Bunuel wrote:
If \(p\) is a prime number, what is the value of \(p\)?
(1) \(\sqrt[3]{1-p^2}=-n\), where \(n\) is a prime number.
(2) \(8p^2+1=m\), where \(m\) is a prime number.
I wrote this question, so below is my solution:
(1) \(\sqrt[3]{1-p^2}=-n\), where \(n\) is a prime number.
It's clear that \(p\neq{2}\) as in this case \(\sqrt[3]{1-p^2}\neq{integer}\), so \(p\) is a prime more than 2, so odd. Then \(1-p^2=odd-odd=even\) --> cube root from even is either an even integer or not an integer at all, we are told that \(\sqrt[3]{1-p^2}=-n=integer\), so \(-n=even=-prime\) --> \(n=2\) (the only even prime) --> \(\sqrt[3]{1-p^2}=-2\) --> \(p=3\). Sufficient.
(2) \(8p^2+1=m\), where \(m\) is a prime number.
\(p\) can not be any prime but 3 for \(8p^2+1\) to be a prime number: because if \(p\) is not 3, then it's some other prime not divisible by 3, but in this case \(p^2\) (square of a not multiple of 3) will yield the remainder of 1 when divided by 3 (not a multiple of 3 when squared yields remainder of 1 when divided by 3), so \(p^2\) can be expressed as \(3k+1\) --> \(8p^2+1=8(3k+1)+1=24k+9=3(8k+3)\) --> so \(8p^2+1\) becomes multiple of 3, so it can not be a prime. So \(p=3\) and in this case \(8p^2+1=73=m=prime\). Sufficient.
Answer: D.
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