The sum of the squares of two consecutive natural numbers is 421. The numbers are

12 and 13

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13 and 14

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14 and 15

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15 and 16

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Solution

The correct option is C
14 and 15

Let the consecutive natural no's be x, x+1
$x^{2} + (x + 1)^{2} = 421 x^{2} + x^{2} + 2 x + 1 = 421 2 x^{2} + 2 x - 420 = 0 x^{2} + x - 210 = 0 x^{2} + 15 x - 14 x - 210 = 0 x (x + 15) - 14 (x + 15) = 0 \Rightarrow x = - 15 a n d x = 14 S i n c e x i s a n a t u r a l n u m b e r x = - 15 i s n o t p o s s i b l e ∴ x = 14 & x + 1 = 15 N u m b e r s a r e 14 a n d 15$

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