Problem with simplification of Legendre symbol Could anyone explain why this is true:
If $p$ is an odd prime and $ a' $ is defined by $aa'\equiv 1 \, \text{(mod p)}$ then $(a(a+1)/p)=((1+a')/p) $ 
Doesn't this by the properties of Legendre symbol mean that $a(a+1) \equiv 1+a' \, \text{(mod p)}$ ? I cant see how this is true. I guess it follows from the fact that $aa'\equiv 1 \, \text{(mod p)}$, but cant really make the connection.
 A: We have $aa'=1+jp$, now assume that $(a(a+1)/p)=1$ then there is $x$ such that 
\begin{eqnarray*}
a(a+1)=x^2+kp \\
\end{eqnarray*}
Multiply by $a'^2$
\begin{eqnarray*}
aa'(aa'+a')=(xa')^2+kpa'^2 \\
(1+jp)(a'+1+jp)= a'+1+jp(a'+2)+j^2p^2=(xa')^2+kpa'^2 \\
a'+1=(xa')^2+p(ka'^2-j(a'+2)-j^2p) \\
\end{eqnarray*}
So $((1+a')/p)=1$. There are the two other cases to consider.
A: 
Doesn't this by the properties of Legendre symbol mean that $a(a+1) \equiv 1+a' \, \text{(mod p)}$

No, it does not. The property of the Legendre symbol which you're referring to only works in one direction, i.e. $a \equiv b \mod p \implies (a/p) = (b/p)$. If it worked in the opposite direction it would mean all quadratic residues are $\equiv$ mod p, e.g. $1 \equiv 4 \mod 7$, clearly not true.
With this misconception out of the way, consider the proof of $$aa'\equiv 1 \, \text{(mod p)} \implies (a(a+1)/p)=((1+a')/p) $$
First we prove a Lemma that $(a/p) = (a'/p)$.
Case 1: $(a / p) = 1 \implies r^2 \equiv a \mod p$ for some $r$. $r$ has an inverse mod p (since p is prime), call it $r'$. Multiplying both sides by $r'$
$$ r^2 r'^2 \equiv a r'^2 \mod p $$
$$\implies 1 \equiv a r'^2 \mod p $$
but since a has a unique inverse mod p, that means
$$ r'^2 \equiv a' \mod p $$
and so $(a'/p) = 1$
Case 2: $(a/p) = -1$ Suppose BWOC that $(a'/p) \neq -1$ then $(a'/p) = 1$ which means from the previous case that $(a/p) = 1$, a contradiction. So $(a/p) = -1$ 
P.S. There is no case $(a/p) = 0$ because then $a = 0$ which has no multiplicative inverse.
We know that $(a/p) = (a'/p)$, so
$$(a(a+1)/p)=(a/p)((1+a)/p) $$$$= (a'/p)((1+a)/p) $$$$= (a'(1+a)/p) $$$$ = ((a'+1)/p)$$
