If (p + q)-1(p-1 + q-1) = paqb, prove that a + b +2 = 0, where p and q are different positive primes.
50 Likes
Given,
⇒(p+q)−1(p−1+q−1)=paqb⇒1(p+q)(1p+1q)=paqb⇒1(p+q)×p+qpq=paqb⇒1pq=paqb⇒p−1q−1=paqb∴a=−1 and b=−1.\Rightarrow (p + q)^{-1}(p^{-1} + q^{-1}) = p^aq^b \\[1em] \Rightarrow \dfrac{1}{(p + q)}\Big(\dfrac{1}{p} + \dfrac{1}{q}\Big) = p^aq^b \\[1em] \Rightarrow \dfrac{1}{(p + q)} \times \dfrac{p + q}{pq} = p^aq^b \\[1em] \Rightarrow \dfrac{1}{pq} = p^aq^b \\[1em] \Rightarrow p^{-1}q^{-1} = p^aq^b \\[1em] \therefore a = -1 \text{ and } b = -1.⇒(p+q)−1(p−1+q−1)=paqb⇒(p+q)1(p1+q1)=paqb⇒(p+q)1×pqp+q=paqb⇒pq1=paqb⇒p−1q−1=paqb∴a=−1 and b=−1.
Substituting values of a and b in L.H.S. of a + b + 2 = 0 we get,
a + b + 2 = -1 + (-1) + 2 = -2 + 2 = 0.
Hence, proved that a + b + 2 = 0.
Answered By
22 Likes
If x4y2z3 = 49392, find the values of x, y and z, where x, y and z are different positive primes.
If a6b−43=ax.b2y\sqrt[3]{a^6b^{-4}} = a^x.b^{2y}3a6b−4=ax.b2y, find x and y, where a, b are different positive primes.
If (p−1q2p2q−4)7÷(p3q−5p−2q3)−5=pxqy\Big(\dfrac{p^{-1}q^{2}}{p^2q^{-4}}\Big)^7 ÷ \Big(\dfrac{p^3q^{-5}}{p^{-2}q^{3}}\Big)^{-5} = p^xq^y(p2q−4p−1q2)7÷(p−2q3p3q−5)−5=pxqy, find x + y, where p and q are different positive primes.
Solve the following equation for x:
52x + 3 = 1
