<s>[INST] 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!
 [/INST] 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.
</s>