3-4. Find all numbers by which the fraction \(\frac{5 l+6}{8 l+7}\) can be reduced when \(l\) is an integer.
Solution 4. We will compute \(\text{GCD}(5l+6, 8l+7)\), using the fact that \(\text{GCD}(a, b) = \text{GCD}(a-b, b)\). As a result, we get \(\text{GCD}(5l+6, 8l+7) = \text{GCD}(5l+6, 3l+1) = \text{GCD}(2l+5, 3l+1) = \text{GCD}(2l+5, l-4) = \text{GCD}(l+9, l-4) = \text{GCD}(13, l-4)\). The number 13 is prime, so the fraction can only be reducible by 13. When \(l=4\), we get the fraction \(\frac{26}{39}\), which is indeed reducible by 13.