How many values of $k$ satisfy $\left (\frac{k}{p}\right )=\left (\frac{k+1}{p}\right)=1$ where p is a odd prime? The values of $k$ must be between $1$ and $p-1$ this means :
$$k\in\left\{1,2,\cdots,p-1\right\}$$                   

The question: Given an odd prime $p$ What is the number of elements $k\in\left\{1,2,\cdots,p-1\right\}$ such that
   $$\left (\frac{k}{p}\right )=\left (\frac{k+1}{p}\right)=1$$

I used trial and error to find values of $k$ with various prime numbers and I couldn't find a pattern. 
$p = 3: \left(\frac{a}{3}\right)=1$ when
$a = 1 $
Thus, no values of $k$ satisfies the equation. 
$p = 5: \left(\frac{a}{5}\right) = 1$ when
$a=1, 4$
Thus, no values of k satisfies the equation. 
$p = 7: (a/7) = 1$ when
$a = 1, 2 ,4$
Since (a/7) = 1 when a = 1 and 2, then k = 1 satisfies the equation. 
I kept of doing this for each prime until 17 but I can't find a pattern of finding the number of values of k that will satisfy  $\left (\frac{k}{p}\right )=\left (\frac{k+1}{p}\right)=1$ for each odd prime. 
 A: This was a very hard and amazing question I have ever solved (If I made any mistake please mention it)
Let $p$ be an odd prime the most important is $n$ and $n+1$ have the same Jacobi symbol if and only if $\frac{n+1}{n}=1+n^{-1}$ is quadratic residues $\mod p$, so we have 
$$|A|=\left |\left\{n\Big / \left(\frac{n}{p}\right)=\left(\frac{n+1}{p}\right) 1\leq n\leq p-2 \right\}\right|=\frac{p-3}{2} $$
( count the number of squares of the form $1+a$ with $a$ is invertible), now let $f(x)=\frac{(x-1)^2}{4x}\mod p$ it's clear that the range of $f$ is exactly the set $A$ why?, because $f(x)=f(y)$ if and only $x=y^{-1}$ or $x=y$ so $|Range(f)|=\frac{p-3}{2}$ and it's clear also that $Range(f)\subset A$ because $f(x)$ is a quadratic residue if and only if $f(x)+1$ is quadratic residue (or just compute $1+f(x)^{-1}=\left(\frac{x+1}{x-1}\right)^2$)
Now we have to determine the number of elements $f(x)$ which are quadratic residues,let $g$ be a primitive root of $\mathbb{Z_p}$, first (to get rid of the inverse) we have:
$$A=\{f(g),\cdots,f(g^{\frac{p-3}{2}})\}$$
and now observe that $f(g^{k})$ is a quadratic residue if and only if $g^k$ is a quadratic residue, and this if and only if $k$ is even, finally:
$$\left |\left\{n\Big/ \left(\frac{n}{p}\right)=\left(\frac{n+1}{p}\right)=1\ \ \  , 1\leq n\leq p-2 \right\}\right|=\left\lfloor \frac{p-3}{4}\right\rfloor $$
This was very amazing and we can also conclude that:
$$\left |\left\{n\Big/ \left(\frac{n}{p}\right)=\left(\frac{n+1}{p}\right)=-1\ \ \  , 1\leq n\leq p-2 \right\}\right|=\left\lfloor \frac{p-1}{4}\right\rfloor $$
