Question

Find two consecutive positive integers, sum of whose squares is 365.m. ​

Answer 1

Let us say, the two consecutive positive integers be x and x + 1.

Therefore, as per the given statement,

x2 + (x + 1)2 = 365

⇒ x2 + x2 + 1 + 2x = 365

⇒ 2x2 + 2x – 364 = 0

⇒ x2 + x – 182 = 0

⇒ x2 + 14x – 13x – 182 = 0

⇒ x(x + 14) -13(x + 14) = 0

⇒ (x + 14)(x – 13) = 0

Thus, either, x + 14 = 0 or x – 13 = 0,

⇒ x = – 14 or x = 13

since, the integers are positive, so x can be 13, only.

So, x + 1 = 13 + 1 = 14

Therefore, the two consecutive positive integers will be 13 and 14.

Answer 2

${\huge{\blue{\underline{\underline{Answer:}}}}}$

Let the consecutive positive integer be

x and ( x + 1 )

Sum of the square of the number is = 365

x² + ( x + 1 )2 = 365

x² + x² + 2x + 1 = 365

2x² + 2x + 1 = 365

2x² + 2x + 1 - 365 = 0

2x² + 2x - 364 = 0

Divide LHS and RHS by 2

2x² / 2 + 2x / 2 - 364 / 2 = 0 / 2

x² + x - 182 = 0

x² + 14x - 13x - 182 = 0

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

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

if x + 14 = 0 and x - 13 = 0

So

x = -14 ( minus neglected )

x = 13

So : x = 13

other number = x + 1 = 13 + 1 = 14

$\mathbb\blue{So...x = 13, 14}$