Try this problem.

$n+1 = 2006^2+2007^2$

$\Rightarrow n = 2006^2 + 2007^2 -1$

$\quad \quad = 2006^2 + (2007+1)(2007-1)$

$\quad \quad = 2006^2 + 2008(2006)$

$\quad \quad = 2006(2006+2008)$

$\quad \quad = 2006(4014)$

$\Rightarrow 2n + 1 = 4012(4014) + 1$

$\quad \quad \quad \quad = 4012(4012+2)+1$

$\quad \quad \quad \quad = 4012^2 + 2(4012)+1$

$\quad \quad \quad \quad = (4012+1)^2 = 4013^2$

$\Rightarrow \sqrt{2n + 1} = \boxed{4013}$

I've got a solution for the general case. Let us suppose a =2006, (a+1 = 2007). Thus $n = a^2+(a+1)^2 -1$ Thus $2n+1 = 2( a^2+(a+1)^2 -1) +1 = 4a^2+4a+1 = (2a+1)^2$ By applying the sqrt $\sqrt{(2n+1 )} = 2a+1$ leading us to the solution 4013. Notice that this formula is valid for any real a.

$n+1$ = $2006^{2} + 2007^{2}$ ,

$n+1$ = $2006^{2} + (2006+1)^{2}$ ,

$n+1$ = $2006^{2} + 2006^{2}+1+2.2006.1$ ,

$n$ = $2006^{2} + 2006^{2}+2.2006$

$2n$ = $4.2006^{2}+4.2006$

$2n+1$ = $4.2006^{2}+4.2006+1$

now make perfect square a nd you will get the answer $\boxed{4013}$

: n+1 = 2006^2 + 2007^2

: n+1 = 2006^2 + (2006+1)^2

: n+1 = 2006^2 + 2006^2 + 2 * 2006 + 1

: n+1 = 2 * 2006^2 + 2 * 2006 + 1

: n = 2 * 2006^2 + 2 * 2006

: 2n = 4 * 2006^2 + 4 * 2006

: 2n+1 = 4 * 2006^2 + 4 * 2006 + 1

: 2n+1 = (2 * 2006+1)^2

: sqrt(2n+1) = 2 * 2006+1

Formula using:
x2 + y2 = (x-y)2 + 2 x y x2 – y2 = (x+y)(x-y) Solution : N + 1 = 20072 +20062 N + 1 = (2007-2006)2 + 2 2007 2006 N +1 = 12 + 2 2007 2006 N = 2 2007 2006 So, √2N +1 = √2 2 2007 2006 +1 √2N +1 = √2 2007 2 2006 +1 √2N +1 = √4014*4012 +1 √2N +1 = √(4013+1)(4013-1) +1 √2N +1 = √40132 -12 + 1 √2N +1 = √40132 -1 + 1 √2N +1 =√40132
√2N +1 = 4013

2n + 1 = 2 \times 2006^{2 } + 2 \times 2007^{2} - 1

2n + 1 = 2 \times 2006^{2 } + 2 \times 2007^{2} - (2007 - 2006)^{2}

2n + 1 = 2006^{2 } + 2007^{2} + 2 \ times 2006 \ times 2007

2n + 1 = (2007 + 2006)^{2}

\sqrt{2n +1} = 4013

Let a = 2006 then a² + (a + 1)² = n + 1

2a² + 2a = n

2n + 1 = 2(2a² + 2a) + 1 = 4a² + 4a + 1 = (2a + 1)²

(2n + 1)^1/2 = 2a + 1 = 2006 × 2 + 1 = 4013

Let x=2006

n+1=x^2 + (x+1)^2

n+1=x^2 + x^2 +2x +1

n=2x^2 + 2x

Sqrt (2n +1)=sqrt (4x^2 +4x+1)

=sqrt (2x+1)^2

=2x+1

=2 (2006)+1

=4013

I just rewrote the 1 in LHS as (2007-2006)^2 and expanded. Then, the two terms in the RHS cancel, resulting in n = 2×2006×2007. Later, I calculated the root as sqrt (n+n+1), where I substituted the original equation and and that obtained before. Thus, sqrt (2006^2 + 2007^2 + 2×2006×2007) = sqrt[(2006+2007)^2] = 2006+2007 = 4013. Done!

first, find n. n=(2006 2006) + (2007 2007) -1 n=8,052,084

then sqrt 2(8,052,084) + 1

sqrt 16,104,169

equals 4013

We have, n+1=2006^2 + 2007^2 & by solving we get, n=8052084.. Now put the value of n ... We get 4013 which is the required answer.

Let x = 2006.

2006^2 + 2007^2 = x^2 + (x + 1)^2

n + 1 = x^2 + x^2 + 2x + 1

n = x^2 + x^2 + 2x = 2x^2 + 2x

2n + 1 = 4x^2 + 4x + 1 = (2x + 1)^2

Sqrt(2n + 1) = 2x + 1 = 2×2006 + 1 = 4013