The sum of squares of 2 consecutive odd natural numbers is 74. What is the smaller odd number of the two?

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The answer is 5.

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We can express odd numbers of the form $(2n \pm x)$ where $x$ is any odd number.

Here I take $x = 1$

So, the odd numbers obtained are $(2n-1)$ and $(2n+1)$

According to the given conditions:-

${(2n-1)}^2 + {(2n+1)}^2 = 74$

$4n^2 - 4n + 1 + 4n^2 + 4n + 1 = 74$

$8n^2 + 2 = 74$

$8n^2 = 74 -2$

$8n^2 = 72$

$n^2 = \dfrac {72}{8}$

$n^2 = 9$

$n = \sqrt{9}$

$n = \pm 3$

Now, as we need only natural numbers, $n$ is not equal to $-3$

$\therefore n = 3$

So, the numbers obtained are $[2(3) - 1]$ and $[2(3) + 1]$

= $5$ and $7$ .

So, the smaller number is $5$ . $_\square$