does this proof for "question 6" have a flaw? 
Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^2 + b^2$. Show that $\frac{a^2 + b^2}{ ab + 1}$ is the square of an integer.

If we just divide $a^2+b^2$ with $ab+1$, we get a remainder in terms of $a$ and $b$. Since it is given that $ab+1$ divides $a^2+b^2$, 
we can equate the remainder to zero and get a relation between $a$ and $b$, which is: $a=b^3$ or $b=a^3$ , which on substitution gives us $b^2$ or $a^2$ hence proving what was to be proved.
 A: Just because $a$ and $b$ are integers so that $ab+1$ divides $a^2+b^2$, that doesn't mean that the expression you get when you carry out the polynomial division $(a^2+b^2)\div(ab + 1)$ is nice in any way. After all, if it were a nice expression in $a$ and $b$ for some $a$ and $b$, then necessarily it would be a nice expression in $a$ and $b$ for all $a$ and $b$.
Also, you can't just equate the remainder with $0$. Take $\frac{a + 2}{a}$, for instance. Polynomial division gives a remainder of $2$, or possibly $\frac2a$, depending on what you mean by "remainder". Am I supposed to believe that $\frac{a+2}{a}$ is an integer as long as $2 = 0$, or $\frac2a = 0$? That doesn't bring me any closer to a solution.
Finally, I'm not sure that I know how to carry out the polynomial division $(a^2+b^2)\div(ab + 1)$. Are you certain that you do?
A: I don't know how you have divided $a^2+b^2$ by $ab+1$, or what remainder you've gotten. But your proof is certainly not correct, since you have "proved" that one number is the cube of the other, which is not necessarily true, e.g. take $a=8$, $b=30$.
