{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

h3 - 1 Homework III 1.78(iii Find all the integers s which...

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

1 Homework III: June 26, 2009 1 . 78 ( iii ) . Find all the integers s which are solutions to the congruence 243 x + 1 101 mod 725 . The Euclidean Algorithm gives 1 = 182 · 243 - 61 · 725. Hence, 243 x +17 101 mod 725 gives 23 x 84 mod 725, and x 182 · 84 = 15288 63 mod 725. Thus, s = 63 + 725 n for all n Z . 1 . 85 . Prove that there are no integers x, y and z such that x 2 + y 2 + z 2 = 999. In light of our earlier example involving squares, we look mod 8. Now 999 7 mod 8. Since a square is congruent mod 8 to either 0, 1, or 4, we should consider all sums u + v + w mod 8 , where u, v, w lie in { 0 , 1 , 4 } to see if 7 can occur. Any such sum involving only 0 and 1 is at most 3, not 7. Therefore, we consider only sums involving at least one 4. Note that no entry in the right hand column 7 mod 8. Since 999 7 0 0 4 4 0 1 4 5 0 4 4 8 1 0 4 5 1 1 4 6 1 4 4 9 4 0 4 8 4 1 4 9 4 4 4 12 mod 8, it cannot be a sum of three squares. Question: Does this show that if a 2 mod 8, then a is not a sum of three squares? No, for it may be a sum of two squares (and we could just set the third square to be 0). What if

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ 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