Question

the sum of squares of two consecutive natural numbers is 41 find the numbers

Answer 1

Let 'x' be one number and 'x+1' be the other number.
Then,
     $x^{2} + (x+1)^{2} = 41$
$x^{2} + x^{2} +2x+1 = 41$
       $2 x^{2} +2x-40 = 0$
           $x^{2} +x-40 = 0$

As you can see, this is not a perfect square equation. So, the solutions will not be whole numbers. 

Please recheck the question.

Answer 2

I hope this answer help for you