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Show that the the only prime of form n^3-1 is 7.
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n^(3-1) or n^(3)-1?
\[n^3-1\]
because it factors
n^3-1= n^3-1^3= (n-1)(n^2-n+1) We know: If n is an integer, n-1 is an integer and n^2-n+1 is also an integer. Thus, (n-1)(n^2-n+1)=ab for integers a and b if ab is prime, either a or b must be 1 (or else, we would have more than two factors). Thus, either: n-1=1 OR n^2-n+1=1 If the first is true: n=2 If the second is true, n=0 and n=1 So, we test n=0, n=1, and n=2 in n^3-1 to see which one(s) are prime: n=0, n^3-1=0-1=-1 not prime n=1, n^3-1=1-1=0 not prime n=2, n^3-1=8-1=7 prime Thus, 7 is the only prime in the form of n^3-1.
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