Let p and q be different primes greater than 2. Prove that pq can be written as difference of two squares in exactly two different ways.
Without loss of generality claim p > q.Now, we try to write pq in the form pq = a2-b2 where a and b are non-negative integers.If b = 0, then pq = a2. But p and q are different, so pq is not a square. So b>0Factorising, we get: pq = (a-b)(a+b) ()So a+b must divide pq. Since p and q are prime, we can write out the factors of pq: 1, p, q, pq. Now we try the possible choices:1) a+b = 1This is false since both a and b are greater than 02)a+b = qThen, substituting in (), we get that a-b = p. But we claimed p>q so a-b > a+b, which is false since b > 03)a+b = pSubstituting in () we get that a-b = q. Adding these 2 equations we get that a = (p+q)/2 and hence b = (p-q)/2So pq = [(p+q)/2]2 - [(p-q)/2]24) a+b = pqFrom () we get that a-b =1Adding these two we get that a = (pq+1)/2And consequently b = (pq-1)/2so pq = [(pq+1)/2]2-[(pq-1)/2]2Since these were all the possible cases, we conclude that pq can be written as difference of two squares in exactly 2 different ways.
