Show that the sum of the squares of 3,4,5 and 6 consecutive numbers can not be a square $\textbf{Edit:}$
Thank you so far for the answers. I still do not understand how to prove that the sum of 6 consecutive squares is not a square.
I've tried ${6d^2+30d+55 \ne n^2 \Rightarrow 6(d^2 + 5d +9)+1 \ne n^2}$ and would somehow like to elaborate this, but (yet again) I get stuck.
Is is correct to say that ${6(d^2 + 5d +9)+1}$ is on the form
${6k+1}$? Because then I could possibly stop there, having
${6k+1 \ne n^2}$
Thanks!
${\rule{8cm}{0.4pt}}$
\begin{equation}
\end{equation}
I've gotten this far:
$\textbf{Sum of 3 consecutive squares}$
Rewrite of problem:
\begin{equation}
              \begin{aligned}
                  (d-1)^2+d^2+(d+1)^2 &\ne n^2\\
                  3d^2+2 &\ne n^2 \\
                  \label{p1_1}
              \end{aligned}
          \end{equation}
Assume ${n=d\cdot3+r}$, where ${r=0,1}$ or $2$.
Then,
\begin{equation}
n^2 \equiv_3 r^2 \equiv_3 0 \vee 1
\end{equation}
Conclusion:
Since ${r=2}$ in ${3d^2+2}$ and ${r=0 \vee 1}$ in ${n^2 \equiv_3 r^2}$ we have shown that
${3d^2+2 \ne n^2}$.
Question: Is my conclusion and reasoning valid and "good enough"?
$\textbf{Sum of 4 consecutive squares}$
Rewrite of problem:
\begin{equation}
              \begin{aligned}
                  d^2+(d+1)^2 + (d+2)^2 + (d+3)^2 &\ne n^2\\
                  4(d^2+3d+3)+2&\ne n^2\\
                  \label{p1_b}
              \end{aligned}
          \end{equation}
${\Rightarrow} {r=2}$.
Assume ${n=d\cdot4+r}$, where ${r=0,1,2}$ or $3$.
Then,
\begin{equation}
n^2 \equiv_4 r^2 \equiv_4 0 \vee 1
\end{equation}
Conclusion:
Same reasoning as in conclusion of sum of 3 consecutive squares.
$\textbf{Sum of 5 consecutive squares}$
\begin{equation}
               \begin{aligned}
                   (d-2)^2+(d-1)^2+d^2+(d+1)^2+(d+2)^2 &\ne n^2\\
                   5d^2+10 &\ne n^2\\
                   5(d^2+2) &\ne n^2\\
                   \label{p1_c}
               \end{aligned}
           \end{equation}
${\Rightarrow r=2 \vee 10}$ (is this correct?)
${n^2 \equiv_5 r^2 \equiv_5 0 \vee \pm 1}$
Conclusion:
Same reasoning as in conclusion of sum of 3 consecutive squares.
$\textbf{Sum of 6 consecutive squares}$
\begin{equation}
              \begin{aligned}
                  d^2+(d+1)^2+(d+2)^2+(d+3)^2+(d+4)^2+(d+5)^2 &\ne n^2\\
                  6d^2+30d+55&\ne n^2\\
                  \label{p1_d}
              \end{aligned}
          \end{equation}
Here I'm stuck.
I feel pretty unsure about all my reasonings and would therefore appreciate your help. Please, feel free to be verbose in your answers.
Thank you!

Let us write the sum of six consecutive squares as
$$\sum_{i=-2}^3(d+i)^2=6d^2+6d+19=n^2.$$
Then $n$ is odd, say $n=2m+1$, and we have
$3(d^2+d+3)=2m^2+2m.$
But this equates an odd number to an even number, and so no such $n$ can exist.
Your reasoning is correct for the other cases.
