Eigenvalues of symmetric $\mathbb{R}^{p\times p}$ matrix I want to prove that 
$$A^{(p)}                                                                      
=                                                                            
\begin{pmatrix}                                                              
a & 1 & 1 & \dots & 1 & 1 & 1 \\                                           
1 & a & 1 & \dots & 1 & 1 & 1 \\                                           
1 & 1 & a & \dots & 1 & 1 & 1 \\                                           
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\             
1 & 1 & 1 & \dots & 1 & a & 1 \\                                           
1 & 1 & 1 & \dots & 1 & 1 & a                                              
\end{pmatrix}       
\in\mathbb{R}^{p\times p}$$
has eigenvalues $\lambda_+=a+p-1$ and $\lambda_-=a-1$ with degeneracy $p-1$ by using mathematical induction. Evidence that these eigenvalues could be correct were found 'empiricaly' by Mathematica.

Induction Base Case $(p=2)$
$$\det(A^{(2)}-\lambda E_2)                                                    
=                                                                            
\det\begin{pmatrix}a-\lambda & 1\\1 & a-\lambda\end{pmatrix}                 
=(\lambda-(a-1))(\lambda-(a+1))      $$
Induction Hypothesis $(p\in\mathbb{N})$
$$\det(A^{(p)}-\lambda E_p)                                                    
=(\lambda-(a-1))^{p-1}(\lambda-(a+p-1))$$
Induction Step $(p\to p+1)$
$$\det(A^{(p+1)}-\lambda E_{p+1})                                              
=                                                                           
\det\begin{pmatrix}                                                          
a-\lambda & 1 & 1 & \dots & 1 & 1 & 1 \\                                   
1 & a-\lambda & 1 & \dots & 1 & 1 & 1 \\                                   
1 & 1 & a-\lambda & \dots & 1 & 1 & 1 \\                                   
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\             
1 & 1 & 1 & \dots & 1 & a-\lambda & 1 \\                                   
1 & 1 & 1 & \dots & 1 & 1 & a-\lambda                                      
\end{pmatrix}$$
We can subtract the last row from the first row
$$\det\begin{pmatrix}                                                          
a-\lambda-1 & 0 & 0 & \dots & 0 & 0 & 1-a+\lambda \\                       
1 & a-\lambda & 1 & \dots & 1 & 1 & 1 \\                                   
1 & 1 & a-\lambda & \dots & 1 & 1 & 1 \\                                   
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\             
1 & 1 & 1 & \dots & 1 & a-\lambda & 1 \\                                   
1 & 1 & 1 & \dots & 1 & 1 & a-\lambda                                      
\end{pmatrix}$$
and apply Laplace's formula to the first row. The first entry (upper left corner) gives
$$(a-\lambda-1)                                                               
\det\begin{pmatrix}                                                          
a-\lambda & 1 & \dots & 1 & 1 & 1 \\                                       
1 & a-\lambda & \dots & 1 & 1 & 1 \\                                       
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\                      
1 & 1 & \dots & 1 & a-\lambda & 1 \\                                       
1 & 1 & \dots & 1 & 1 & a-\lambda                                          
\end{pmatrix}$$
which is just the case of our induction hypothesis, thus
$$-(\lambda-(a-1))^p(\lambda-(a+p-1))  $$
is the first contribution to Laplace's formula. The second contribution reads
$$(1-a+\lambda)                                                               (-1)^{p+2}                                                                   
