Question 1123424: find the gcd of 3+4i and 4+3i in the ring Z[i]
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! look at the norms
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N(3+4i) = 9 + 16 = 25
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N(4+3i) = 16 + 9 = 25
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the primes dividing each have to divide the rational prime 5, since 5^2 = 25
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these are (2+i), (2-i)
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check if (2+i) divides (3+4i)
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(3+4i)/(2+i) = (3+4i)(2-i)/(2+i)(2-i) = (10+5i)/5
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so (2+i) divides (3+4i)
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now check if (2+i) divides (4+3i)
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(4+3i)/(2+i) = (4+3i)(2-i)/(2+i)(2-i) = (11+2i)/5
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(2+i) does not divide (4+3i) and it follow that (2-i) divides (4+3i)
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check if (2-i) divides (3+4i)
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(3+4i)/(2-i) = (3+4i)(2+i)/(2-i)(2+i) = (2+8i)/5
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(2-i) does not divide (3+4i)
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therefore we can conclude that the gcd is 1
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