There are infinitely many sets of distinct primes whose squares add up to a square number and, presumably, sets of any size (https://mathoverflow.net/questions/501745/primes-whose-squares-add-up-to-another-square).
Find one of these that can be used to perfectly tile a square, like Duijvestijn's famous order-21 perfect squared square, but with all its squares of prime side, or show this is not possible.
Answer: No.
Proof: Consider the Smith diagram of a squaring:
Source: Wikipedia
This is a planar graph where vertices represent horizontal segments, edges represent squares, and faces represent vertical segments. Suppose we have a prime perfect squaring whose Smith diagram has v vertices, e edges, and f faces.
Consider any vertex besides the two poles. The sum of its incoming edges equals the sum of its outgoing edges. If it had only 1 of each, then they would be equal, and the squaring would be imperfect, so its degree must be at least 3. If its adjacent edges are all odd primes, then there must be an even number of them, so its degree must be at least 4, unless it is one of at most 2 vertices adjacent to an even prime (2). For the poles themselves, both must have degree at least 2 (else our squaring would only have 1 square), and one must have degree at least 3 (else 2 squares would overlap in the center of our squaring). In summary, v − 4 vertices have degree at least 4, 3 of the others have degree at least 3, and the last has degree at least 2, so the sum of the degrees of the vertices is at least 4v − 5. This double counts each edge, so 2e ≥ 4v − 5 ⇒ 4v ≤ 2e + 5.
Similarly, consider any face. The sum of the edges on its left equals the sum of the edges on its right. For perfection, it must border at least 3 edges. For primality, it must border at least 4 edges, unless it is one of at most 2 faces bordering a 2. Therefore, the sum over all faces of the number of edges bordering them is at least 4f − 2. This double counts each edge, so 2e ≥ 4f − 2 ⇒ 4f ≤ 2e + 2.
Because we have a connected plane graph, Euler's formula gives v − e + f = 2. Multiplying by 4 and substituting the two inequalities gives 8 = 4v − 4e + 4f ≤ (2e + 5) - 4e + (2e + 2) = 7 — a contradiction. Therefore, no prime perfect squaring exists.
