Al(bc), then either alb or alc

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SUMMARY

The discussion centers on the mathematical statement regarding divisibility: if \( a \) divides \( bc \), then \( a \) must divide either \( b \) or \( c \) only if \( a \) is a prime number. The user provides a counterexample using the integers 4, 12, and 6, demonstrating that while 4 divides 12, it does not divide 6, thus invalidating the statement for composite numbers. The conclusion emphasizes the necessity of \( a \) being prime for the divisibility condition to hold true.

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Homework Statement


al(bc), then either alb or alc
I'm trying to explain why this is true.


Homework Equations





The Attempt at a Solution


a divides bc.
a divides b or a divides c is true because b is just a multiple of c and a will still be a factor.
 
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This statement is true only if a is prime. Use the fact that if gcd (a,b) = c then there is some integers x,y such that ax+by=c
 
It's not true. Counterexample: 4 | 12, but 4 doesn't divide 6 and 4 doesn't divide 2.
 

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