Question

f (p+q)^-1 (p-1+q^-1) = p^aq^b proved that a+b+2=0 where p and q are different positive primes​

Answer 1

SOLUTION

GIVEN

$\displaystyle \sf{ {(p + q)}^{ - 1} ( { p}^{ - 1} + {q}^{ - 1} ) = {p}^{a} \: {q}^{b} }$

TO PROVE

a+b+2=0

EVALUATION

Here the given expression

$\displaystyle \sf{ {(p + q)}^{ - 1} ( { p}^{ - 1} + {q}^{ - 1} ) = {p}^{a} \: {q}^{b} }$

$\displaystyle \sf{ \implies \frac{1}{(p + q)} \bigg( \frac{1}{p} + \frac{1}{q} \bigg) = {p}^{a} \: {q}^{b} }$

$\displaystyle \sf{ \implies \frac{1}{(p + q)} \bigg( \frac{q + p}{pq}\bigg) = {p}^{a} \: {q}^{b} }$

$\displaystyle \sf{ \implies \frac{1}{(p + q)} \bigg( \frac{p + q}{pq}\bigg) = {p}^{a} \: {q}^{b} }$

$\displaystyle \sf{ \implies \frac{1}{pq} = {p}^{a} \: {q}^{b} }$

$\displaystyle \sf{ \implies {p}^{ - 1} \: {q}^{ - 1} = {p}^{a} \: {q}^{b} }$

Comparing both sides we get

$\sf{a = - 1 \: \: \: \: and \: \: \: b = - 1}$

$\therefore \: \sf{a + b + 2}$

$= - 1 - 1 + 2$

$= 0$

Hence proved

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