Question

Find all positive integers a, b for which the number (√2+√a)/(√3+√b) is a rational number

Answer 1

@ganeshie8 @ikram002p

Answer 2

Let, m = (√a + √2)/(√b + √3) - (1)

=> m = (√a + √2)(√b - √3)/(b - 3); if b - 3 is non zero

=> m = {√(ab) - √(3a) + √(2b) - √(6)}/(b-3) - (2)

Now, for m to be a rational number:

(i) √6 must cancel out.
(ii) No square roots must be present.

Lets first see the (i) point:

For √6 to cancel out, one of √2b and √(ab) must be equal to √6

Thus, √(2b) = √6 or √(ab) = √6

Case - 1:

√(2b) = √6

=> 2b = 6

=> b = 3 - (3)

Substituing the value of b in equation (1).

m = (√a + √2)/(2√3)

=> m = (√(3a) + √6)/6 - (4)

Now, m can be a rational number only if √(3a) + √(6) is a rational number, which is not possible for any value of a.

Thus, if b = 3, then there is no solution to equation (1).

Case - 2:

√(ab) = √(6)

=> ab = 6

Since a and b are integers and must be non negative. Thus, possible solutions to above equation are:

(a,b) = (1,6); (2,3); (3,2); (6,1)

Now, we check each solution by substituting the values of a and b in equation (1).

Only (3,2) and (6,1) satisfy the equation.


It was what I attempted.

I realise that there is a mistake in this thing:

"For √6 to cancel out, one of √2b and √(ab) must be equal to √6"


Any idea about how to solve it ?

Answer 3

@hartnn

Answer 4

@Kainui

Answer 5

\[ \sqrt a + \sqrt2 \over \sqrt b + \sqrt 3\]

rationalize the denominatior\[={\left( \sqrt a+ \sqrt2 \right)\left( \sqrt b-\sqrt 3\right) \over b-3}\]

Answer 6

Ok

Answer 7

*

Answer 8

now looking only at numertor:\[=\sqrt{ab}+\sqrt{2b}-\sqrt{3a}-\sqrt6\]

Answer 9

Okay.

Answer 10

a=3, b=2 works

Answer 11

b can't be 3

Answer 12

Yes

Answer 13

I got that. But, what I can't get is that whether (3,2) and (6,1) are the only possible solutions or are there any more ?

Answer 14

=the only other solution could be \[\sqrt{2b}-\sqrt{3a}=\sqrt6\] and ab is perfect square...
both are not possible at the same time. so you have all the answers

Answer 15

Okay.

Thanks a lot @PaxPolaris :)

Answer 16

:O nice question :)