Are there numbers $N$ such that $N^2=n_1^2+n_2^2+\cdots+n_k^2+1$ In the past few hours I spent some time looking for numbers of the form
$$
N^2=n_1^2+n_2^2+\cdots+n_k^2+1
$$
where $n_i \gt 1 \wedge n_i \neq n_j \wedge k\ge 2$ , and I found none. I searched up to $100$. I wonder if this kind of numbers exist at all...
So I came here for an answer. Can you provide some instances or rules to generate them, or on the contrary they simply can't exist for some reason.
Thanks.
Note that possible numbers are $N^2=n_1^2+n_2^2+1$ or $N^2=n_1^2+n_2^2+n_3^2+1$ etc... we can have any number of squares on the right side...

$n^2=n1^2+n2^2+n3^2\tag{1}$
Substitute $n1=x, n2=b, n3=c, n=x+k$ to equation $(1).$
We get  $$x = 1/2\,{\frac {-{k}^{2}+{b}^{2}+{c}^{2}}{k}}$$
Thus, we get a parametric solution.
$(n,n1,n2,n3) = (k^2+b^2+c^2, -k^2+b^2+c^2, 2bk, 2ck)$ 
Choose (b,c,k) so that $(n,n1,n2)$ is divisible by $n3$, then
we get the solution of $n^2=n1^2+n2^2+1$.  
                         [n, n1, n2, n3] 
                         [3, 2, 2, 1]
                         [9, 8, 4, 1]
                         [19, 18, 6, 1]
                         [33, 32, 8, 1]
                         [51, 50, 10, 1]
                         [73, 72, 12, 1]
                         [99, 98, 14, 1]
                         [129, 128, 16, 1]
                         [163, 162, 18, 1]

After simple algebra, we get a parametric solution below. 
$(n,n1,n2,n3)=(1+2p^2, 2p^2, 2p, 1)$
$p$ is arbitrary.  
$n^2=n1^2+n2^2+n3^2+n4^2\tag{2}$
Substitute $n1=x, n2=b, n3=c, n4=d, n=x+k$ to equation $(2).$
Thus, we get a parametric solution.  
$(n,n1,n2,n3,n4) = (k^2+b^2+c^2+d^2, -k^2+b^2+c^2+d^2, 2bk, 2ck, 2dk)$ 
Choose (b,c,d,k) so that $(n,n1,n2,n3)$ is divisible by $n4$, then
we get the solution of $n^2=n1^2+n2^2+n3^2+1$.  
$(n,n1,n2,n3,n4)=(1+2q^2+2p^2, 2q^2+2p^2, 2q, 2p, 1)$
$p,q$ are arbitrary.  
