Mathematical Reflections 2016, Issue 1 - Problem O361

Problem:
Determine the smallest natural number $n>2$ such that there exist $n$ consecutive integers whose sum of their squares is a perfect square.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Observe that $$(-4)^2+(-3)^2+(-2)^2+(-1)^2+0^2+1^2+2^2+3^2+4^2+5^2+6^2=121=11^2,$$ so $n=11$ works. We prove that this is the smallest natural number greater than $2$ that satisfies the given condition.

(i) $n=3$. Let $m \in \mathbb{Z}$. We have $$(m-1)^2+m^2+(m+1)^2=3m^2+2 \equiv 2 \pmod{3},$$ so the sum of the squares of $3$ consecutive integers cannot be a perfect square.

(ii) $n=4$. Let $m \in \mathbb{Z}$. We have $$(m-1)^2+m^2+(m+1)^2+(m+2)^2=4m^2+4m+6 \equiv 2 \pmod{4},$$ so the sum of the squares of $4$ consecutive integers cannot be a perfect square.

(iii) $n=5$. Let $m \in \mathbb{Z}$. We have $$(m-2)^2+(m-1)^2+m^2+(m+1)^2+(m+2)^2=5(m^2+2),$$ but $m^2+2 \not \equiv 0 \pmod{5}$, so the sum of the squares of $5$ consecutive integers cannot be a perfect square.

(iv) $n=6$. Let $m \in \mathbb{Z}$. We have $$(m-2)^2+(m-1)^2+m^2+(m+1)^2+(m+2)^2+(m+3)^2=6m(m+1)+19.$$ Since $m(m+1)$ is even, then $6m(m+1) \equiv 0 \pmod{4}$ and $6m(m+1)+19 \equiv 3 \pmod{4}$. So the sum of the squares of $6$ consecutive integers cannot be a perfect square.

(v) $n=7$. Let $m \in \mathbb{Z}$. We have $$(m-3)^2+(m-2)^2+(m-1)^2+m^2+(m+1)^2+(m+2)^2+(m+3)^2=7(m^2+4),$$ but $m^2+4 \not \equiv 0 \pmod{7}$, so the sum of the squares of $7$ consecutive integers cannot be a perfect square.

(vi) $n=8$. Let $m \in \mathbb{Z}$. We have $$(m-3)^2+(m-2)^2+\ldots+(m+3)^2+(m+4)^2=4[2m(m+1)+11].$$ Since $m(m+1)$ is even, then $2m(m+1) \equiv 0 \pmod{4}$ and $2m(m+1)+11 \equiv 3 \pmod{4}$. So the sum of the squares of $8$ consecutive integers cannot be a perfect square.

(vii) $n=9$. Let $m \in \mathbb{Z}$. We have $$(m-4)^2+(m-3)^2+\ldots+(m+3)^2+(m+4)^2=3(3m^2+20),$$ but $3m^2+20 \not \equiv 0 \pmod{3}$, so the sum of the squares of $9$ consecutive integers cannot be a perfect square.

(viii) $n=10$. Let $m \in \mathbb{Z}$. We have $$(m-4)^2+(m-3)^2+\ldots+(m+4)^2+(m+5)^2=5[2m(m+1)+17],$$ but $2m(m+1)+17 \not \equiv 0 \pmod{5}$, so the sum of the squares of $10$ consecutive integers cannot be a perfect square.

The conclusion follows.

Note:
The solution is not unique. For example, $$18^2+19^2+20^2+21^2+22^2+23^2+24^2+25^2+26^2+27^2+28^2=77^2.$$
It's easy to see why. We must find $m \in \mathbb{Z}$ such that $$(m-5)^2+(m-4)^2+\ldots+(m+4)^2+(m+5)^2=11(m^2+10)$$ is a perfect square. So, $m \equiv \pm 1 \pmod{11}$ and one can check if $11(m^2+10)$ is a perfect square.

Conjecture:An easy case by case analysis shows that there are no solutions for $11<n \leq 22$. For $n=23$ there is a solution given by $$17^2+18^2+19^2+\ldots+27^2+28^2+29^2=138^2.$$ (Observe that $(m-11)^2+\ldots+(m+11)^2=23(m^2+44)$ and it must be $m \equiv \pm 5 \pmod{23}$ and $m \equiv 0 \pmod{2}$)

If $n>1$ is a natural number $n$ such that there exist $n$ consecutive integers whose sum of their squares is a perfect square, then $n$ is prime.