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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.

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,bc 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.)


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