Find d=GCD(a,b) and integers x and y such that d=ax+by....

Related BrainMass Solutions

• Congruences : Primes, Inverse Modulo, GCD and Wilson's theorem

  (ii) only if: suppose ax is congruent to b mod n has a solution x in integers.

• Abstract Algebra : Prime Factors and Greatest Common Divisor

  (greatest common devisor) There exits two integers x, y such that a x + c y = 1 (1) and (b, c) = 1 then there exits two integers u, v such that b u + c v = 1 (2)

• Diophantine equation

  Prove that the full set of solutions to this equation is given by x = x_0 + (mb/gcd(a,b)) y = y_0 - (ma/gcd(a,b)) as m ranges over the integers.

• Proofs, Diophantine equations, and sequences

  If gcd (a, b)|c, does the equation always have an integer solution? Solution: Suppose that (x1, y1) is an integral solution to ax^2 + by^2 = c and that d = gcd(a,b).

• Greatest Common Divisors

  Suppose that d' = gcd(b, r), we know that d'|b and d'|r So, there exist integers m and n such that b=d'm and r=d'n So, a=qb+r=qd'm+d'n=d'(qm+n) So, d'|a Since d'|b, we have d'|d= gcd(a, b) Similarly, we can prove

• Equality of gcd's

  Second, since x=gcd(ac,b), then x|b, x|ac. From the condition, we know gcd(a,b)=1, then we can find some integers m and n, such that am+bn=1. Then x|(bnc+acm)=c(bn+am)=c. Then we get x|b and x|c. So x|gcd(b,c)=y. therefore, x=y.

• Congruence Functions Solved

  Solution: We have gcd (42, 90) = 6, so by Theorem Given below, there is a solution since 6 is a factor of 12. The congruence ax b (mod n) has a solution if and only if b is divisible by d, where d = (a, n).

• Greatest common divisor

  (a) Determine gcd(987, 320). (b) Determine integers x and y such that 987x + 320y = gcd(987, 320). 3. (a) Determine gcd(−2193, 1008).

• Greatest Common Divisor of Polynomials

  Let f(x) = x^4+2x^3−x^2−4x−2 and g(x) = x^4+x^3−x^2−2x−2. Find the greatest common divisor d(x) of f(x) and g(x) in Q[x]. Find polynomials a(x), b(x) in Q[x] such that d(x) = a(x)f(x) + b(x)g(x).