Find two consecutive positive integers, sum of whose squares is 365. [ncert 10th.]

Let the two consecutive positive integers

are  x and x+1.

Given that x²+(x+1)² = 365

Using (a+b)=a²+b²+2ab, we get,

=> x²+x²+1+2x=365.  

 => 2x²+2x+1-365=0

=> 2x²+2x-364=0

Dividing both sides by 2 (non zero), we get,

=> x²+x-182=0

=> x²+14x-13x-182=0

=> (x²+14x)-(13x+182)=0

=> x(x+14)-13(x+14)=0

=> (x+14)(x-13)=0

=> Either x+14=0 or x-13=0

=> Either x= -14 , 13

But we have to find positive integer, so -14 must be rejected.

So x= 13 is the required positive integer.

Now two consecutive positive integers shall be x, x+1 

=> 13 and 13+1=14

So 13 and. 14 are the required positive consecutive integers. Ans.