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

☯ Let the two consecutive numbers be x and x + 1.

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$\qquad\boxed{\bf{\mid{\overline{\underline{\purple{\bigstar\: According\: to \: the \: Question :}}}}}\mid}$

$:\implies\sf x^2 + (x + 1)^2 = 365\qquad\qquad\bigg\lgroup\bf Given \bigg\rgroup$

$:\implies\sf x^2 + x^2 + 2x + 1 = 365$

$:\implies\sf 2x^2 + 2x + 1 - 365 = 0$

$:\implies\sf 2x^2 + 2x - 364 = 0$

$:\implies\sf 2(x^2 + x - 182) = 0$

$:\implies\sf x^2 + x - 182 = 0$

$:\implies\sf x^2 + 14x - 13x - 182 = 0$

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

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

$:\implies\sf x - 13 = 0\;or\;x + 14 = 0$

$:\implies\bf x = 13\;or\;x = - 14$

⠀━━━━━━━━━━━━━━━━━━━━━━━━━

☯ Since the consecutive numbers are positive.

Therefore,

• x = 13

• x + 1 = 14

$\therefore$ The two consecutive positive integers are 13 and 14.