In this question it was shown that collections of $8k+1$ odd squares can be found that sum to an odd square. In his answer, Denis Shatrov provided an algorithm by which $8k$ odd numbers $\{a_1,a_2,\dots,a_{8k}\}$ are chosen arbitrarily, and an odd number $a_{8k+1}$ can be derived such that the sum of the squares of the $a_i$ is also an odd square $b^2$, viz:
$$N=\sum_{i=1}^{8k}a_i^2$$ $$a_{8k+1}=\left (\frac{N-4}{4}\right )$$ $$b=\left (\frac{N+4}{4}\right )$$ $$b^2=\sum_{i=1}^{8k+1}a_i^2$$
Interestingly, $a_{8k+1}$ and $b$ generated by this algorithm are consecutive odd numbers.
Since the initial $8k$ numbers can be chosen arbitrarily, they can be odd primes (going forward, I will assume that they are distinct odd primes). In that case, can $a_{8k+1}$ and $b$ also be primes? If so, then those generated numbers would be twin primes. A quick empirical survey showed that for $k=1$, it is relatively easy to find collections of $8$ odd squares of primes that give rise to solution in which $a_9$ is a prime, but examples of $b$ being prime were not found.
This observation can be understood by analyzing matters $\bmod 3$. Odd primes other than $3$ have squares $\equiv 1 \bmod 3$. For $k\equiv 1 \bmod 3$, the sum of $8k+1$ odd numbers is either $\equiv 0 \bmod 3$ (when all $a_i \ne 3$) or $\equiv 2 \bmod 3$ (if $a_1 = 3$). In the first case, $b\equiv 0 \bmod 3$ means that $b$ cannot be prime. In the second case, there are no integers $b$ such that $b^2 \equiv 2 \bmod 3$. So when $k\equiv 1 \bmod 3$, it is not possible to find collections of $\{a_i\}$ in which $b$ is the larger member of a pair of twin primes.
Repeating this analysis for $k\equiv 2 \bmod 3$ and $k\equiv 0 \bmod 3$ indicates there is no such absolute barrier to obtaining examples where $b$ is the larger member of a pair of twin primes. In the case $k\equiv 2 \bmod 3$, it is required that $3$ be one of the primes in the initial collection of $a_i$, whereas in the case $k\equiv 0 \bmod 3$, it is required that $3$ not be one of the primes in the initial collection of $a_i$.
My questions are these: Independent of the Shatrov algorithm, is it possible to obtain collections of $8k+1$ distinct odd primes such that the sum of their squares is the square of a prime? Within the Shatrov algorithm, is it possible to obtain collections of $8k+1$ distinct odd primes such that they sum to the square of an odd prime AND $(a_{8k+1},b)$ constitute a twin prime pair?
I realize that the collections of odd prime square summands will have to contain $(17,25,41,49,\dots)$ members, so either theoretical or computational approaches may be challenging.
Best Answer
$$3^2 + 5^2 + 7^2 + 11^2 + 13^2 + 17^2 + 19^2 + 23^2 + 29^2 + 31^2 + 37^2 + 41^2 + 43^2 + 47^2 + 53^2 + 67^2 + 71^2 = 151^2$$
A solution can be found with dynamic programming. In more detail, we can define $f(i, t, c)$ to be true if $t$ is obtainable as a sum of squares of $c$ of the first primes, and then we can find a solution of length $k$ involving the first $n$ primes or confirm one doesn't exists in time $O(k n^3 \log^2(n))$.
$$3^2 + 5^2 + 7^2 + 11^2 + 13^2 + 17^2 + 19^2 + 23^2 + 29^2 + 31^2 + 37^2 + 41^2 + 43^2 + 47^2 + 53^2 + 73^2 + 4649^2 = 4651^2$$
A solution to the second problem can be found in a similar way - choose a pair of twin primes $p, p+2$, and find $k-1$ primes whose squares sum to $4(p+1)$.
sets with 25 primes: $$5^2 + 7^2 + 11^2 + 13^2 + 17^2 + 19^2 + 23^2 + 29^2 + 31^2 + 37^2 + 41^2 + 43^2 + 47^2 + 53^2 + 61^2 + 67^2 + 71^2 + 73^2 + 79^2 + 83^2 + 89^2 + 97^2 + 101^2 + 107^2 + 113^2 = 311^2$$
$$5^2 + 7^2 + 11^2 + 13^2 + 17^2 + 19^2 + 23^2 + 29^2 + 31^2 + 37^2 + 41^2 + 43^2 + 47^2 + 53^2 + 59^2 + 61^2 + 67^2 + 71^2 + 73^2 + 79^2 + 83^2 + 89^2 + 107^2 + 113^2 + 20147^2 = 20149^2$$
41: $$3^2 + 5^2 + 7^2 + 11^2 + 13^2 + 17^2 + 19^2 + 23^2 + 29^2 + 31^2 + 37^2 + 41^2 + 43^2 + 47^2 + 53^2 + 59^2 + 61^2 + 67^2 + 71^2 + 73^2 + 79^2 + 83^2 + 89^2 + 97^2 + 101^2 + 103^2 + 107^2 + 109^2 + 113^2 + 127^2 + 131^2 + 137^2 + 139^2 + 149^2 + 151^2 + 157^2 + 163^2 + 167^2 + 181^2 + 197^2 + 211^2 = 659^2$$
$$3^2 + 5^2 + 7^2 + 11^2 + 13^2 + 17^2 + 19^2 + 23^2 + 29^2 + 31^2 + 37^2 + 41^2 + 43^2 + 47^2 + 53^2 + 59^2 + 61^2 + 67^2 + 71^2 + 73^2 + 79^2 + 83^2 + 89^2 + 97^2 + 101^2 + 103^2 + 107^2 + 109^2 + 113^2 + 127^2 + 131^2 + 137^2 + 139^2 + 149^2 + 151^2 + 157^2 + 167^2 + 173^2 + 179^2 + 181^2 + 96587^2 = 96589^2$$