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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 Oct 18 at 5:27
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    $\begingroup$ 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 Oct 18 at 12:01
  • $\begingroup$ I've awarded +1 but this Answer is not complete. $\endgroup$ Commented Oct 21 at 20:28
  • $\begingroup$ @WlodAA Why isn't the answer complete? :) $\endgroup$ Commented Oct 21 at 20:30
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    $\begingroup$ The OT's Question stated "another" hence $\ p^2=p^2\ $ does not count. Perhaps(?) it's not hard to address the \$\ x^2\ $ case but it still needs a bit of doing. $\endgroup$ Commented Oct 21 at 21:20

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