MATH535_HW1

# MATH535_HW1 - Serge Ballif MATH 535 Homework 1 1(a Find...

• Notes
• 3

This preview shows pages 1–2. Sign up to view the full content.

Serge Ballif MATH 535 Homework 1 August 31, 2007 1) (a) Find integers x 0 and y 0 such that 4 x + 11 y = 1 . (b) Explain how all pairs ( x, y ) of integers 4 x + 11 y = 1 are related to the pair you found. (a) The ordered pair ( x 0 , y 0 ) = (3 , - 1) satisfies the equation. (b) The line 4 x +11 y = 1 (or y = - 4 11 x + 1 11 ) goes through the point (3 , - 1), but as x varies from one integer to the next, we see that y is increasing (or decreasing) by 4 11 . Since 4 and 11 are relatively prime, the only way for the y -value to be an integer at the same time as the x -value is when the x -value is of the form 3+11 m for some integer m . Thus all pairs of integer solutions of 4 x + 11 y = 1 are of the form (3 + 11 m, - 1 - 4 m ). 2) Let P ( X ) and Q ( X ) [ X ] be polynomials. Prove that if F ( X ) is a gcd of P ( X ) and Q ( X ) in [ X ] , then for some non-zero α , α · F ( X ) [ X ] . F ( X ) is a gcd of P ( X ) and Q ( X ) all divisors of both P ( X ) and Q ( X ) divide F ( X ). Let G ( X ) [ X ] be a gcd of P ( X ) and Q ( X ) in [ X ]. Since [ X ] [ X ], we know that G ( X ) divides F ( X ). However, F ( X ) divides both P ( X ) and Q ( X ), so it divides any sum of multiples of P ( X ) and Q ( X ). Since G ( X ) is a gcd of P ( X ) and Q ( X ) in [ X ], then G ( X ) = r ( X

This preview has intentionally blurred sections. Sign up to view the full version.

This is the end of the preview. Sign up to access the rest of the document.
• Fall '08
• BROWNAWELL,WOODRO

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern