Question

I am not getting what is written here

Answer 1

http://www.algebra.com/algebra/homework/playground/test.faq.question.432963.html

Answer 2

this part
Now p and q cannot both be odd primes at the same time, because by then p-q and p+q will both be even, and their product will be even, not prime. (Neither can we assume that p = q = 2. why?)
Hence we can assume that q is equal to 2 (the only even prime),

Answer 3

p and q are primes... that means they are odd(except 2)...
again p^2-q^2=(p+q)*p-q)
both p+q and p-q are even
and their multiplication will always be even

again even prime no is 2 and thus find it urself
hpoe u got it

Answer 4

Since r is odd, then it can't be even unless r=2, since r is prime. Basically all this means is that (p+q) and (p-q) can't be an even number otherwise r is an even number greater than 2... meaning it's not prime.

Answer 5

then you should find the sol for which r is prime

Answer 6

I think I said that wrong because I'm kind of tired at the beginning there. r is prime. X_X

Answer 7

why we asuume q as 2

Answer 8

There are essentially 4 cases we need to consider based on the pairity of p and q. They can either both be even, odd, or opposites.

Suppose they're both odd, then they'll look like this:

p=2n+1
q=2m+1

so (p+q)=2(m+n)+2 which is always an even number.

What if they're both even? Then obviously we we also get an even number from the sum.

Remember, r is a prime number, so r can't be even, unless it's 2.

Ok last two cases is when they're opposite. So if one is even and one is odd, which do we pick though? Well since 2 is the smallest prime number, we know q must be 2 since r is a prime number, and primes are positive.

(p-2)(p+2)=r

See how if we chose p=2 we would be wrong because then it would imply r is negative.