Question

let a,b be natural numbers such that 2a-b , a-2b, a+b are all distinct squares. what is the smallest possible value of b

Answer 1

Least valueof b is 21

Answer 2

Answer:

Least value of b=21

Step-by-step explanation:

Let $2a-b=n^{2}$,                                                          (1)

$a-2b=p^{2}$,                                                               (2)

$a+b=k^{2}$,                                                                 (3)

Adding (2) and (3), we get

$2a-b=p^{2}+k^{2}$

From equation (1), we have  $2a-b=n^{2}$

therefore, $p^{2}+k^{2}=n^{2}$ (p<k as a+b<a-2b)

For b to be the smallest,$k^{2}$  and $p^{2}$ are also small and

must be multiple of 3 (as $3b=k^{2}-p^{2}$)

For smallest possible value of b, the least value of k and p be 12 and

9 respectively.

Therefore, the smallest possible value of b is 21.