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Congruence of squares
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===Factorize 1649=== Using ''n'' = 1649, as an example of finding a congruence of squares built up from the products of non-squares (see [[Dixon's factorization method]]), first we obtain several congruences :<math>41^2 \equiv 32 = 2^5 \pmod{1649},</math> :<math>42^2 \equiv 115 = 5 \cdot 23 \pmod{1649},</math> :<math>43^2 \equiv 200 = 2^3 \cdot 5^2 \pmod{1649}.</math> Of these, the first and third have only small primes as factors, and a product of these has an [[parity (mathematics)|even]] power of each small prime, and is therefore a square :<math>32 \cdot 200 = 2^{5+3} \cdot 5^2 = 2^8 \cdot 5^2 = (2^4 \cdot 5)^2 = 80^2</math> yielding the congruence of squares :<math>32 \cdot 200 = 80^2 \equiv 41^2 \cdot 43^2 \equiv 114^2 \pmod{1649}.</math> So using the values of 80 and 114 as our ''x'' and ''y'' gives factors :<math>\gcd( 114-80, 1649 ) \cdot \gcd( 114+80, 1649 ) = 17 \cdot 97 = 1649.</math>
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