Show that this equation is true

Question

It is given that p is a prime number and p > 5.

a) Show that p= 6n + 1 or p = 6n + 5 for some positive integer n.

b) Hence, show that p² + 2 is divisible by 3.

 

How can I 'show that'? With examples? When they say 'some integer' so they mean use several integers as n to 'show that' it is true?

Answer 1

 

p is prime and p>5, show p=6n+1 or p=6n+5 for some positive integer n

for n=0,1,2,3,4,... all whole numbers can be written in the form:

6n, 6n+1, 6n+2, 6n+3, 6n+4, or 6n+5

since n is positive, we are starting with 6,7,8,9,10,11,...

p>5 means we are talking about primes greater than 5

getting back to 6n, 6n+1, 6n+2, 6n+3, 6n+4, and 6n+5, and don't forget that all whole numbers can be expressed in one of these forms

6n is divisible by 2 so it is composite

6n+2 is divisible by 2 so it is composite (6n+2=2(3n+1))

6n+3 is divisible by 3 so it is composite (6n+3=3(2n+1))

6n+4 is divisible by 2 so it is composite (6n+4=2(3n+2))

that leaves 6n+1 and 6n+5 for all the other whole numbers and a whole number, other than 0 and 1, are either composite or prime; therefore 6n+1, not being composite, must be prime and 6n+5, not being composite, must be

prime; also so there's no confusion, if 6n+1 becomes composite, then 6n+5 is prime and if 6n+5 becomes composite then 6n+1 is prime; if n=4, then 6n+1=25 and 6n+5=29; if n=5, then 6n+5=35 and 6n+1=31

 

the second part...

p^2+2 must be divisible by 3

(6n+1)^2=36n^2+12n+1, now add 2 to get 36n^2+12n+1+2=36n^2+12n+3=3(12n^2+4n+1) which is divisible by 3

(6n+5)^2=36n^2+60n+25, now add 2 to get 36n^2+60n+25+2=36n^2+60n+27=3(12n^2+20n+9) which is divisible by 3

 

 

Answer 2

Hi Marietta,

That's exactly what they mean. So for instance, take the integer 4 as your n value and try it out on p=6n+1 and p=6n+5.

Then, take one of the p values you get and show how (p²+2)/3 equals an integer.

 

~Kehley