<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] here's  an expansion of my comment:
$$0+a\equiv a\pmod r$$ means that since $$a=b^2\equiv 0,1\pmod 4$$ we can ignore the $a\equiv 0\pmod 4$ because they don't change the remainders modulo 4.
By similar logic, since the square they sum to $n^2$ has the same remainder possibilities modulo 4,  we get that the number of $a\equiv 1\pmod 4$ that exist, must be $$c\equiv 0,1\pmod 4$$   and since the number of $a\equiv 1\pmod 4$, is also $$\lfloor \frac{k}{2}\rfloor$$ or  $$\lceil\frac{k}{2}\rceil$$ we know one of these must be congruent to $c$ .
If we take $k=4,5,6$ we have $2$ for the first form for the first 2 of those, and $3$ for the last, or we get $2,3,3$ and all these values aren't congruent to $c$, so they don't work in $k$. In fact adding 8 to $k$ results in the same remainder for the fractions so all $k\equiv 4,5,6\pmod 8$ are out. The only case spared is $k=3$ but with 1 odd value which was previously ruled out.  So we are Done.
</s>