Proof: This is straightforward modular arithmetic. We have 2x+3y≡0mod 17⇔13(2x+3y)=26x+39y≡9x+5y≡0mod 17, since 13 is a unit in ℤ/17ℤ. Generally, when z ⧸| a,b, we have ax+by is divisible by z a prime for the same integral values as cx+dy iff −a−1b≡−c−1dmod z, and when z | a but z ⧸| c then (1,0) is a value for ax+by but not for cx+dy, similarly for z | c but z ⧸| a, and when z | a,b we have the same values when b and d coincide in divisibility or lack of divisibility by z.
Friday, April 11, 2014
Problem 7
Proof: This is straightforward modular arithmetic. We have 2x+3y≡0mod 17⇔13(2x+3y)=26x+39y≡9x+5y≡0mod 17, since 13 is a unit in ℤ/17ℤ. Generally, when z ⧸| a,b, we have ax+by is divisible by z a prime for the same integral values as cx+dy iff −a−1b≡−c−1dmod z, and when z | a but z ⧸| c then (1,0) is a value for ax+by but not for cx+dy, similarly for z | c but z ⧸| a, and when z | a,b we have the same values when b and d coincide in divisibility or lack of divisibility by z.
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