Where p and q are different positive primes.

If (p + q)^-1 (p^-1 + q^-1) = p^aq^b, prove that a + b + 2 = 0, where p and q are different positive primes.

Solution:

It is given that

(p + q)^-1 (p^-1 + q^-1) = p^aq^b

We can write it as

By cross multiplication

p^-1q^-1 = p^aq^b

By comparing the powers

a = – 1 and b = – 1

Here

LHS = a + b + 2

Substituting the values

= – 1 – 1 + 2

= 0

= RHS

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