<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] I don't know if this may interest you or answer your question but if you play around with field different from $\mathbb{R}$, special things happens. For example working with field $\mathbb{F}$ with characteristic equal to 2 it's always true that
$$(a+b)^2 = a^2 + b^2, \space \forall a,b \in \mathbb{F}$$
And if we take another finite sum $(a + b+ c+ d)^2$ in order to obtain the claim what we want is
$$2(ab + ac + ad + ...) = 0$$
and since that term is even we can choose a field with proper even characteristic and we obtain a series of examples that are non trivial.
</s>