Question

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

Answer 1

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Let us say, the two consecutive positive integers be x and x + 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 or x – 13 = 0,

⇒ x = – 14 or x = 13

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

∴ x + 1 = 13 + 1 = 14

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

Answer 2

a^2 + (a+1)^2 = 365
=> a^2 + a^2 + 2a + 1 = 365
=> 2a^2 + 2a -364 = 0
=> a^2 + a - 182 = 0
Now 182 = 2 x 91 = 2 x 7 x 13 = 13 x 14 and 14-13=1
=> a^2 + 14a - 13a - 182 = 0
=> a(a+14) -13(a + 14) = 0
=> (a-13)(a+14) = 0
Integer is positive, so a is not equal to -14
=> a =13
=> a+1 = 14