Question

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

Answer 1

$\huge\underline\mathrm{SOLUTION:-}$

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

Therefore, as per the given questions,

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.

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Answer 2

Let one number be 'x' then the

other consecutive integer will be 'x+1'.

According to the question,

${x}^{2} + {(x + 1)}^{2} = 365$

${x}^{2} + {x}^{2} + 1 + 2x = 365$

$2 {x}^{2} + 2x + 1 = 365 \\ 2 {x}^{2} + 2x + 1 - 365 = 0 \\ 2 {x}^{2} + 2x - 364 = 0 \\ 2( {x}^{2} + x - 182) = 0 \\ {x}^{2} + x - 182 = 0 \\ {x}^{2} + 14x - 13x - 182 = 0 \\ x(x + 14) - 13(x + 14) = 0 \\ (x + 14)(x - 13) = 0$

$(x + 14) = 0 \: or \: (x - 13) = 0$

$x = - 14 \: or \: x = 13$

But as it is given that the numbers are positive integers -14 is rejected.

Therefore the consecutive positive integers are 13 and 14.

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