $a\geq 1,$ $ab^2+b+7>b^2>b^2-7a>0,$ so $b^2-7a\not\equiv 0\pmod{ab^2+b+7},$ a contradiction. Therefore, $b^2-7a\leq 0$ and $7a-b^2\geq 0.$

If $7a-b^2=0,$ then $a=\dfrac{b^2}{7},$ so $7\mid b$ and $b=7c$ for some integers $c.$ Then, $a=\dfrac{b^2}{7}=\dfrac{49c^2}{7}=7c^2,$ so $(a,b)=(7c^2,7c).$ Thus, all of the solutions in this case are of this form.

If $7a-b^2>0,$ then since $ab^2+b+7\mid 7a-b^2,$ we must have $ab^2+b+7\leq 7a-b^2.$ However, if $b\geq 3,$ then $ab^2+b+7\geq 9a+10>7a-b^2,$ so $b=1$ or $b=2.$ If $b=1,$ then $a+8=ab^2+b+7\mid 7a-b^2=7a-1,$ so $a+8\mid 7(a+8)-(7a-1)=57.$ Since $a$ is a positive integer, $a+8=19$ or $a+8=57,$ so $a=11$ or $a=49,$ yielding $(a,b)=(11,1)$ or $(a,b)=(49,1).$

If $b=2,$ then $4a+9=ab^2+b+7\mid 7a-b^2=7a-4.$ However, since $2(4a+9)=8a+18>7a-4>0,$ we must have $4a+9=7a-4,$ so $a=\dfrac{13}{3},$ which is not an integer. Thus, there are no solutions in this case.

As such, the only solutions are $(11,1),$ $(49,1),$ and $(7c^2, 7c)$ for all positive integers $c,$ as desired. $\Box$

beautiful! :first: