73rd Problem 2016

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 $5$ . $_\square$

$\LARGE let:\quad x=1st\quad number\\ \LARGE \quad \quad \quad x+2=2nd\quad number\\ \\ \LARGE \left( x \right) ^{ 2 }+\left( x+2 \right) ^{ 2 }=74\\ \LARGE { x }^{ 2 }+{ x }^{ 2 }+4x+4=74\\ \LARGE \qquad \quad { 2x }^{ 2 }+4x=70\\ \LARGE { \quad \qquad \quad x }^{ 2 }+2x=35\\ \LARGE { \quad \quad x }^{ 2 }+2x-35=0\\ \LARGE \quad (x-5)(x+7)=0\\ \LARGE \quad \quad x=5\quad |\LARGE \quad x=-7\\ \LARGE x+2=7\quad |\quad x+2=-5\\ \\ \LARGE \boxed { \quad x=5\\ x+2=7 } \\$

$\begin{aligned} (2x+1)^2 + (2x+3)^2 &= 74 \\ 4x^2 + 4x + 1 + 4x^2 + 12x + 9 &= 74 \\8x^2 + 16x - 64 &= 0 \\ 8(x+4)(x-2) &= 0 \\ \implies x&= 2 \end{aligned}$

$2x+1 = \boxed{5}$