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Unformatted text preview: Assignment 6 – Solutions 1. Suppose that a,b ∈ Z and m,n ∈ P . Prove that an ≡ bn (mod mn ) if and only if a ≡ b (mod m ). Solution: “= ⇒ ” Suppose that an ≡ bn (mod mn ). Then an = bn + kmn for some k ∈ Z . Since n > 0, then we can divide by n to obtain a = b + km , whence a ≡ b (mod m ). “ ⇐ =” Suppose that a ≡ b (mod m ). Then a = b + qm for some q ∈ Z . Thus, an = bn + qmn (multiplying by n ). Thus, an ≡ bn (mod mn ). 2. Prove that if n ≡ 3 (mod 4) then n is not the sum of two squares. Solution: Suppose that n is the sum of two squares, say n = x 2 + y 2 . We make a table of squares modulo 4. x 0 1 2 3 x 2 0 1 0 1 From the table we see that x 2 ≡ 0 or 1 (mod 4). Similarly we have y 2 ≡ 0 or 1 (mod 4). It follows that n = x 2 + y 2 ≡ 0 + 0, 0 + 1, 1 + 0 or 1 + 1 (mod 4), that is n ≡ 0, 1 or 2 (mod 4). Thus n 6≡ 3 (mod 4). 3. Find all possible pairs of digits ( a,b ) such that 99 38 a 91 b ....
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This note was uploaded on 07/24/2011 for the course MATH 135 taught by Professor Andrewchilds during the Winter '08 term at Waterloo.
 Winter '08
 ANDREWCHILDS
 Math

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