689

$\sqrt { x\left( x+1 \right) \left( x+2 \right) \left( x+3 \right) +1 } =\sqrt { \left( { x }^{ 2 }+3x \right) \left( { x }^{ 2 }+3x+2 \right) +1 } =\sqrt { \left( { x }^{ 2 }+3x \right) \left( { x }^{ 2 }+3x+1 \right) +{ x }^{ 2 }+3x+1 } =$ $=\sqrt { \left( { x }^{ 2 }+3x+1 \right) \left( { x }^{ 2 }+3x+1 \right) } =\sqrt { { \left( { x }^{ 2 }+3x+1 \right) }^{ 2 } } =\left| { x }^{ 2 }+3x+1 \right| \overset { x=689 }{ = } \left| { 689 }^{ 2 }+3\cdot 689+1 \right| =\left| 474721+2067+1 \right| =\boxed { 476789 }$ $$

$\begin{aligned} N & = (689)(690)(691)(692) + 1 & \small \color{#3D99F6} \text{Let }a = 690.5 \\ & = \left(a-\frac 32 \right) \left(a-\frac 12 \right) \left(a+\frac 12 \right) \left(a+\frac 32 \right) + 1 \\ & = \left(a^2 -\frac 94 \right) \left(a^2 -\frac 14 \right) + 1 \\ & = a^4 -\frac 52 a^2 +\frac {25}{16} \\ & = \left(a^2 - \frac 54\right)^2 \\ & = 476789^2 \end{aligned}$

Therefore, $\sqrt N = \boxed{476789}$ .

$(m-1)(m)(m+1)(m+2)+1=a^2$

Take 1 to RHS and combine (1,4),(2,3) terms

$(m^2+m)(m^2+m-2)=(a+1)(a-1)$

Difference between the terms in LHS and RHS both is $2$ . So we can conclude that :

$(m^2+m=a+1),(m^2+m-2=a-1)$

In question, $m=690$ and $a$ is to be found

So, $(690)^2+690=a+1$

$476790=a+1$ Or,

$\boxed{a=476789}$

$(a-2)(a-1)a(a+1)+1=(a^2-2a)(a^2-1)+1=a^4-2a^3-a^2+2a+1=(a^2-a-1)^2$ . For $a=691$ , the value of the square root of this expression is $(691^2)-691-1=476789$ .