Can the sum of the first n squares of primes be a perfect square?

Question

So far, all I have found is for $n=1$, for which the square is 4. (Rather trivial)

To give some context, I recently watched the Numberphile video Squared Squares.

Basically, a squared square is a square that is entirely made up of other integral squares (preferably all of different sizes). I wanted to see whether there exists a squared square made of $n$ squares that is entirely composed of squares with the side-lengths of all primes up to the $n$th prime.

By finding an example that satisfies the question, it will be shown that there is at least the possibility for one of these squares existing. If it can be proven that there are no examples other than $n=1$, then it follows that no squared square with conditions exist.

Answer 1

There are no others up to $n= 10^6$. I strongly suspect there are none. The sum of the first $n$ squares of primes is on the order of $n^3 \log^2(n)$, and heuristically the probability that it is a square is on the order of $n^{-3/2} \log(n)^{-1}$. The series $\sum_n n^{-3/2}/\log(n)$ converges, so we should expect to find only finitely many squares. This is not a proof, but together with the empirical evidence it is highly suggestive.

EDIT: Here is a Maple program which uses all the primes $< 3 \times 10^7$ (there are $1857859$ of them). Still no more found.

P:=select(isprime, [2,seq(i,i=3..3*10^7,2)]):
S:= ListTools:-PartialSums(map(t -> t^2, P)):
select(t -> issqr(S[t]), [$1..nops(S)]);

$$ [1] $$

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P.S. By the way, if you use the primes $p_n$ themselves rather than squares of primes, then solutions to,

$$\sum_{n=1}^k p_n = s^2$$

are OEIS sequences A033997 (for $k$) and A061888 (for $s$) the largest known so far is,

$$\sum_{n=1}^{126789311423} p_n = 468726713734^2$$

but which is likely to be infinite.