### 7.1.1 GCD

The gcd command finds the greatest common divisor (GCD) of a set of integers or polynomials. (See also Section 11.2.5 for polynomials.) It can be called with one or two arguments.

- gcd can take one argument: seq, a sequence or list of integers or polynomials.
- gcd(seq) returns the GCD of the elements of seq.
- Alternatively, gcd can take two arguments: s and t, two lists of the same length containing integers or polynomials (alternatively, a matrix m with two rows whose elements are integers or polynomials).
- gcd(s,t) (or gcd(m)) returns the list whose kth element is the GCD of the kth elements of s and t (or the kth column of m).

The Gcd command is the inert form of gcd; namely, it evaluates to gcd, for later evaluation.

#### Examples

With one argument:

gcd(18,15)

3 |  |  |  |  |  |  |  |  |  |

gcd(18,15,21,36)

3 |  |  |  |  |  |  |  |  |  |

gcd([18,15,21,36])

3 |  |  |  |  |  |  |  |  |  |

gcd(-5-12*i,11-10*i)

3+2 i |  |  |  |  |  |  |  |  |  |

With two arguments:

gcd([6,10,12],[21,5,8])

or:

gcd([[6,10,12],[21,5,8]])

| ⎡ ⎣ | 3,5,4 | ⎤ ⎦

Find the greatest common divisor of 4n+1 and 5n+3 when n ∈ ℕ.

f(n):=gcd(4*n+1,5*n+3)

Then (in a program editor level):

essai(n):={ local j,a,L; L:=NULL; for (j:=-n;j<n;j++) { a:=f(j); if (a!=1) { L:=L,[j,a]; } } return L; }

Now:

essai(20)

| ⎡ ⎣ | −16,7 | ⎤ ⎦ | , | ⎡ ⎣ | −9,7 | ⎤ ⎦ | , | ⎡ ⎣ | −2,7 | ⎤ ⎦ | , | ⎡ ⎣ | 5,7 | ⎤ ⎦ | , | ⎡ ⎣ | 12,7 | ⎤ ⎦ | , | ⎡ ⎣ | 19,7 | ⎤ ⎦

From this information, a reasonable conjecture would be that gcd(4n+1,5n+3)=7 if n=7k−2 for some k ∈ ℤ and gcd(4n+1,5n+3)=1 otherwise.

Since gcd(a,b)=gcd(a,b−c a) for integers a, b and c; we have gcd(4n+1,5n+3)=gcd(4n+1,5n+3−(4n+1))=gcd(4n+1,n+2)= gcd(4n+1−4(n+2),n+2)=gcd(−7,n+2)=gcd(7,n+2), and so gcd(4n+1,5n+3)=7 if 7 divides n+2, namely n+2=7k or n=7k−2, and gcd(4n+1,5n+3)=1 otherwise. This proves the conjecture.

An inert form of GCD:

Gcd(18,15)

gcd | ⎛ ⎝ | 18,15 | ⎞ ⎠

eval(Gcd(18,15))

3 |  |  |  |  |  |  |  |  |  |

(See Section 9.1.1.)