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3 2 .2 0 0 = 28 . 5'
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(a4 . 51)2 = 80'.
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This example illustrates that we can obtain a perfect square from non-square congruence relations such as those in (7.2). To find these perfect squares, we only need t o concern oiirselves with the powers in the prime factorization of the relations under consideration. Also, by properties of exponentiation, the corresponding powers add when we multiply terms. Furthermore, any time we multiply terms and all of the resulting powers arc even, we obtain a perfect square. Consequently, we only need t o be concerned with whether the powers are even or odd, that is, we only need t o know the powers modulo 2. We can associate each number with a vector of t h e powers in its prime factorization, and we obtain a perfect square by multiplying corresponding t'crins whenever the siirri of these vectors contain only even numbers. In the exarnple above we have 32+
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(mod 2),
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where the numbers in the vector represent the powers of the prinie factors 2 and 5, respectively, in the prime decomposition of 32. Similarly.
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The product therefore satisfies
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Since the powers are all even, we know that 32 . 200 is a perfect square modulo 1649. Furthermore, from (7.2), we know t h a t this square is equal to (41 . 43)2 (mod 1649), giving us the desircd congruence of squares. While the actual powers are required to determine the value that is squared (80 in this example), to determine whether or not we have a congruence of squares, wc only require the mod 2 vectors of powers. We can multiply any number of relations to obtain a perfect square. Also, the number of distinct primes in the factorizations of the numbers under
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consideration determines the size of the vectors. Since we ultimately want to factor large numbers, it is imperative that we keep the vectors small. Consequently, we will choose a bound B and a set of primes, where each prime is less than B. This set of primes is our factor base. While all primes in the factor base are less than B, generally not every such prime will be included in the factor base (see Problem 7). A number that factors completely over the given factor base is said to be B-smooth. Smooth relations (or, more precisely, B-smooth relations) are those relations that factor completely over the factor base. By restricting our attention to B-smooth relations, we restrict the size of the vectors of powers. The fewer elements in the factor base, the smaller the vectors that we need to deal with (which is good), but the harder it is to find B-smooth values (which is bad). An example should illustrate the idea. Suppose we want to factor the integer N = 1829 and we choose B = 13, as in the example given in [92]. Since we want numbers with small factors, it is advantageous to deal with modular numbers between - N/2 and N/2, instead of in the range 0 to N - 1. This creates a slight complication, since we must include -1 in our factor base, but this is easily managed. In this example, B = 15 and we take -1,2,3,5,7,11,13 as our factor base. Now we could simply select a random r and check whether r (mod N ) ' is B-smooth, repeating this until we have obtained a sufficient number of B-smooth values. Instead, we use a slightly more systematic approach. We select the values and I 1for k = 1,2,3,4, and test whether the m , square of each, modulo 1829, is B-smooth. For this example we obtain
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