tut12sols - 2 The University of Sydney MATH2068 Number...

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The University of Sydney MATH2068 Number Theory and Cryptography (http://www.maths.usyd.edu.au/u/UG/IM/MATH2068/) Semester 2, 2008 Lecturer: R. Howlett Tutorial 12 1. You are given that 941 is prime. ( i ) Working with residue arithmetic mod 941, how many multiplications are needed to compute 6 235 ? (For example, three multiplications are needed to compute 6 8 .) ( ii ) Using a calculator, check (at least part of) the following table of powers of 6 (reduced mod 941). i 2 4 8 16 32 64 128 192 224 232 234 235 6 i 36 355 872 56 313 105 674 195 811 501 157 1 ( iii ) Solve x 2 6 (mod 941). [Hint: find an even n with 6 n 6.] (For example, to find the residue of 1296 mod 941 using your calculator, divide 1296 by 941, subtract off the integer part, and multiply back by 941.) Solution. ( i ) Twelve multiplications are needed: one for each entry in the table above. One multiplication gives you 6 2 , you square that to get 6 4 , square that to get 6 8 , and so on up to 6 128 , by which time you have done seven multiplications. Since you have computed 6 64 and 6 128 you can multiply them to get 6 192 , multiply that by 6 32 (already calculated) to get 6 224 , then multiply that by 6 8 , then by 6 2 , then by 6. No solution is provided, or (I hope) needed for Part ( ii ). However, you are hereby warned that such calculator calculations may be required in the exam. ( iii ) From the table we see that 6 470 = (6 235 ) 2 = 1; so 6 is a square mod 941. So a solution definitely exists. But – better still – the table shows us that 6 raised to an odd power (namely, 235) equals 1. So 6 236 = 6, and the square roots of 6 must be ± 6 118 . From the table we see that 6 118 = 105 × 313 × 56 × 355 × 36, and on calculating this we find that it is 299. So the square roots of 6 are 299 and - 299 = 642. 2.
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tut12sols - 2 The University of Sydney MATH2068 Number...

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