Question

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

Answer 1

Answer:

Hey there

Let two consecutive positive integer be n and n+1,

then,

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

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Answer 2

Answer:

$13 \: and \: 14$

Step-by-step explanation:

It is the correct answer.

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