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

Dear Student

Let us say, the two consecutive positive integers bexandx+ 1.

Therefore, as per the given questions,

x^2+ (x+ 1)^2= 365

⇒x^2+x^2+ 1 + 2x= 365

⇒2x^2+ 2x - 364 = 0

⇒x^2+x- 182 = 0

⇒x^2+ 14x- 13x- 182 = 0

⇒x(x+ 14) -13(x+ 14) = 0

⇒(x+ 14)(x- 13) = 0

Thus, either,x+ 14 = 0 orx- 13 = 0,

⇒x= - 14 orx= 13
since, the integers are positive, soxcan be 13, only.

x+ 1 = 13 + 1 = 14
two consecutive positive integers will be 13 and 14.

Thanks