For how many prime pairs (p, q) does there exist an integer n such that
(p2 + 1)*(q2 + 1) = n2 + 1 ?	Choose one:	A) 2	B) 4	C) 6	D) 8	E) infinitely many
First, we show that one of p and q is 2. If both p and q are odd, then n2 + 1 = (p2 + 1)(q2 + 1)   is even so that n is also odd. Hence, there is an integer t such that n = 2t + 1 so that n2 + 1 =   4t2+4t+2. But this means n2+1 is not divisible by 4 whereas (p2+1)(q2+1) clearly is. Thus, we   must have one of p or q is 2. We now have n2 + 1 = 5(x2 + 1) where either x = p or x = q. Note   that x 3. We deduce that 5x2 = n2 − 4 = (n − 2)(n + 2). Since x is prime and n + 2 &gt; n − 2,   we obtain that n−2 2 {1, 5, x} and n+2 = 5x2/(n−2). Since also (n+2)−(n−2) = 4, we get   that one of 5x2 − 1, x2 − 5 and 4x equals 4. As x is prime, we obtain x = 3. Thus, (p, q) = (2, 3)   or (3, 2). Each pair gives a solution to the equation n2+1 = (p2+1)(q2+1), implying the answer.   correct answer 2
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For how many prime pairs (p, q) does there exist an integer n such that
(p2 + 1)*(q2 + 1) = n2 + 1 ?