Why no pair $(a,b)$ of odd positive integers $a>1$ and $b>1$ exist that make the quotient $\frac{(3a+1)(3b+1)}{a\cdot b}$ an integer?

Question

Let $a>1$ and $b>1$ be two odd positive integers.

How we can show that there do not exist a pair $(a,b)$ that makes the quotient $\frac{(3a+1)(3b+1)}{a\cdot b}$ an integer?

Answer 1

$\frac {(3a+1)(3b+1)}{ab} = \frac {9ab + 3a + 3b + 1}{ab}=9 + \frac {3a+3b+1}{ab}$

For this to be an integer we need $ab \le 3a + 3b + 1$. Wolog $a\le b$ and we need $a\le 3\frac ab + 3 + \frac 1b \le 6+\frac 1b$ so we must have $a \le 5$.

So $a = 3$ or $5$. If $a = 3$ then $3|ab$ and so $ab|3a+3b+1$ would imply $3|3a+3b+1$ which is not possible.

If $a = 5$ then we have $5b|15 + 3b+1$ which means $5b \le 15+3b +1$ which mean $b \le 8$ so $b \le 7$ so $b = 5$ or $7$.

So we either have $25|31$ or $35|37$ neither of which are true.