Question

If the diff of the squares of two natural number is 19 find sum of squares of these numbers

Answer 1

Let x and y be two natural numbers.

Then according to the question

either x^2-y^2= 19 or y^2-x^2= 19.

it's possible only when either x= 10 and y= 9 or x=9 and y=10

Therefore, x^2+y^2= 100+81=181.

Answer 2

The sum of squares of these numbers is 181.

Explanation:

Let a and b and a> b are the two natural numbers such that

$a^2-b^2=19$

$\Rightarrow\ (a+b)(a-b) =19$   [∵ a²-b²=(a+b)(a-b)]

We can write 19 as 1 x 19 (∵ 19 is a prime number)

Then, $(a+b)(a-b) =1\times 19$

Since a and b are natural numbers and a> b , then

$a+b=19-----(1)\\ a-b=1-----(2)$

Add (1) and (2) , we get

$2a=20\Rightarrow\ a=10$

Subtract (2) from (1) , we gte

$2b=18\Rightarrow\ b=9$

So a= 10 and b= 9

Then, $a^2+b^2= 10^2+9^2=100+81=181$

Hence, the sum of squares of these numbers is 181.

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