Question

find two consecutive positive natural numbers sum of whose squares is
$365$

Answer 1

Let the numbers be x, x+1
${x}^{2} + {(x + 1)}^{2} = 365 \\ = > {x}^{2} + x {}^{2} + 1 + 2x = 365 \\ = > 2 {x}^{2} + 2x = 364 \\ = > 2 {x}^{2} = 364 - 2x \\ = > x = \sqrt{364 - 2x} \: this \: is \: the \: first \: number \\ the \: second \: number \: is \: = \sqrt{364 - 2x} + 1$

Answer 2

$\textbf {Hey there, here is the solution.}$

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STEP 1: Define x:

Let one number be x

The other number is x + 1

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STEP 2: Form the equation:

The sum is their square 365

x² + (x + 1)² = 365

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STEP 3: Solve x:

x² + (x + 1)² = 365

x² + x² + 2x + 1 = 365

2x² + 2x - 364 = 0

x² + x - 182 = 0

(x - 13) (x + 14) = 0

x = 13 or x = -14 (rejected, because it is negative)

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STEP 4: Find the numbers:

one number = x = 13

the other number = x + 1 = 13 + 1 = 14

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Answer: The numbers are 13 and 14

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$\textbf {Cheers}$