Show that the sum of the squares of 3,4,5 and 6 consecutive numbers can not be a square

Question

$\textbf{Edit:}$ 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!

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\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!

Answer 1

Of six consecutive numbers, three are even and three are odd. The square of the even numbers are congruent to $0$ mod $4$; the squares of the odds are congruent to $1$ mod $4$. Consequently the sum of six consecutive squares is congruent to $3$ mod $4$, which cannot be a square.

Answer 2

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.

Answer 3

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.