Simplifiable

Number Theory Level 3

Let us take two fractions $\dfrac{a}{b}$ and $\dfrac{c}{d}$ , where $a,b,c$ and $d$ are pairwise coprime positive integers .

When is $\dfrac{ad+cb}{bd}$ simplifiable ? That is, when the numerator and denominator are divisible by the same number? Find the number of all such quadruplets $(a,b,c,d)$ where $a,b,c,d<16$ .

The answer is 0.

1 solution

Stanislava Sojáková
Jul 11, 2016

If the $ad+cb$ and $bd$ are both divisible by some integer $k$ , that is $k|ad+cb$ and $k|bd$ , they are both divisible by a prime number $p$ such that $p|k$ . This is because $k$ must be divisible by some prime number $p$ , and if $p|k$ and $k|ad+cb$ and $k|bd$ , then $p|ad+cb$ and $p|bd$ .

Now, if $p|bd$ , then without loss of generality we have $p|b$ . If $p|b$ and $p|ad+cb$ , then $p|ad$ . From this, either $p|a$ or $p|d$ . However, because of the coprimeness of $a,b$ and $b, d$ we have a contradiction, and there are no such quadruplets $(a,b,c,d)$ .