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Are there infinitely many sets of distinct primes whose squares add up to another square?

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Yes.

In the paper Sums of squares of primes, by Robert E. Dressler, Louis Pigno, and Robert Young, they present the following theorem.

Theorem. 17163 is the largest integer which is not representable as a sum of distinct squares of primes.

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  • $\begingroup$ This is a related thread. I wondered whether there is a uniform bound on the number of squares needed and from the discussion there I conclude that 1) yes, there is, and 2) it is $3$, $4$, or $5$. $\endgroup$ Commented 10 hours ago
  • $\begingroup$ On this uniform bound, we have $p^2 \equiv 1 \pmod{24}$ for all primes $p > 3$. So if $n = 24m = p_1^2 + \ldots + p_k^2$, with all $p_i$ distinct, then $k \ge 11$. In fact, for $1 \le i \le 4$, if $n = 24m + i$ is not the sum of $i$ squares of primes, while $n = p_1^2 + \ldots + p_k^2$, with all $p_i$ distinct, then $k \ge 11 + i$. On the other hand, we do have the result by Hua (mentioned by Timothy Chow and GH from MO in the question linked by TLo) that all large enough $n = 24m + 5$ can be written as the sum of squares of $5$ primes. $\endgroup$ Commented 3 hours ago

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