Adv Alegbra HW Solutions 34

Adv Alegbra HW Solutions 34 - 34 b 5 mod 10. Hence, b = 5 +...

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34 b 5 mod 10. Hence, b = 5 + 10 q . and so b 2 ( 5 + 10 q ) 2 mod 100. But ( 5 + 10 q ) 2 = 25 + 2 · 5 · q + 100 q 2 25 mod 100 , and so the last two digits of b 2 are 25, not 35. Therefore, no such b exists. (More is true. We have proved that if the last digit of a perfect square is 5, then its last two digits are 25.) 1.87 If x is an odd number not divisible by 3, prove that x 2 1 mod 24. Solution. Here are two ways to proceed. The odd numbers < 24 not divisible by 3 are 1, 5, 7, 11, 13, 17, 19, 23; square each mod 24. Alternatively, Example 1.161 says that the squares mod 8 are 0, 1, and 4. Now x 2 1 is divisible by 24 if and only if it is divisible by 3 and by 8 (as 3 and 8 are relatively prime). If x is to be odd, then x 0mod3or x 2 mod 3; looking at x mod 8, the hypothesis eliminates those x with x 2 0mod8or x 2 4 mod 8. 1.88 Prove that if p is a prime and if a 2 1mod p , then a ≡± 1mod p . Solution.
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