stne wrote:
But where does it say that the given exp. is equal to zero ? The exp. can at max.be reduced to (n-2)(n-1), this does not necessarily mean that n =2 or n=1 , the variable n can take an infinite number of values. Unless and until we know the RHS of the exp. we cannot deduce the value of n. Hope you agree.
The method given above by
EgmatQuantExpert uses a better logic,I suppose.
Yes, you're right that the expression is not equal to zero. But the question asks if the number is prime, and so we care whether we can factor the number. If our number is equal to (n-2)(n-1), and if those factors are positive, our number could only be prime if one of those factors is equal to 1 -- otherwise we would have written our number as a product of two smaller numbers, which means our number is not prime. And if one of those factors is equal to 1, it must be the smaller of the two factors, so n -2 = 1, and n = 3. The product would also be prime if n = 0, but we're told n is positive here.
It is true that you can reach the answer to this question by noticing that the expression is always even, but that won't work in general. When you're asked if a number is prime, you really care if you can factor the number, not whether it's even or odd. If the question had asked instead about n^2 + 4n + 3, say, then an even/odd analysis wouldn't get you very close to an answer.
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