\det\begin{pmatrix}                                                          
1 & a-\lambda & 1 & \dots & 1 & 1 \\                                       
1 & 1 & a-\lambda & \dots & 1 & 1 \\                                       
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\                      
1 & 1 & 1 & \dots & 1 & a-\lambda\\                                        
1 & 1 & 1 & \dots & 1 & 1                                                  
\end{pmatrix}$$
and we can subtract the first column from all other columns yielding
$$\det\begin{pmatrix}                                                          
0 & a-\lambda-1 & 0 & \dots & 0 & 0 \\                                     
0 & 0 & a-\lambda-1 & \dots & 0 & 0 \\                                     
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\                      
0 & 0 & 0 & \dots & 0 & a-\lambda-1\\                                      
1 & 1 & 1 & \dots & 1 & 1                                                  
\end{pmatrix}$$
on which we can reapply Laplace's formula and use that the determinant of a diagonal matrix equals the product of the diagonal entries, hence
$$(-1)^{p+1}(\lambda-(a-1))^{p+1}  $$
is the second contribution to the first application of Laplace's formula. Now combining both I find
$$\det(A^{(p+1)}-\lambda E_{p+1})                                              
=                                                                           
(-1)(\lambda-(a-1))^p(\lambda-(a+p-1))+(-1)^{p+1}(\lambda-(a-1))^{p+1}
=(-1)(\lambda-(a-1))^p(\lambda-a-p+1+(-1)^p(\lambda-a+1))   $$
which seems faulty: on the one hand side the factor $(-1)^{p}$ should not exist as for odd $p$ the characteristic polynomial should varnish which has been shown to be not the case for i.e. $p=3$ (see Mathematica). On the other hand even if that factor would be wrong the final expression would yield
$$(-2)(\lambda-(a-1))^p(\lambda-(a-p/2+1))$$
which would contradict the induction hypothesis.
Did I made a mistake in evaluating the determinant or is the induction hypothesis wrong?

This is an answer not by induction, in response to a comment/answer by OP. 
$$
\newcommand{\bu}{ {\mathbf u}}
\newcommand{\bv}{ {\mathbf v}}
\newcommand{\bM}{ {\mathbf M}}
\newcommand{\bI}{ {\mathbf I}}
\newcommand{\bU}{ {\mathbf U}}
\newcommand{\bQ}{ {\mathbf Q}}
$$
Your matrix $\bM$ is 
$$
\bM = \bU + (a-1)\bI
$$
where $\bU$ is a matrix of all $1$s. It's clear that 
$$
(n-1) + a
$$
is an eigenvalue, for the vector $\bu'$ consisting of all $1$s is a corresponding eigenvector. We can normalize this to a vector $\bu$ consisting of all $\frac{1}{\sqrt{n}}$ entries.   Furthermore, we can write $\bU$ as $n \bu \bu^t$, which shows that $U$ has rank $1$, which is no surprise, because all its columns are the same. 
Now let $\bv_1, \ldots, \bv_{n-1}$ be an orthonormal basis for the space 
$$
H = \{ \bv \mid \bu \cdot \bv = \bu^t \bv = 0 \}
$$
of vectors orthogonal to $\bu$. Let's compute $\bM\bv_i$ for any one of these. It's
\begin{align}
\bM\bv_i 
&= ( \bU + (a-1)\bI ) \bv_i \\
&= ( n\bu \bu^t + (a-1)\bI ) \bv_i \\
&=  (n\bu \bu^t) \bv_i  + (a-1)\bI \bv_i \\
&=  n\bu (\bu^t \bv_i)  + (a-1)\bv_i \\
&=  n\bu (0)  + (a-1)\bv_i \\
&=  (a-1)\bv_i \\
\end{align}
In other words, each of the vectors $\bv_i$ is an eigenvector of eigenvalue $a-1$. 
So the vectors $\{\bu, \bv_1, \bv_2, \ldots, \bv_{n-1}\}$ constitute an orthonormal set of eigenvectors of $\bM$, and if we form a matrix $\bQ$ with those as its columns, we get
$$
\bQ^t \bM \bQ = \pmatrix{(n-1)+ a & & & & & \\
  & a-1 & & & & \\
  &  & a-1 & & & \\  
&  &  & \ldots& & \\
&  &  & & a-1 & \\
&  &  &  & & a-1}
$$
which is pretty much a complete eigenanalysis of $\bM$.
