# tut07sols - The University of Sydney MATH2068 Number Theory...

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Unformatted text preview: 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 7 1. You are told that n = 63120357457 and ( n ) = 63119854960. ( i ) Applying the Euclidean Algorithm to n and ( n ), the first few steps are n = ( n ) + 502497 , ( n ) = 125612 502497 + 201796 , 502497 = 2 201796 + 98905 , 201796 = 2 98905 + 3986 , 98905 = 24 3986 + 3241 . Complete this, and use the result to show that n is square-free. ( ii ) Use the formula for ( n ) proved in lectures to show that if p is a prime divisor of n then ( n ) n- 1 p n , and so p n/ ( n- ( n )). ( iii ) Show that n is the product of two primes p and q that both exceed 125612. ( iv ) Use the fact that ( pq ) = ( p- 1)( q- 1) to find p + q (where p, q are as in Part ( iii )). ( v ) Find p and q by solving the equation x 2- ( p + q ) x + pq = 0. Solution. ( i ) 3986 = 3241 + 745, 3241 = 4 745 + 261, 745 = 2 261 + 223, 261 = 223 + 38, 223 = 5 38 + 33, 38 = 33 + 5, 33 = 6 5 + 3, 5 = 3 + 2 and 3 = 2 + 1. Thus gcd( n, ( n )) = 1. Now if n is not square-free there must be a prime divisor p of n such that p k | n for some k &gt; 1. Then p k- 1 ( p- 1) = ( p k ) | ( n ), and so p | ( n ). Since also p | n , this contradicts gcd( n, ( n )) = 1. So n is square-free. ( ii ) Let p = p 1 , p 2 , . . . , p k be all the prime divisors of n . We have ( n ) = n (1- 1 p 1 )(1- 1 p 2 ) (1- 1 p k ) n (1- 1 p 1 ), since 0 &lt; 1- 1 p i &lt; 1 for all i . Thus ( n ) n- 1 p n . Rearranging gives 1 p n &lt; n- ( n ), whence 1 p &lt; ( n- ( n )) /n , and p &gt; n/ ( n- ( n )). 2 ( iii ) In view of the Euclidean Algorithm calculations above we see that n- ( n ) = 502497 and when n is divided by this number the quotient is 125612. So n/ ( n- ( n )) &gt; 125612, and so by Part (...
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## tut07sols - The University of Sydney MATH2068 Number Theory...

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