Given two or more positive integers, the **greatest common divisor (gcd)**, or **highest common factor (hcf)**, is the largest positive integer that divides the numbers without a remainder.

If we are given the prime factorization of the numbers, finding the greatest common divisor is straightforward. Let $A = p_1 ^{q_1} p_2 ^ {q_2} \ldots p_n^{q_n}$ and $B= p_1 ^{r_1} p_2 ^ {r_2} \ldots p_n^{r_n}$, where $p_i$ are positive prime integers and $q_i \geq 1$. Then

$\gcd{(A, B)} = p_1 ^ { \min(q_1, r_1)} p_2 ^ {\min (q_2, r_2)} \ldots p_n^{\min (q_n, r_n) }.$

However, in general, factorizing numbers is a very difficult problem, so instead we use the **Euclidean Algorithm**.

To find the greatest common divisor using the Euclidean Algorithm, we perform long division several times. Starting with a pair of numbers $(A, B)$, we obtain $A = Q\times B + R$, where $Q$ is the quotient and $0 \leq R < B$ is the remainder. We then repeat the sequence with the pair $(B, R)$.

As a concrete example, the following is the Euclidean Algorithm performed to calculate $\gcd( 16457, 1638 )$:

$\begin{aligned}16457 & = &1638 \times 10 & + &77 \\1638 & = &77 \times 21 & + &21\\ 77 & = & 21 \times 3 & + &14 \\21 & = & 14 \times 1 & + &7\\14 & = & 7 \times 2 & +& 0 \end{aligned}$

The process stops since we reached 0, and we obtain

$7 = \gcd (7, 14) = \gcd(14, 21) = \gcd (21, 77) = \gcd (77, 1638) = \gcd( 1638, 16457) .$

## 1. Show that if $\gcd(A,N)=1$, then $\gcd(A,B) = \gcd(A, BN)$.

Solution: Let $P$ be the set of primes that divide either $A$ or $B$. Let $P'$ be the set of primes that divide either $A$, $B$ or $N$. Let $r_i '$ be the exponent of $p_i '$ in the term $BN$. If $p_i ' \mid A$, then $p_i \not | N$, hence $r_i ' = r_i$, so $\min(q_i, r_i ' ) = \min (q_i, r_i)$. If $p_i ' \not | A$, then $q_i=0$, so $\min(q_i, r_i) = 0 = \min(q_i , r_i')$. Hence, $\gcd(A,B) = \gcd(A,BN)$.

## 2. (IMO '59) Prove that $\frac {21n+4} {14n+3}$ is irreducible for every natural number $n$.

Solution: $\frac {21n+4} {14n+3}$ is irreducible if and only if the numerator and denominator have no common factor, i.e. their greatest common divisor is 1. Applying the Euclidean Algorithm,

$\begin{aligned} 21n+4 & = &1 \times (14n+3) & + &7n+1 \\ 14n+3 & = &2 \times (7n+1) & + &1\\ 7n+1 & = & (7n+1) \times 1 & + &0. \\ \end{aligned}$

Hence, $\gcd(21n+4, 14n+3) =1$, which shows that the fraction is irreducible.

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TopNewestthanks ! a lot for posting such notes post some more on different topics

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