<s>[INST] Examples where $(a+b+\cdots)^2 = (a^2+b^2+\cdots)$ Consider the two infinite series
$$
    \frac{\pi}{\sqrt{8}} = 1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \cdots
$$
and
$$
    \frac{\pi^2}{8} = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{9^2} + \frac{1}{11^2} + \cdots
$$
(Notice that the first series has signs that go two-by-two rather than every-other.)
Squaring the first equality also gives $\pi^2/8$ and so these two, when put together, satisfy the 'highschooler's dream' for squaring a sum: just square each term and sum,
$$
    (a + b + c + \cdots)^2 = (a^2 + b^2 + c^2 + \cdots)
$$
with nothing like $2ab + 2ac + 2bc + \cdots$ needed.
A trivial example of this would be
$$
    (a + 0)^2 = a^2 + 2a0 + 0^2 = a^2 + 0^2
$$
but it only succeeds because one addend is zero.
My questions are

*

*Are there any other simple nontrivial examples?  I believe any other nontrivial example must be an infinite sum. edit: John Omielan provides the simple finite example $(1+1-\frac{1}{2})^2 = 1^2 + 1^2 + \frac{1}{2^2}$.

*Is there an "obvious" demonstration that the above sum (other than the direct evaluation) satisfies the highschooler's dream?  Put another way, is there a simple demonstration that the infinite sum of "cross terms" vanishes?

 [/INST] The first term
$$
\eqalign{
  & \left( {a_1  + a_2  +  \cdots  + a_n } \right)^2  = R^2 \quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left( {{{a_1 } \over R} + {{a_2 } \over R} +  \cdots
  + {{a_n } \over R}} \right)^2  - 1 = 0\quad  \Rightarrow   \cr 
  & \left( {\left( {b_1  + b_2  +  \cdots  + b_n } \right) + 1} \right) \cdot
 \left( {\left( {b_1  + b_2  +  \cdots  + b_n } \right) - 1} \right) = 0 \cr} 
$$
is the equation (in $b_k$) of two diagonal planes, symmetric wrt the  origin, with normal vector $(1,1, \ldots , 1)$,
through the point
$$
 \pm \left( {{1 \over n},{1 \over n}, \ldots ,{1 \over n}} \right)
$$
and thus each at a distance from the origin of
$$
{1 \over {\sqrt n }}
$$
The second term
$$
\eqalign{
  & a_1 ^2  + a_2 ^2  +  \cdots  + a_n ^2  = R^2 \quad  \Rightarrow   \cr 
  &  \Rightarrow \quad b_1 ^2  + b_2 ^2  +  \cdots  + b_n ^2  = 1 \cr} 
$$
is a unitary sphere centered at the origin.
Therefore the equality
$$
\eqalign{
  & \left( {a_1  + a_2  +  \cdots  + a_n } \right)^2  = a_1 ^2  + a_2 ^2  +  \cdots  + a_n ^2  = R^2
 \quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left( {b_1  + b_2  +  \cdots  + b_n } \right)^2
  = b_1 ^2  + b_2 ^2  +  \cdots  + b_n ^2  = 1 \cr} 
$$
is satisfied whenever the points $b_k$ lie on one of the two circles resulting from the intersection,
and the points $a_k$ on any dilation of those circles, i.e. on the conic surface with vertex at the origin, axis $(1,1, \ldots , 1)$, cross-section defined by the  above circle on the unitary sphere.
</s>