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secondary 4 | A Maths

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= (n - 2)(n -1)

for n² - 3n + 2 to be prime, one of its factors has to be 1 since a prime number is only divisible by itself and 1.

However, it is also possible for both factors above to be negative, which results in a positive prime number. So is it possible for one of the factors to be -1 as well.

n - 2 = 1 or n - 1 = 1

n = 3 or n = 2

check :

When n = 3, (n - 2)(n -1)

= (3 - 2)(3 - 1)

= 2 (prime)

When n = 2, (n - 2)(n -1)

= (2 - 2)(2 - 1)

= 0 (not prime. So n = 2 is rejected)

n - 2 = -1 or n - 1 = -1

n = 1 or n = 0

check :

When n = 1, (1 - 2)(1 - 1)

= 0 (not prime. n = 1 is rejected)

When n = 0, (0 - 2)(0 -1)

= 2 (prime)

So, n = 3 or n = 0.

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First, we factorise the expression using cross-factorisation.

Factorising the expression yields (n - 1) (n - 2).

Now, (n - 1) represents a number; so does (n - 2), so (n - 1) (n - 2) is just simply the product of two numbers. For example, if n = 5, then (n - 1) equals 4 and (n - 2) equals 3, so (n - 1) (n - 2) is the product of 4 and 3.

A prime number is a positive integer which can be divided by 1 and itself (the definition does not include 0 and 1 as a prime number).

For the product (n - 1) (n - 2) to be a prime number, either of the bracketed numbers must equal to 1 (or -1 in some cases). It's alright for one of the numbers in the bracket to be -1, so long as the other number also has a negative sign (so that the product of two negative numbers is a positive number). However, obtaining 1 or -1 does not automatically mean that multiplication with the other number yields a prime number, so we must check accordingly.

There are four cases to consider by setting (n - 1) = 1, (n - 1) = -1, (n - 2) = 1 and (n - 2) = -1.

CASE 1 -----> (n - 1) = 1

Setting (n - 1) = 1, we get n = 2. This gets us (n - 2) = 0, and (n - 1) (n - 2) = 1 x 0 = 0 is not a prime number. So, n cannot be 2, and this case is rejected.

CASE 2 -----> (n - 1) = -1

Setting (n - 1) = -1, we get n = 0. This gets us (n - 2) = -2, and (n - 1) (n - 2) = (-1) x (-2) = 2 is a prime number. So, n can be 0.

CASE 3 -----> (n - 2) = 1

Setting (n - 2) = 1, we get n = 3. This gets us (n - 1) = 2, and (n - 1) (n - 2) = 2 x 1 = 2 is a prime number. So, n can be 3.

CASE 4 -----> (n - 2) = -1

Setting (n - 2) = -1, we get n = 1. This gets us (n - 1) = 0, and (n - 1) (n - 2) = 0 x (-1) = 0 is not a prime number. So, n cannot be 1, and this case is rejected.

All other values of n will not get you a prime number. For example, if n = 5, then (n - 1) (n - 2) = 4 x 3 which obviously does not equate to a prime number (which is the basis of why we must set either of the bracketed terms to 1 or -1).

Hence, n can be 0 or 3.