Question

If a and b are any two odd primes, show that (a^2-b^2) is composite.

Answer 1

$$a^2-b^2=(a+b)(a-b)$$... done?

Answer 2

or, another way to think about it, is that any 2 odd primes you choose will differ by more than 1. At the very least, they must differ by 2.

example:
11-13
17-19

obviously you cannot have consecutive prime numbers (besides 2&3) because if you did have consecutive prime numbers, then 1 must be even, which no longer makes it prime.

So, case in point, because (a^2 - b^2) = (a+b)(a-b), one of the factors a+b or (a-b) must be at least greater than or equal to 2.

and if one of the terms is even, that means the resulting number will not be prime.

Answer 3

i guess to re-iterate to make more sense:

the only way to get a prime number out of (a+b)(a-b) is if one of the terms is equal to 1. Because prime numbers only factors are themselves, and 1.

So one term of (a+b)(a-b) must be prime, and one term must equal 1.

however, I just showed that you can never have a term equal to 1, therefore, must always be composite.

Answer 4

Thankyou