# 5 which of the following congruences have solutions a

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5. Which of the following congruences have solutions a. x 2 11 (mod 29 ) b. x 2 23 (mod 31 ) c. x 2 5 (mod 19 )
d. x 2 7 (mod 13 ) Since each modulus is prime, we know that x a ( mod p ) will have a solution if and only if a p 1 2 1 ( mod p ) . Thus we have to compute 11 14 ( mod 29) , 23 15 ( mod 31) , 5 9 ( mod 19) , 7 6 ( mod 13) . Use the successive squaring algorithm for computing these (see web page). For example, the last of these is just (49) . (49) . (49) ( mod 13) . Since 49 10 ( mod 13) , this simplifies to (10) . (10) . (10) ( mod 13) or ( 3) . ( 3) . ( 3) ( mod 13) or 27 ( mod 13) , that is 1 . Thus x 2 7 ( mod 13) has no solution. [This will not be covered on the exam; I include it as it is the next thing we will cover in class.] As an alternative method, we can compute the Legendre symbol a n . For the first of these, we need 11 29 . Note both 11 and 29 are odd and in fact both are prime so that gcd (11 , 29) = 1 . Then, 11 29 = 29 11 = 7 11 = 11 7 = 4 7 = 2 7 2 = 1 . Thus we can conclude that 11 is not a square root ( mod 29) . For the second problem we have: 23 31 = 31 23 = 8 23 = 2 23 3 . We can compute 2 23 directly from the theorem: since 23 7 ( mod 8) it is +1 and so 23 31 = 1 . This shows there is no solution to this congruence. Finally, we compute 5 19 : 5 19 = 19 5 = 4 5 = 2 5 2 = 1 . Thus 5 9 ( mod 19)
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