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# Euclidean Algorithm

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.

## Technique

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

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

## Worked Examples

### 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.

Note by
1 year, 6 months ago

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Achal Jain
3 months, 3 weeks ago

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