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Question
Find two consecutive natural numbers whose squares have the sum 221.
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Solution
Let the number be x, x + 1
Then x2 + ( x + 1)2 = 221
⇒ x2 + x2 + 1 + 2x - 221 = 0
⇒ 2x2 + 2x - 220 = 0
⇒ x2 + x - 110 = 0
⇒ x2 + 11x - 10x - 110 = 0
⇒ x(x + 11) - 10(x + 11) = 0
⇒ (x = 11) (x - 10) = 0
⇒ x = -11 or x = 10
But x = -11 is rejected ...[∵ It cannot been as its is a natural number]
∴ x = 10
Hence, required numbers are 10, 10 + 1.
i.e., 10 and 11.
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