h3 - 1 Homework III: June 26, 2009 1.78(iii). Find all the...

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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
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This note was uploaded on 11/15/2010 for the course MATH 417 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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h3 - 1 Homework III: June 26, 2009 1.78(iii). Find all the...

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