Solve $a_1^2 + 2a_2^2 + 3a_3^2 + ... + na_n^2 = a_{n+1}^2$ for natural numbers

Question

Solve the equation $\sum_{k=1}^nk\cdot a_k^2 = a_{n+1}^2$ in natural numbers (i.e. $a_n \in \mathbb N~~ \forall n \in \mathbb N$).

I really have no idea as to how to approach the issue. A starter hint would be nice! Thanks in advance!

Answer 1

Here is ONE non-zero solution for the special case that $a_1=a_2=\dots=a_n=A$. In that case $\sum ka_k^2=A^2\sum n=A^2\frac{n(n+1)}{2}$. If that sum equals a perfect square, then $\frac{n(n+1)}{2}$ must be a perfect square. This occurs when the even member of $n,(n+1)$ is twice a square, and the odd member is itself a square. Such numbers are called square triangular numbers. The first occurrence of that (for $n>1$) is $n=8$.

$$\sum_{k=1}^8 kA^2=36A^2$$ In this case that results in $a_{n+1}=6A$.

There may be more general solutions, and if I come across one, I will post it here.

Answer 2

A second general solution occurs when $a_k=k\cdot A$ $$\sum_{k=1}^n k(k\cdot A)^2=A^2\sum_{k=1}^n k^3=A^2\frac{n^2(n+1)^2}{4}$$ The final term is a perfect square for all values of $n$.

In this case, $a_{n+1}=A\frac{n(n+1)}{2}$

Answer 3

$a_1^2+2a^2_2+\cdots+na^2_n=a^2_{n+1}\tag{1}$

Substitute $a_1=1+m_1t, a_2=0+m_2t, a_3=0+m_3t, \cdots,a_n=0+m_nt,a_{n+1}=1+m_{n+1}t$ to equation $(1)$, then we get $$t=-2(m_1-m_{n+1})/(m_1^2+2m_2^2+3m_3^2+\cdots+nm_n^2-m_{n+1}^2).$$ Thus, we get a parametric solution.
$m_n$ is arbitrary.

Example for $n=3:$
\begin{eqnarray} &a_1& = -m_1^2+2m_2^2+3m_3^2-m_4^2+2m_1m_4 \\ &a_2& =-2m_2(m_1-m_4) \\ &a_3& = -2m_3(m_1-m_4) \\ &a_4& = m_1^2+2m_2^2+3m_3^2+m_4^2-2m_1m_4 \\ \end{eqnarray}

$(m_1,m_2,m_3,m_4)=(1,1,2,2): (a_1,a_2,a_3,a_4)=(13, 2, 4, 15)$ $(m_1,m_2,m_3,m_4)=(1,2,3,4): (a_1,a_2,a_3,a_4)=(13, 6, 9, 22)$ $(m_1,m_2,m_3,m_4)=(4,3,2,1): (a_1,a_2,a_3,a_4)=(7, 6, 4, 13)$

Answer 4

All solutions in natural numbers can be found as follows.

Let $a_2,a_3,...,a_n$ be any natural numbers such that $$T= 2a_2^2 + 3a_3^2 + … + na_n^2$$ is either odd or divisible by $4$. Let $T=D\times E$ be any factorisation of $T$ with $D-E$ even.

Then $a_1=\frac{|D-E|}{2}$ and $a_{n+1}=\frac{D+E}{2}$ solve the equation because $a_{n+1}^2-a_1^2=DE$.

EXAMPLE Starting with the 'arbitrary' choice $ 2\times4^2+3\times3^2+4\times1^2=63$.

We have the three factorisations $63=63\times 1$, $63=9\times 7$ and $63=21\times 3$. So there are three solutions.

$$(a_1,a_5)\text { is any one of } (31,32),(1,8),(9,12). $$