A Pythagorean triple is a triple of integers $$(a,b,c)$$ forming the side lengths of a right triangle and satisfying the Pythagorean equation $$a^2 + b^2 = c^2$$. There's some evidence that these triples were known to the ancient Babylonians, for whom the regular numbers (having only $$2$$, $$3$$, and $$5$$ as prime factors) were especially important. So, what are the regular Pythagorean triples? We might as well limit our attention to primitive triples (those having no common divisor) because we can form any other triple by multiplying all the numbers in a primitive triple by the same scalar.

Theorem: The only regular primitive Pythagorean triple is $$(3,4,5)$$.

Proof: Any primitive Pythagorean triple has the form $$(|p^2 - q^2|, 2pq, p^2 + q^2)$$ where $$p$$ and $$q$$ are relatively prime and exactly one of the two (say $$p$$) is even; for instance, $$(3,4,5)$$ is generated in this way by $$p = 2$$, $$q = 1$$. In order for $$2pq$$ to be regular, $$p$$ and $$q$$ must themselves be regular; in order for $$p^2 - q^2$$ to be regular, $$p + q$$ and $$p - q$$ must also be regular. But these four numbers $$p$$, $$q$$, $$p - q$$, and $$p + q$$ are all relatively prime, so the prime factors $$2$$, $$3$$, and $$5$$ must be partitioned among them. By the pigeonhole principle, one of the four, necessarily either $$q$$ or $$p − q$$, must be $$\pm 1$$. If $$q = 1$$ then $$p$$ and $$p + q$$ are consecutive regular numbers; if $$p − q = \pm 1$$ then $$p + q$$ and $$(p + q) + (p − q) = 2p$$ are consecutive regular numbers. For a consecutive pair $$(m, m + 1)$$ of regular numbers, if $$m$$ is even we get a possible solution $$p = m$$, $$q = 1$$ and (if $$m$$ is divisible by $$4$$) $$p = m/2$$, $$q = m/2 + 1$$, while if $$m$$ is odd we only get the possible solution $$p = (m + 1)/2$$, $$q = (m − 1)/2$$. Additionally we need only consider the case when $$m$$ and $$m + 1$$ are prime powers, for if the four numbers $$p$$, $$q$$, $$p − q$$, and $$p + q$$ are distinct then each of them can have only one prime factor while if they are not distinct then $$q = p − q = 1$$, $$p = 2$$, $$2p = 4$$, and $$p + q = 3$$ are all prime powers.

Størmer's theorem shows that there are only ten pairs of consecutive regular numbers to examine, and only for four of these are both numbers nontrivial prime powers: $$(2,3)$$, $$(3,4)$$, $$(4,5)$$, and $$(8,9)$$. The $$(2,3)$$ case leads (with $$p = 2$$ and $$q = 1$$) to the $$(3,4,5)$$ triangle. The case $$(3,4)$$ leads again to the $$(3,4,5)$$ triangle. The $$(4,5)$$ case leads (with $$p = 4$$, $$q = 1$$) to the $$(15,8,17)$$ triangle and (with $$p = 2$$, $$q = 3$$) to the $$(5,12,13)$$ triangle, and the $$(8,9)$$ case leads (with $$p = 8$$ and $$q = 1$$) to the $$(63,16,65)$$ triangle and (with $$p = 4$$, $$q = 5$$) to the $$(9,40,41)$$ triangle. Of these, only the $$(3,4,5)$$ triangle is regular.

Because every Pythagorean triple includes numbers divisible by $$2$$, $$3$$, and $$5$$, this shows more generally that $$(3,4,5)$$ and its multiples are the only Pythagorean triples with only three prime factors. This also shows that there are only four primitive Pythagorean triples $$(a,b,c)$$ for which $$a$$ and $$b$$ are regular, ignoring the regularity of $$c$$: $$(3,4,5)$$, $$(5,12,13)$$, $$(8,15,17)$$, and $$(9,40,41)$$. Partial results of Stewart and Tijdeman on the abc conjecture can be used to show that, for any set $$P$$ of primes, there are only finitely many Pythagorean triples $$(a,b,c)$$ for which $$a$$ and $$b$$ are both $$P$$-smooth, but I don't know of an efficient method for listing all such triples in general.