Crux Mathematicorum 2016, Issue 1 - Problem 4108

Problem:
Write $2010$ as a sum of consecutive squares. Is it possible to write $2014$ as the sum of several consecutive squares?

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
(a) Let $k$ be the number of consecutive perfect squares which satisfy the conditions and let $n \in \mathbb{N}$.  Then, $$(n+1)^2+(n+2)^2+\ldots+(n+k)^2=2010.$$ Since $$\begin{array}{lll}(n+1)^2+\ldots+(n+k)^2&=&kn^2+2n(1+2+\ldots+k)+(1^2+2^2+\ldots+k^2)\\&=&kn^2+2n\cdot\dfrac{k(k+1)}{2}+\dfrac{k(k+1)(2k+1)}{6}\\&=&\dfrac{k}{6}[6n^2+6n(k+1)+(k+1)(2k+1)], \end{array}$$ we get
$$k[6n^2+6n(k+1)+(k+1)(2k+1)]=12060.    \qquad          (1)
$$
Moreover, since $\dfrac{k(k+1)(2k+1)}{6}=1^2+2^2+\ldots+k^2 \leq (n+1)^2+(n+2)^2+\ldots+(n+k)^2=2010$, we get $k \leq 17$. So, $k \ | \ 12060$ and $k \leq 17$, which gives $k \in \{1,2,3,4,5,6,9,10,12,15\}$. Observe that if $k=4h+2$, where $h \in \mathbb{N}$, then the left-hand side in (1) is not divisible by $4$, but the right-hand side is divisible by $4$. So, $k \in \{1,3,5,9,12,15\}$. If $k=15$, equation (1) gives $$6n(n+16)+496=804 \implies 6n(n+16)=308,$$ a contradiction. If $k=12$, equation (1) gives $$6n(n+13)+325=1005 \implies 6n(n+13)=680,$$ a contradiction. If $k=9$, equation (1) gives $$6n(n+10)+190=1340 \implies 6n(n+10)=1150,$$ a contradiction. If $k=5$, equation (1) gives $$6n(n+6)+66=2412 \implies n(n+6)=391,$$ which gives $n=17$. Therefore, $18^2+19^2+20^2+21^2+22^2=2010$, and the maximum number of consecutive perfect squares which satisfy the condition is $5$.

(b) The answer is no. Indeed, if there exist $k,n \in \mathbb{N}$ such that $$(n+1)^2+(n+2)^2+\ldots+(n+k)^2=2014,$$ then as we have seen in (a),
$$
k[6n^2+6n(k+1)+(k+1)(2k+1)]=2014\cdot6=12084   \qquad        (2)
$$
and $\dfrac{k(k+1)(2k+1)}{6} \leq 2014$. So, $k \ | \ 12084$, $k \leq 17$ and $k \neq 4h+2$ for any $h \in \mathbb{N}$. It follows that $k \in \{1,3,4,12\}$. Since $2014$ is not a perfect square, $k \neq 1$. As $24^2+25^2+26^2=1877<2014<25^2+26^2+27^2=2030$, then $k \neq 3$. As $20^2+21^2+22^2+23^2=1854<2014<21^2+22^2+23^2+24^2=2030$, then $k \neq 4$. Finally, if $k=12$ equation (2) gives $6n(n+13)+325=1007$, i.e. $6n(n+13)=682$, a contradiction. Therefore there does not exist $k \in \mathbb{N}$ which satisfy the condition, and the conclusion follows.