a and b do not have to be prime for this to hold. As long as a and b are integers, a^2 - b^2 will have two factors, a + b and a - b (since a^2 - b^2 factors to (a + b)(a - b) by difference of squares). Assuming that a + b ≠ 1 and a - b ≠ 1, a^2 - b^2 will be composite.
To show that a + b ≠ 1 and a - b ≠ 1 for all odd a and b (including prime a and b), note that a + b and a - b are both even if a and b are both odd. Since an even number cannot equal an odd number (1 in this case), a + b and a - b cannot equal 1. Hence, a^2 - b^2 is composite.
I hope this helps!
To show that a + b ≠ 1 and a - b ≠ 1 for all odd a and b (including prime a and b), note that a + b and a - b are both even if a and b are both odd. Since an even number cannot equal an odd number (1 in this case), a + b and a - b cannot equal 1. Hence, a^2 - b^2 is composite.
I hope this helps!
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a²-b² factors into (a+b)(a-b). Since a and b are defined to be odd primes neither can be 2. a+b is obviously >1 and a-b cannot be = 1 (since if a-b = 1 then a and b are coprime and one must be even, since 2 is the only even prime, their difference cannot be 1).
Hence if a-b >< 1 and a+b > 1 then (a+b)(a-b) is a composite integer.
Hence if a-b >< 1 and a+b > 1 then (a+b)(a-b) is a composite integer.
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Well, for example if a is 13 and b is 5 13^2= 169 and that is divisible by 13. Similarly 5 squared is 25 and can be divided by more than itself and 1. Does that make sense??